Are there any other methods to apply to solving simultaneous equations? The 2019 Stack...

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Are there any other methods to apply to solving simultaneous equations?



The 2019 Stack Overflow Developer Survey Results Are InSolving matrix equations of the form $XA = XB$Simultaneous EquationsSolving a set of linear equations for variables with non-constant valuesMistake in my NLP using Lagrange Multipliers?Solving equations system: $xy+yz=a^2,xz+xy=b^2,yz+zx=c^2$Solving simultaneous equations involving a quadraticMethods for solving a $4$ system of equationFunctions for fixed-point iteration on a nonlinear system of equationsSystem of simultaneous equations involving integral part (floor)Different ways of solving simultaneous equations












18












$begingroup$


We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:




$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$




I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.



I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$begin{align}3x+3y - y &= 36 tag{1a}\ 5x + 5y - y &= 64tag{1b}end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.



I can even use matrices!



$(1)$ and $(2)$ could be written in matrix form:



$$begin{align}begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}begin{bmatrix} x \ yend{bmatrix}&=begin{bmatrix}36 \ 64end{bmatrix}tag3 \ begin{bmatrix} x \ yend{bmatrix} &= {begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}}^{-1}begin{bmatrix}36 \ 64end{bmatrix} \ &= frac{1}{2}begin{bmatrix}4 &-2 \ -5 &3end{bmatrix}begin{bmatrix}36 \ 64end{bmatrix} \ &=frac12begin{bmatrix} 16 \ 12end{bmatrix} \ &= begin{bmatrix} 8 \ 6end{bmatrix} \ \ therefore x&=8 \ therefore y&= 6end{align}$$





Question



Are there any other methods to solve for both $x$ and $y$?










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  • 5




    $begingroup$
    you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
    $endgroup$
    – Doug M
    yesterday






  • 2




    $begingroup$
    This is a linear system of equations, which some believe it is the most studied equation in all of mathematics. The reason being that it is so widely used in applied mathematics that there's always reason to find faster and more robust methods that will either be generic or suit the particularities of a given problem. You might roll your eyes at my claim when thinking of your two variable system, but soem engineers need to solve such systems with hundreds of variables in their jobs.
    $endgroup$
    – Mefitico
    yesterday






  • 3




    $begingroup$
    I hope someone performs GMRES by hand on this system and reports the steps.
    $endgroup$
    – Rahul
    yesterday








  • 2




    $begingroup$
    Since linear systems are so well studied, there are many approaches (that are essentially equivalent - but maybe not the iterative solution). As such, does this question essentially boil down to a list of answers, which is not technically on topic for this site?
    $endgroup$
    – Teepeemm
    yesterday






  • 2




    $begingroup$
    There is an entire subject called Numerical Linear Algebra which studies efficient ways to solve $Ax = b$. There are many notable algorithms. For example, you could use an iterative algorithm such as the Jacobi method or Gauss-Seidel or, as @Rahul noted, GMRES. There are other direct methods also. For example, you could find the QR factorization $A = QR$, where $Q$ is orthogonal and $R$ is upper triangular, and solve $Rx = Q^T b$ using back substitution.
    $endgroup$
    – littleO
    yesterday
















18












$begingroup$


We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:




$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$




I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.



I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$begin{align}3x+3y - y &= 36 tag{1a}\ 5x + 5y - y &= 64tag{1b}end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.



I can even use matrices!



$(1)$ and $(2)$ could be written in matrix form:



$$begin{align}begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}begin{bmatrix} x \ yend{bmatrix}&=begin{bmatrix}36 \ 64end{bmatrix}tag3 \ begin{bmatrix} x \ yend{bmatrix} &= {begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}}^{-1}begin{bmatrix}36 \ 64end{bmatrix} \ &= frac{1}{2}begin{bmatrix}4 &-2 \ -5 &3end{bmatrix}begin{bmatrix}36 \ 64end{bmatrix} \ &=frac12begin{bmatrix} 16 \ 12end{bmatrix} \ &= begin{bmatrix} 8 \ 6end{bmatrix} \ \ therefore x&=8 \ therefore y&= 6end{align}$$





Question



Are there any other methods to solve for both $x$ and $y$?










share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
    $endgroup$
    – Doug M
    yesterday






  • 2




    $begingroup$
    This is a linear system of equations, which some believe it is the most studied equation in all of mathematics. The reason being that it is so widely used in applied mathematics that there's always reason to find faster and more robust methods that will either be generic or suit the particularities of a given problem. You might roll your eyes at my claim when thinking of your two variable system, but soem engineers need to solve such systems with hundreds of variables in their jobs.
    $endgroup$
    – Mefitico
    yesterday






  • 3




    $begingroup$
    I hope someone performs GMRES by hand on this system and reports the steps.
    $endgroup$
    – Rahul
    yesterday








  • 2




    $begingroup$
    Since linear systems are so well studied, there are many approaches (that are essentially equivalent - but maybe not the iterative solution). As such, does this question essentially boil down to a list of answers, which is not technically on topic for this site?
    $endgroup$
    – Teepeemm
    yesterday






  • 2




    $begingroup$
    There is an entire subject called Numerical Linear Algebra which studies efficient ways to solve $Ax = b$. There are many notable algorithms. For example, you could use an iterative algorithm such as the Jacobi method or Gauss-Seidel or, as @Rahul noted, GMRES. There are other direct methods also. For example, you could find the QR factorization $A = QR$, where $Q$ is orthogonal and $R$ is upper triangular, and solve $Rx = Q^T b$ using back substitution.
    $endgroup$
    – littleO
    yesterday














18












18








18


8



$begingroup$


We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:




$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$




I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.



I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$begin{align}3x+3y - y &= 36 tag{1a}\ 5x + 5y - y &= 64tag{1b}end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.



I can even use matrices!



$(1)$ and $(2)$ could be written in matrix form:



$$begin{align}begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}begin{bmatrix} x \ yend{bmatrix}&=begin{bmatrix}36 \ 64end{bmatrix}tag3 \ begin{bmatrix} x \ yend{bmatrix} &= {begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}}^{-1}begin{bmatrix}36 \ 64end{bmatrix} \ &= frac{1}{2}begin{bmatrix}4 &-2 \ -5 &3end{bmatrix}begin{bmatrix}36 \ 64end{bmatrix} \ &=frac12begin{bmatrix} 16 \ 12end{bmatrix} \ &= begin{bmatrix} 8 \ 6end{bmatrix} \ \ therefore x&=8 \ therefore y&= 6end{align}$$





Question



Are there any other methods to solve for both $x$ and $y$?










share|cite|improve this question











$endgroup$




We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:




$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$




I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.



I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$begin{align}3x+3y - y &= 36 tag{1a}\ 5x + 5y - y &= 64tag{1b}end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.



I can even use matrices!



$(1)$ and $(2)$ could be written in matrix form:



$$begin{align}begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}begin{bmatrix} x \ yend{bmatrix}&=begin{bmatrix}36 \ 64end{bmatrix}tag3 \ begin{bmatrix} x \ yend{bmatrix} &= {begin{bmatrix} 3 &2 \ 5 &4end{bmatrix}}^{-1}begin{bmatrix}36 \ 64end{bmatrix} \ &= frac{1}{2}begin{bmatrix}4 &-2 \ -5 &3end{bmatrix}begin{bmatrix}36 \ 64end{bmatrix} \ &=frac12begin{bmatrix} 16 \ 12end{bmatrix} \ &= begin{bmatrix} 8 \ 6end{bmatrix} \ \ therefore x&=8 \ therefore y&= 6end{align}$$





Question



Are there any other methods to solve for both $x$ and $y$?







linear-algebra systems-of-equations






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share|cite|improve this question













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share|cite|improve this question








edited yesterday









Rodrigo de Azevedo

13.2k41962




13.2k41962










asked yesterday









user477343user477343

3,62031245




3,62031245








  • 5




    $begingroup$
    you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
    $endgroup$
    – Doug M
    yesterday






  • 2




    $begingroup$
    This is a linear system of equations, which some believe it is the most studied equation in all of mathematics. The reason being that it is so widely used in applied mathematics that there's always reason to find faster and more robust methods that will either be generic or suit the particularities of a given problem. You might roll your eyes at my claim when thinking of your two variable system, but soem engineers need to solve such systems with hundreds of variables in their jobs.
    $endgroup$
    – Mefitico
    yesterday






  • 3




    $begingroup$
    I hope someone performs GMRES by hand on this system and reports the steps.
    $endgroup$
    – Rahul
    yesterday








  • 2




    $begingroup$
    Since linear systems are so well studied, there are many approaches (that are essentially equivalent - but maybe not the iterative solution). As such, does this question essentially boil down to a list of answers, which is not technically on topic for this site?
    $endgroup$
    – Teepeemm
    yesterday






  • 2




    $begingroup$
    There is an entire subject called Numerical Linear Algebra which studies efficient ways to solve $Ax = b$. There are many notable algorithms. For example, you could use an iterative algorithm such as the Jacobi method or Gauss-Seidel or, as @Rahul noted, GMRES. There are other direct methods also. For example, you could find the QR factorization $A = QR$, where $Q$ is orthogonal and $R$ is upper triangular, and solve $Rx = Q^T b$ using back substitution.
    $endgroup$
    – littleO
    yesterday














  • 5




    $begingroup$
    you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
    $endgroup$
    – Doug M
    yesterday






  • 2




    $begingroup$
    This is a linear system of equations, which some believe it is the most studied equation in all of mathematics. The reason being that it is so widely used in applied mathematics that there's always reason to find faster and more robust methods that will either be generic or suit the particularities of a given problem. You might roll your eyes at my claim when thinking of your two variable system, but soem engineers need to solve such systems with hundreds of variables in their jobs.
    $endgroup$
    – Mefitico
    yesterday






  • 3




    $begingroup$
    I hope someone performs GMRES by hand on this system and reports the steps.
    $endgroup$
    – Rahul
    yesterday








  • 2




    $begingroup$
    Since linear systems are so well studied, there are many approaches (that are essentially equivalent - but maybe not the iterative solution). As such, does this question essentially boil down to a list of answers, which is not technically on topic for this site?
    $endgroup$
    – Teepeemm
    yesterday






  • 2




    $begingroup$
    There is an entire subject called Numerical Linear Algebra which studies efficient ways to solve $Ax = b$. There are many notable algorithms. For example, you could use an iterative algorithm such as the Jacobi method or Gauss-Seidel or, as @Rahul noted, GMRES. There are other direct methods also. For example, you could find the QR factorization $A = QR$, where $Q$ is orthogonal and $R$ is upper triangular, and solve $Rx = Q^T b$ using back substitution.
    $endgroup$
    – littleO
    yesterday








5




5




$begingroup$
you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
$endgroup$
– Doug M
yesterday




$begingroup$
you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
$endgroup$
– Doug M
yesterday




2




2




$begingroup$
This is a linear system of equations, which some believe it is the most studied equation in all of mathematics. The reason being that it is so widely used in applied mathematics that there's always reason to find faster and more robust methods that will either be generic or suit the particularities of a given problem. You might roll your eyes at my claim when thinking of your two variable system, but soem engineers need to solve such systems with hundreds of variables in their jobs.
$endgroup$
– Mefitico
yesterday




$begingroup$
This is a linear system of equations, which some believe it is the most studied equation in all of mathematics. The reason being that it is so widely used in applied mathematics that there's always reason to find faster and more robust methods that will either be generic or suit the particularities of a given problem. You might roll your eyes at my claim when thinking of your two variable system, but soem engineers need to solve such systems with hundreds of variables in their jobs.
$endgroup$
– Mefitico
yesterday




3




3




$begingroup$
I hope someone performs GMRES by hand on this system and reports the steps.
$endgroup$
– Rahul
yesterday






$begingroup$
I hope someone performs GMRES by hand on this system and reports the steps.
$endgroup$
– Rahul
yesterday






2




2




$begingroup$
Since linear systems are so well studied, there are many approaches (that are essentially equivalent - but maybe not the iterative solution). As such, does this question essentially boil down to a list of answers, which is not technically on topic for this site?
$endgroup$
– Teepeemm
yesterday




$begingroup$
Since linear systems are so well studied, there are many approaches (that are essentially equivalent - but maybe not the iterative solution). As such, does this question essentially boil down to a list of answers, which is not technically on topic for this site?
$endgroup$
– Teepeemm
yesterday




2




2




$begingroup$
There is an entire subject called Numerical Linear Algebra which studies efficient ways to solve $Ax = b$. There are many notable algorithms. For example, you could use an iterative algorithm such as the Jacobi method or Gauss-Seidel or, as @Rahul noted, GMRES. There are other direct methods also. For example, you could find the QR factorization $A = QR$, where $Q$ is orthogonal and $R$ is upper triangular, and solve $Rx = Q^T b$ using back substitution.
$endgroup$
– littleO
yesterday




$begingroup$
There is an entire subject called Numerical Linear Algebra which studies efficient ways to solve $Ax = b$. There are many notable algorithms. For example, you could use an iterative algorithm such as the Jacobi method or Gauss-Seidel or, as @Rahul noted, GMRES. There are other direct methods also. For example, you could find the QR factorization $A = QR$, where $Q$ is orthogonal and $R$ is upper triangular, and solve $Rx = Q^T b$ using back substitution.
$endgroup$
– littleO
yesterday










11 Answers
11






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oldest

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16












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Is this method allowed ?



$$begin{pmatrix}3&2&36\5&4&64 end{pmatrix} sim begin{pmatrix}1& 2/3&12\5&4&64 end{pmatrix} sim begin{pmatrix}1&2/3&12\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&1&6 end{pmatrix}$$



Which yields $x=8$ and $y=6$





The first step is $R_1 to R_1 times frac{1}{3}$



The second step is $R_2 to R_2 - 5R_1$



The third step is $R_1 to R_1 -R_2$



The fourth step is $R_2 to R_2times frac{3}{2}$



Here $R_i$ denotes the $i$ -th row.






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  • $begingroup$
    I have never seen that! What is it? :D
    $endgroup$
    – user477343
    yesterday






  • 1




    $begingroup$
    elementary operations!
    $endgroup$
    – Chinnapparaj R
    yesterday






  • 1




    $begingroup$
    I assume $R$ stands for Row.
    $endgroup$
    – user477343
    yesterday






  • 23




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    It's also called Gaussian elimination.
    $endgroup$
    – YiFan
    yesterday






  • 3




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    See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
    $endgroup$
    – Eric Towers
    yesterday



















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How about using Cramer's Rule? Define $Delta_x=left[begin{matrix}36 & 2 \ 64 & 4end{matrix}right]$, $Delta_y=left[begin{matrix}3 & 36\ 5 & 64end{matrix}right]$
and $Delta_0=left[begin{matrix}3 & 2\ 5 &4end{matrix}right]$.



Now computation is trivial as you have: $x=dfrac{detDelta_x}{detDelta_0}$ and $y=dfrac{detDelta_y}{detDelta_0}$.






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  • 1




    $begingroup$
    Wow! Very useful! I have never heard of this method, before! $(+1)$
    $endgroup$
    – user477343
    yesterday






  • 1




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    You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
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    – Paras Khosla
    yesterday






  • 10




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    Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
    $endgroup$
    – alephzero
    yesterday






  • 2




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    @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
    $endgroup$
    – mlk
    yesterday






  • 3




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    @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
    $endgroup$
    – user1717828
    yesterday





















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By false position:



Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification



$$3x'+2y'=0,\5x'+4y'=2.$$



We easily eliminate $y'$ (using $4y'=-6x'$) and get



$$-x'=2.$$



Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.






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  • 1




    $begingroup$
    This is a great method. +1 :)
    $endgroup$
    – Paras Khosla
    yesterday



















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Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.



plot of simultaneous equations






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  • $begingroup$
    That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
    $endgroup$
    – user477343
    yesterday





















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Any method you can come up with will in the end amount to Cramer's rule, which gives explicit formulas for the solution. Except special cases, the solution of a system is unique, so that you will always be computing the ratio of those determinants.



Anyway, it turns out that by organizing the computation in certain ways, you can reduce the number of arithmetic operations to be performed. For $2times2$ systems,
the different variants make little difference in this respect. Things become more interesting for $ntimes n$ systems.



Direct application of Cramer is by far the worse, as it takes a number of operations proportional to $(n+1)!$, which is huge. Even for $3times3$ systems, it should be avoided. The best method to date is Gaussian elimination (you eliminate one unknown at a time by forming linear combinations of the equations and turn the system to a triangular form). The total workload is proportional to $n^3$ operations.





The steps of standard Gaussian elimination:



$$begin{cases}ax+by=c,\dx+ey=f.end{cases}$$



Subtract the first times $dfrac da$ from the second,



$$begin{cases}ax+by=c,\0x+left(e-bdfrac daright)y=f-cdfrac da.end{cases}$$



Solve for $y$,



$$begin{cases}ax+by=c,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



Solve for $x$,



$$begin{cases}x=dfrac{c-bdfrac{f-cdfrac da}{e-bdfrac da}}a,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



So written, the formulas are a little scary, but when you use intermediate variables, the complexity vanishes:



$$d'=frac da,e'=e-bd',f'=f-cd'to y=frac{f'}{e'}, x=frac{c-by}a.$$



Anyway, for a $2times2$ system, this is worse than Cramer !



$$begin{cases}x=dfrac{ce-bf}{Delta},\y=dfrac{af-cd}{Delta}end{cases}$$ where $Delta=ae-bd$.





For large systems, say $100times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)






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  • 2




    $begingroup$
    +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
    $endgroup$
    – Surb
    yesterday



















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Fixed Point Iteration



This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $begin{bmatrix} x\ yend{bmatrix}$ on the left hand side.



Define
$fbegin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} (36-2y)/3 \ (64-5x)/4end{bmatrix}$



Start with an intial guess of $begin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} 0\ 0end{bmatrix}$



The result is $fbegin{bmatrix} 0\ 0end{bmatrix}=begin{bmatrix} 12\ 16end{bmatrix}$



Now plug that back into f



The result is $fbegin{bmatrix} 12\ 6end{bmatrix}=begin{bmatrix} 4/3\ 1end{bmatrix}$



Keep plugging the result back in. After 100 iterations you have:



$begin{bmatrix} 7.9991\ 5.9993end{bmatrix}$



Here is a graph of the progression of the iteration:
iteration path






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  • 2




    $begingroup$
    So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
    $endgroup$
    – user477343
    yesterday





















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$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$



From $(1)$, $x=frac{36-2y}{3}$, substitute in $(2)$ and you'll get $5(frac{36-2y}{3})+4y=64 implies y=6$ and then you can get that $x=24/3=8$



Another Method
From $(1)$, $x=frac{36-2y}{3}$



From $(2)$, $x=frac{64-4y}{5}$



But $x=x implies frac{36-2y}{3}=frac{64-4y}{5}$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) implies y=6$ and substitute to get $x=8$






share|cite|improve this answer











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  • 1




    $begingroup$
    Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
    $endgroup$
    – user477343
    yesterday





















5












$begingroup$

Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
$$ mathrm{GB}({3x+2y-36, 5x+4y-64}) = {y-6, x-8} $$
and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this problem. However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.





Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$. A sequence of reductions (similar to the Euclidean algorithm for GCDs): begin{align*}
(5,4,-64) - (3,2,-36) &= (2,2,-28) \
(3,2,-36) - (2,2,-28) &= (1,0,-8) tag{1} \
(2,2,-28) - 2(1,0,-8) &= (0,2,-12) tag{2} \
end{align*}

From (1), we have $x=8$. From (2), $2y = 12$, so $y = 6$. (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components. One implementation of the LLL algorithm, terminates with the shorter vectors ${(-1,2,4), (-2,2,4)}$, but we would continue to manipulate these to get the desired zeroes.)






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$endgroup$





















    4












    $begingroup$

    Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.





    Method $1$: (multiplicity of $y$)




    Let $y=kx$ for some $kinBbb R$. Then $$3x+2y=36implies x(2k+3)=36implies x=frac{36}{2k+3}\5x+4y=64implies x(4k+5)=64implies x=frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)implies (144-128)k=(192-180)implies k=frac34.$$ Now $$x=frac{64}{4k+5}=frac{64}{4cdotfrac34+5}=8implies y=kx=frac34cdot8=6.quadsquare$$






    Method $2$: (use this if you really like quadratic equations :P)




    How about we try squaring the equations? We get $$3x+2y=36implies 9x^2+12xy+4y^2=1296\5x+4y=64implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
      $endgroup$
      – user477343
      yesterday








    • 1




      $begingroup$
      Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
      $endgroup$
      – TheSimpliFire
      yesterday










    • $begingroup$
      So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
      $endgroup$
      – user477343
      yesterday








    • 1




      $begingroup$
      No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
      $endgroup$
      – TheSimpliFire
      yesterday










    • $begingroup$
      Ok. Thank you for clarifying!
      $endgroup$
      – user477343
      yesterday



















    2












    $begingroup$

    It is clear that:





    • $x=10$, $y=3$ is an integer solution of $(1)$.


    • $x=12$, $y=1$ is an integer solution of $(2)$.


    Then, from the theory of Linear Diophantine equations:




    • Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.

    • Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.


    Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that



    $$10+2t=x_0=12+4tqquadtext{and}qquad 3-3t=y_0=1-5t.$$



    Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
    $$x_0=12+4(-1)=8,; y_0=1-5(-1)=6.$$



    Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Consider the three vectors $textbf{A}=(3,2)$, $textbf{B}=(5,4)$ and $textbf{X}=(x,y)$. Your system could be written as $$textbf{A}cdottextbf{X}=a\textbf{B}cdottextbf{X}=b$$ where $a=36$, $b=64$ and $textbf{A}_{perp}=(-2,3)$ is orthogonal to $textbf{A}$. The first equation gives us $textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+lambdatextbf{A}_{perp}$. Now to find $lambda$ we use the second equation, we get $lambda=dfrac{b}{textbf{A}_{perp}cdottextbf{B}}-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2timestextbf{A}_{perp}cdottextbf{B}}$. Et voilà :
      $$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$






      share|cite|improve this answer









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        11 Answers
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        active

        oldest

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        16












        $begingroup$

        Is this method allowed ?



        $$begin{pmatrix}3&2&36\5&4&64 end{pmatrix} sim begin{pmatrix}1& 2/3&12\5&4&64 end{pmatrix} sim begin{pmatrix}1&2/3&12\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&1&6 end{pmatrix}$$



        Which yields $x=8$ and $y=6$





        The first step is $R_1 to R_1 times frac{1}{3}$



        The second step is $R_2 to R_2 - 5R_1$



        The third step is $R_1 to R_1 -R_2$



        The fourth step is $R_2 to R_2times frac{3}{2}$



        Here $R_i$ denotes the $i$ -th row.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          I have never seen that! What is it? :D
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          elementary operations!
          $endgroup$
          – Chinnapparaj R
          yesterday






        • 1




          $begingroup$
          I assume $R$ stands for Row.
          $endgroup$
          – user477343
          yesterday






        • 23




          $begingroup$
          It's also called Gaussian elimination.
          $endgroup$
          – YiFan
          yesterday






        • 3




          $begingroup$
          See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
          $endgroup$
          – Eric Towers
          yesterday
















        16












        $begingroup$

        Is this method allowed ?



        $$begin{pmatrix}3&2&36\5&4&64 end{pmatrix} sim begin{pmatrix}1& 2/3&12\5&4&64 end{pmatrix} sim begin{pmatrix}1&2/3&12\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&1&6 end{pmatrix}$$



        Which yields $x=8$ and $y=6$





        The first step is $R_1 to R_1 times frac{1}{3}$



        The second step is $R_2 to R_2 - 5R_1$



        The third step is $R_1 to R_1 -R_2$



        The fourth step is $R_2 to R_2times frac{3}{2}$



        Here $R_i$ denotes the $i$ -th row.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          I have never seen that! What is it? :D
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          elementary operations!
          $endgroup$
          – Chinnapparaj R
          yesterday






        • 1




          $begingroup$
          I assume $R$ stands for Row.
          $endgroup$
          – user477343
          yesterday






        • 23




          $begingroup$
          It's also called Gaussian elimination.
          $endgroup$
          – YiFan
          yesterday






        • 3




          $begingroup$
          See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
          $endgroup$
          – Eric Towers
          yesterday














        16












        16








        16





        $begingroup$

        Is this method allowed ?



        $$begin{pmatrix}3&2&36\5&4&64 end{pmatrix} sim begin{pmatrix}1& 2/3&12\5&4&64 end{pmatrix} sim begin{pmatrix}1&2/3&12\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&1&6 end{pmatrix}$$



        Which yields $x=8$ and $y=6$





        The first step is $R_1 to R_1 times frac{1}{3}$



        The second step is $R_2 to R_2 - 5R_1$



        The third step is $R_1 to R_1 -R_2$



        The fourth step is $R_2 to R_2times frac{3}{2}$



        Here $R_i$ denotes the $i$ -th row.






        share|cite|improve this answer











        $endgroup$



        Is this method allowed ?



        $$begin{pmatrix}3&2&36\5&4&64 end{pmatrix} sim begin{pmatrix}1& 2/3&12\5&4&64 end{pmatrix} sim begin{pmatrix}1&2/3&12\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&2/3&4 end{pmatrix} sim begin{pmatrix}1&0&8\0&1&6 end{pmatrix}$$



        Which yields $x=8$ and $y=6$





        The first step is $R_1 to R_1 times frac{1}{3}$



        The second step is $R_2 to R_2 - 5R_1$



        The third step is $R_1 to R_1 -R_2$



        The fourth step is $R_2 to R_2times frac{3}{2}$



        Here $R_i$ denotes the $i$ -th row.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Chinnapparaj RChinnapparaj R

        6,2772929




        6,2772929












        • $begingroup$
          I have never seen that! What is it? :D
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          elementary operations!
          $endgroup$
          – Chinnapparaj R
          yesterday






        • 1




          $begingroup$
          I assume $R$ stands for Row.
          $endgroup$
          – user477343
          yesterday






        • 23




          $begingroup$
          It's also called Gaussian elimination.
          $endgroup$
          – YiFan
          yesterday






        • 3




          $begingroup$
          See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
          $endgroup$
          – Eric Towers
          yesterday


















        • $begingroup$
          I have never seen that! What is it? :D
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          elementary operations!
          $endgroup$
          – Chinnapparaj R
          yesterday






        • 1




          $begingroup$
          I assume $R$ stands for Row.
          $endgroup$
          – user477343
          yesterday






        • 23




          $begingroup$
          It's also called Gaussian elimination.
          $endgroup$
          – YiFan
          yesterday






        • 3




          $begingroup$
          See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
          $endgroup$
          – Eric Towers
          yesterday
















        $begingroup$
        I have never seen that! What is it? :D
        $endgroup$
        – user477343
        yesterday




        $begingroup$
        I have never seen that! What is it? :D
        $endgroup$
        – user477343
        yesterday




        1




        1




        $begingroup$
        elementary operations!
        $endgroup$
        – Chinnapparaj R
        yesterday




        $begingroup$
        elementary operations!
        $endgroup$
        – Chinnapparaj R
        yesterday




        1




        1




        $begingroup$
        I assume $R$ stands for Row.
        $endgroup$
        – user477343
        yesterday




        $begingroup$
        I assume $R$ stands for Row.
        $endgroup$
        – user477343
        yesterday




        23




        23




        $begingroup$
        It's also called Gaussian elimination.
        $endgroup$
        – YiFan
        yesterday




        $begingroup$
        It's also called Gaussian elimination.
        $endgroup$
        – YiFan
        yesterday




        3




        3




        $begingroup$
        See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
        $endgroup$
        – Eric Towers
        yesterday




        $begingroup$
        See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
        $endgroup$
        – Eric Towers
        yesterday











        15












        $begingroup$

        How about using Cramer's Rule? Define $Delta_x=left[begin{matrix}36 & 2 \ 64 & 4end{matrix}right]$, $Delta_y=left[begin{matrix}3 & 36\ 5 & 64end{matrix}right]$
        and $Delta_0=left[begin{matrix}3 & 2\ 5 &4end{matrix}right]$.



        Now computation is trivial as you have: $x=dfrac{detDelta_x}{detDelta_0}$ and $y=dfrac{detDelta_y}{detDelta_0}$.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Wow! Very useful! I have never heard of this method, before! $(+1)$
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
          $endgroup$
          – Paras Khosla
          yesterday






        • 10




          $begingroup$
          Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
          $endgroup$
          – alephzero
          yesterday






        • 2




          $begingroup$
          @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
          $endgroup$
          – mlk
          yesterday






        • 3




          $begingroup$
          @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
          $endgroup$
          – user1717828
          yesterday


















        15












        $begingroup$

        How about using Cramer's Rule? Define $Delta_x=left[begin{matrix}36 & 2 \ 64 & 4end{matrix}right]$, $Delta_y=left[begin{matrix}3 & 36\ 5 & 64end{matrix}right]$
        and $Delta_0=left[begin{matrix}3 & 2\ 5 &4end{matrix}right]$.



        Now computation is trivial as you have: $x=dfrac{detDelta_x}{detDelta_0}$ and $y=dfrac{detDelta_y}{detDelta_0}$.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          Wow! Very useful! I have never heard of this method, before! $(+1)$
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
          $endgroup$
          – Paras Khosla
          yesterday






        • 10




          $begingroup$
          Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
          $endgroup$
          – alephzero
          yesterday






        • 2




          $begingroup$
          @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
          $endgroup$
          – mlk
          yesterday






        • 3




          $begingroup$
          @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
          $endgroup$
          – user1717828
          yesterday
















        15












        15








        15





        $begingroup$

        How about using Cramer's Rule? Define $Delta_x=left[begin{matrix}36 & 2 \ 64 & 4end{matrix}right]$, $Delta_y=left[begin{matrix}3 & 36\ 5 & 64end{matrix}right]$
        and $Delta_0=left[begin{matrix}3 & 2\ 5 &4end{matrix}right]$.



        Now computation is trivial as you have: $x=dfrac{detDelta_x}{detDelta_0}$ and $y=dfrac{detDelta_y}{detDelta_0}$.






        share|cite|improve this answer









        $endgroup$



        How about using Cramer's Rule? Define $Delta_x=left[begin{matrix}36 & 2 \ 64 & 4end{matrix}right]$, $Delta_y=left[begin{matrix}3 & 36\ 5 & 64end{matrix}right]$
        and $Delta_0=left[begin{matrix}3 & 2\ 5 &4end{matrix}right]$.



        Now computation is trivial as you have: $x=dfrac{detDelta_x}{detDelta_0}$ and $y=dfrac{detDelta_y}{detDelta_0}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Paras KhoslaParas Khosla

        3,173626




        3,173626








        • 1




          $begingroup$
          Wow! Very useful! I have never heard of this method, before! $(+1)$
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
          $endgroup$
          – Paras Khosla
          yesterday






        • 10




          $begingroup$
          Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
          $endgroup$
          – alephzero
          yesterday






        • 2




          $begingroup$
          @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
          $endgroup$
          – mlk
          yesterday






        • 3




          $begingroup$
          @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
          $endgroup$
          – user1717828
          yesterday
















        • 1




          $begingroup$
          Wow! Very useful! I have never heard of this method, before! $(+1)$
          $endgroup$
          – user477343
          yesterday






        • 1




          $begingroup$
          You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
          $endgroup$
          – Paras Khosla
          yesterday






        • 10




          $begingroup$
          Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
          $endgroup$
          – alephzero
          yesterday






        • 2




          $begingroup$
          @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
          $endgroup$
          – mlk
          yesterday






        • 3




          $begingroup$
          @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
          $endgroup$
          – user1717828
          yesterday










        1




        1




        $begingroup$
        Wow! Very useful! I have never heard of this method, before! $(+1)$
        $endgroup$
        – user477343
        yesterday




        $begingroup$
        Wow! Very useful! I have never heard of this method, before! $(+1)$
        $endgroup$
        – user477343
        yesterday




        1




        1




        $begingroup$
        You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
        $endgroup$
        – Paras Khosla
        yesterday




        $begingroup$
        You must've made a calculation mistake. Recheck your calculations. It does indeed give $(2, 1)$ as the answer. Cheers :)
        $endgroup$
        – Paras Khosla
        yesterday




        10




        10




        $begingroup$
        Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
        $endgroup$
        – alephzero
        yesterday




        $begingroup$
        Cramer's rule is important theoretically, but it is a very inefficient way to solve equations numerically, except for two equations in two unknowns. For $n$ equations, Cramer's rule requires $n!$ arithmetic operations to evaluate the determinants, compared with about $n^3$ operations to solve using Gaussian elimination. Even when $n = 10$, $n^3 = 1000$ but $n! = 3628800$. And in many real world applied math computations, $n = 100,000$ is a "small problem!"
        $endgroup$
        – alephzero
        yesterday




        2




        2




        $begingroup$
        @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
        $endgroup$
        – mlk
        yesterday




        $begingroup$
        @alephzero Just to be technical, there are faster ways to calculate the determinant of large matrices. However the one method I know to do it in n^3 relies on Gaussian elimination itself, which makes it a bit redundant...
        $endgroup$
        – mlk
        yesterday




        3




        3




        $begingroup$
        @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
        $endgroup$
        – user1717828
        yesterday






        $begingroup$
        @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
        $endgroup$
        – user1717828
        yesterday













        11












        $begingroup$

        By false position:



        Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification



        $$3x'+2y'=0,\5x'+4y'=2.$$



        We easily eliminate $y'$ (using $4y'=-6x'$) and get



        $$-x'=2.$$



        Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          This is a great method. +1 :)
          $endgroup$
          – Paras Khosla
          yesterday
















        11












        $begingroup$

        By false position:



        Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification



        $$3x'+2y'=0,\5x'+4y'=2.$$



        We easily eliminate $y'$ (using $4y'=-6x'$) and get



        $$-x'=2.$$



        Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          This is a great method. +1 :)
          $endgroup$
          – Paras Khosla
          yesterday














        11












        11








        11





        $begingroup$

        By false position:



        Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification



        $$3x'+2y'=0,\5x'+4y'=2.$$



        We easily eliminate $y'$ (using $4y'=-6x'$) and get



        $$-x'=2.$$



        Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.






        share|cite|improve this answer









        $endgroup$



        By false position:



        Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification



        $$3x'+2y'=0,\5x'+4y'=2.$$



        We easily eliminate $y'$ (using $4y'=-6x'$) and get



        $$-x'=2.$$



        Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Yves DaoustYves Daoust

        133k676231




        133k676231








        • 1




          $begingroup$
          This is a great method. +1 :)
          $endgroup$
          – Paras Khosla
          yesterday














        • 1




          $begingroup$
          This is a great method. +1 :)
          $endgroup$
          – Paras Khosla
          yesterday








        1




        1




        $begingroup$
        This is a great method. +1 :)
        $endgroup$
        – Paras Khosla
        yesterday




        $begingroup$
        This is a great method. +1 :)
        $endgroup$
        – Paras Khosla
        yesterday











        10












        $begingroup$

        Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.



        plot of simultaneous equations






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
          $endgroup$
          – user477343
          yesterday


















        10












        $begingroup$

        Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.



        plot of simultaneous equations






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
          $endgroup$
          – user477343
          yesterday
















        10












        10








        10





        $begingroup$

        Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.



        plot of simultaneous equations






        share|cite|improve this answer









        $endgroup$



        Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.



        plot of simultaneous equations







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Elements in SpaceElements in Space

        1,28211228




        1,28211228












        • $begingroup$
          That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
          $endgroup$
          – user477343
          yesterday




















        • $begingroup$
          That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
          $endgroup$
          – user477343
          yesterday


















        $begingroup$
        That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
        $endgroup$
        – user477343
        yesterday






        $begingroup$
        That's what my school textbook wants me to do, but it can sometimes be a bit... tiring... but methinks graphing does reveal the essence of simultaneous equations. $(+1)$
        $endgroup$
        – user477343
        yesterday













        9












        $begingroup$

        Any method you can come up with will in the end amount to Cramer's rule, which gives explicit formulas for the solution. Except special cases, the solution of a system is unique, so that you will always be computing the ratio of those determinants.



        Anyway, it turns out that by organizing the computation in certain ways, you can reduce the number of arithmetic operations to be performed. For $2times2$ systems,
        the different variants make little difference in this respect. Things become more interesting for $ntimes n$ systems.



        Direct application of Cramer is by far the worse, as it takes a number of operations proportional to $(n+1)!$, which is huge. Even for $3times3$ systems, it should be avoided. The best method to date is Gaussian elimination (you eliminate one unknown at a time by forming linear combinations of the equations and turn the system to a triangular form). The total workload is proportional to $n^3$ operations.





        The steps of standard Gaussian elimination:



        $$begin{cases}ax+by=c,\dx+ey=f.end{cases}$$



        Subtract the first times $dfrac da$ from the second,



        $$begin{cases}ax+by=c,\0x+left(e-bdfrac daright)y=f-cdfrac da.end{cases}$$



        Solve for $y$,



        $$begin{cases}ax+by=c,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        Solve for $x$,



        $$begin{cases}x=dfrac{c-bdfrac{f-cdfrac da}{e-bdfrac da}}a,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        So written, the formulas are a little scary, but when you use intermediate variables, the complexity vanishes:



        $$d'=frac da,e'=e-bd',f'=f-cd'to y=frac{f'}{e'}, x=frac{c-by}a.$$



        Anyway, for a $2times2$ system, this is worse than Cramer !



        $$begin{cases}x=dfrac{ce-bf}{Delta},\y=dfrac{af-cd}{Delta}end{cases}$$ where $Delta=ae-bd$.





        For large systems, say $100times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
          $endgroup$
          – Surb
          yesterday
















        9












        $begingroup$

        Any method you can come up with will in the end amount to Cramer's rule, which gives explicit formulas for the solution. Except special cases, the solution of a system is unique, so that you will always be computing the ratio of those determinants.



        Anyway, it turns out that by organizing the computation in certain ways, you can reduce the number of arithmetic operations to be performed. For $2times2$ systems,
        the different variants make little difference in this respect. Things become more interesting for $ntimes n$ systems.



        Direct application of Cramer is by far the worse, as it takes a number of operations proportional to $(n+1)!$, which is huge. Even for $3times3$ systems, it should be avoided. The best method to date is Gaussian elimination (you eliminate one unknown at a time by forming linear combinations of the equations and turn the system to a triangular form). The total workload is proportional to $n^3$ operations.





        The steps of standard Gaussian elimination:



        $$begin{cases}ax+by=c,\dx+ey=f.end{cases}$$



        Subtract the first times $dfrac da$ from the second,



        $$begin{cases}ax+by=c,\0x+left(e-bdfrac daright)y=f-cdfrac da.end{cases}$$



        Solve for $y$,



        $$begin{cases}ax+by=c,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        Solve for $x$,



        $$begin{cases}x=dfrac{c-bdfrac{f-cdfrac da}{e-bdfrac da}}a,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        So written, the formulas are a little scary, but when you use intermediate variables, the complexity vanishes:



        $$d'=frac da,e'=e-bd',f'=f-cd'to y=frac{f'}{e'}, x=frac{c-by}a.$$



        Anyway, for a $2times2$ system, this is worse than Cramer !



        $$begin{cases}x=dfrac{ce-bf}{Delta},\y=dfrac{af-cd}{Delta}end{cases}$$ where $Delta=ae-bd$.





        For large systems, say $100times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
          $endgroup$
          – Surb
          yesterday














        9












        9








        9





        $begingroup$

        Any method you can come up with will in the end amount to Cramer's rule, which gives explicit formulas for the solution. Except special cases, the solution of a system is unique, so that you will always be computing the ratio of those determinants.



        Anyway, it turns out that by organizing the computation in certain ways, you can reduce the number of arithmetic operations to be performed. For $2times2$ systems,
        the different variants make little difference in this respect. Things become more interesting for $ntimes n$ systems.



        Direct application of Cramer is by far the worse, as it takes a number of operations proportional to $(n+1)!$, which is huge. Even for $3times3$ systems, it should be avoided. The best method to date is Gaussian elimination (you eliminate one unknown at a time by forming linear combinations of the equations and turn the system to a triangular form). The total workload is proportional to $n^3$ operations.





        The steps of standard Gaussian elimination:



        $$begin{cases}ax+by=c,\dx+ey=f.end{cases}$$



        Subtract the first times $dfrac da$ from the second,



        $$begin{cases}ax+by=c,\0x+left(e-bdfrac daright)y=f-cdfrac da.end{cases}$$



        Solve for $y$,



        $$begin{cases}ax+by=c,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        Solve for $x$,



        $$begin{cases}x=dfrac{c-bdfrac{f-cdfrac da}{e-bdfrac da}}a,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        So written, the formulas are a little scary, but when you use intermediate variables, the complexity vanishes:



        $$d'=frac da,e'=e-bd',f'=f-cd'to y=frac{f'}{e'}, x=frac{c-by}a.$$



        Anyway, for a $2times2$ system, this is worse than Cramer !



        $$begin{cases}x=dfrac{ce-bf}{Delta},\y=dfrac{af-cd}{Delta}end{cases}$$ where $Delta=ae-bd$.





        For large systems, say $100times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)






        share|cite|improve this answer











        $endgroup$



        Any method you can come up with will in the end amount to Cramer's rule, which gives explicit formulas for the solution. Except special cases, the solution of a system is unique, so that you will always be computing the ratio of those determinants.



        Anyway, it turns out that by organizing the computation in certain ways, you can reduce the number of arithmetic operations to be performed. For $2times2$ systems,
        the different variants make little difference in this respect. Things become more interesting for $ntimes n$ systems.



        Direct application of Cramer is by far the worse, as it takes a number of operations proportional to $(n+1)!$, which is huge. Even for $3times3$ systems, it should be avoided. The best method to date is Gaussian elimination (you eliminate one unknown at a time by forming linear combinations of the equations and turn the system to a triangular form). The total workload is proportional to $n^3$ operations.





        The steps of standard Gaussian elimination:



        $$begin{cases}ax+by=c,\dx+ey=f.end{cases}$$



        Subtract the first times $dfrac da$ from the second,



        $$begin{cases}ax+by=c,\0x+left(e-bdfrac daright)y=f-cdfrac da.end{cases}$$



        Solve for $y$,



        $$begin{cases}ax+by=c,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        Solve for $x$,



        $$begin{cases}x=dfrac{c-bdfrac{f-cdfrac da}{e-bdfrac da}}a,\y=dfrac{f-cdfrac da}{e-bdfrac da}.end{cases}$$



        So written, the formulas are a little scary, but when you use intermediate variables, the complexity vanishes:



        $$d'=frac da,e'=e-bd',f'=f-cd'to y=frac{f'}{e'}, x=frac{c-by}a.$$



        Anyway, for a $2times2$ system, this is worse than Cramer !



        $$begin{cases}x=dfrac{ce-bf}{Delta},\y=dfrac{af-cd}{Delta}end{cases}$$ where $Delta=ae-bd$.





        For large systems, say $100times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Yves DaoustYves Daoust

        133k676231




        133k676231








        • 2




          $begingroup$
          +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
          $endgroup$
          – Surb
          yesterday














        • 2




          $begingroup$
          +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
          $endgroup$
          – Surb
          yesterday








        2




        2




        $begingroup$
        +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
        $endgroup$
        – Surb
        yesterday




        $begingroup$
        +1 for the last paragraph which is, I think, of utmost importance. Indeed, our computers solve many, many, linear systems each day (and quite huge ones, not 100x100 but more 100'000 x 100'000). None of them are solved by any the methods discussed in the answers so far.
        $endgroup$
        – Surb
        yesterday











        9












        $begingroup$

        Fixed Point Iteration



        This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $begin{bmatrix} x\ yend{bmatrix}$ on the left hand side.



        Define
        $fbegin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} (36-2y)/3 \ (64-5x)/4end{bmatrix}$



        Start with an intial guess of $begin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} 0\ 0end{bmatrix}$



        The result is $fbegin{bmatrix} 0\ 0end{bmatrix}=begin{bmatrix} 12\ 16end{bmatrix}$



        Now plug that back into f



        The result is $fbegin{bmatrix} 12\ 6end{bmatrix}=begin{bmatrix} 4/3\ 1end{bmatrix}$



        Keep plugging the result back in. After 100 iterations you have:



        $begin{bmatrix} 7.9991\ 5.9993end{bmatrix}$



        Here is a graph of the progression of the iteration:
        iteration path






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
          $endgroup$
          – user477343
          yesterday


















        9












        $begingroup$

        Fixed Point Iteration



        This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $begin{bmatrix} x\ yend{bmatrix}$ on the left hand side.



        Define
        $fbegin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} (36-2y)/3 \ (64-5x)/4end{bmatrix}$



        Start with an intial guess of $begin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} 0\ 0end{bmatrix}$



        The result is $fbegin{bmatrix} 0\ 0end{bmatrix}=begin{bmatrix} 12\ 16end{bmatrix}$



        Now plug that back into f



        The result is $fbegin{bmatrix} 12\ 6end{bmatrix}=begin{bmatrix} 4/3\ 1end{bmatrix}$



        Keep plugging the result back in. After 100 iterations you have:



        $begin{bmatrix} 7.9991\ 5.9993end{bmatrix}$



        Here is a graph of the progression of the iteration:
        iteration path






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
          $endgroup$
          – user477343
          yesterday
















        9












        9








        9





        $begingroup$

        Fixed Point Iteration



        This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $begin{bmatrix} x\ yend{bmatrix}$ on the left hand side.



        Define
        $fbegin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} (36-2y)/3 \ (64-5x)/4end{bmatrix}$



        Start with an intial guess of $begin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} 0\ 0end{bmatrix}$



        The result is $fbegin{bmatrix} 0\ 0end{bmatrix}=begin{bmatrix} 12\ 16end{bmatrix}$



        Now plug that back into f



        The result is $fbegin{bmatrix} 12\ 6end{bmatrix}=begin{bmatrix} 4/3\ 1end{bmatrix}$



        Keep plugging the result back in. After 100 iterations you have:



        $begin{bmatrix} 7.9991\ 5.9993end{bmatrix}$



        Here is a graph of the progression of the iteration:
        iteration path






        share|cite|improve this answer









        $endgroup$



        Fixed Point Iteration



        This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $begin{bmatrix} x\ yend{bmatrix}$ on the left hand side.



        Define
        $fbegin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} (36-2y)/3 \ (64-5x)/4end{bmatrix}$



        Start with an intial guess of $begin{bmatrix} x\ yend{bmatrix}=begin{bmatrix} 0\ 0end{bmatrix}$



        The result is $fbegin{bmatrix} 0\ 0end{bmatrix}=begin{bmatrix} 12\ 16end{bmatrix}$



        Now plug that back into f



        The result is $fbegin{bmatrix} 12\ 6end{bmatrix}=begin{bmatrix} 4/3\ 1end{bmatrix}$



        Keep plugging the result back in. After 100 iterations you have:



        $begin{bmatrix} 7.9991\ 5.9993end{bmatrix}$



        Here is a graph of the progression of the iteration:
        iteration path







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Kelly LowderKelly Lowder

        20515




        20515








        • 2




          $begingroup$
          So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
          $endgroup$
          – user477343
          yesterday
















        • 2




          $begingroup$
          So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
          $endgroup$
          – user477343
          yesterday










        2




        2




        $begingroup$
        So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
        $endgroup$
        – user477343
        yesterday






        $begingroup$
        So we just have $fbegin{bmatrix} 0 \ 0end{bmatrix}$ and then $fbigg(fbegin{bmatrix} 0 \ 0end{bmatrix}bigg)$ and by letting $f^k(cdot ) = f(f(ldots f(f(cdot))ldots )$ $k$ times, this overall goes to $$f^{100}begin{bmatrix} 0 \ 0end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P
        $endgroup$
        – user477343
        yesterday













        6












        $begingroup$

        $$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$



        From $(1)$, $x=frac{36-2y}{3}$, substitute in $(2)$ and you'll get $5(frac{36-2y}{3})+4y=64 implies y=6$ and then you can get that $x=24/3=8$



        Another Method
        From $(1)$, $x=frac{36-2y}{3}$



        From $(2)$, $x=frac{64-4y}{5}$



        But $x=x implies frac{36-2y}{3}=frac{64-4y}{5}$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) implies y=6$ and substitute to get $x=8$






        share|cite|improve this answer











        $endgroup$









        • 1




          $begingroup$
          Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
          $endgroup$
          – user477343
          yesterday


















        6












        $begingroup$

        $$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$



        From $(1)$, $x=frac{36-2y}{3}$, substitute in $(2)$ and you'll get $5(frac{36-2y}{3})+4y=64 implies y=6$ and then you can get that $x=24/3=8$



        Another Method
        From $(1)$, $x=frac{36-2y}{3}$



        From $(2)$, $x=frac{64-4y}{5}$



        But $x=x implies frac{36-2y}{3}=frac{64-4y}{5}$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) implies y=6$ and substitute to get $x=8$






        share|cite|improve this answer











        $endgroup$









        • 1




          $begingroup$
          Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
          $endgroup$
          – user477343
          yesterday
















        6












        6








        6





        $begingroup$

        $$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$



        From $(1)$, $x=frac{36-2y}{3}$, substitute in $(2)$ and you'll get $5(frac{36-2y}{3})+4y=64 implies y=6$ and then you can get that $x=24/3=8$



        Another Method
        From $(1)$, $x=frac{36-2y}{3}$



        From $(2)$, $x=frac{64-4y}{5}$



        But $x=x implies frac{36-2y}{3}=frac{64-4y}{5}$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) implies y=6$ and substitute to get $x=8$






        share|cite|improve this answer











        $endgroup$



        $$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$



        From $(1)$, $x=frac{36-2y}{3}$, substitute in $(2)$ and you'll get $5(frac{36-2y}{3})+4y=64 implies y=6$ and then you can get that $x=24/3=8$



        Another Method
        From $(1)$, $x=frac{36-2y}{3}$



        From $(2)$, $x=frac{64-4y}{5}$



        But $x=x implies frac{36-2y}{3}=frac{64-4y}{5}$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) implies y=6$ and substitute to get $x=8$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Fareed AFFareed AF

        762112




        762112








        • 1




          $begingroup$
          Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
          $endgroup$
          – user477343
          yesterday
















        • 1




          $begingroup$
          Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
          $endgroup$
          – user477343
          yesterday










        1




        1




        $begingroup$
        Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
        $endgroup$
        – user477343
        yesterday






        $begingroup$
        Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
        $endgroup$
        – user477343
        yesterday













        5












        $begingroup$

        Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
        $$ mathrm{GB}({3x+2y-36, 5x+4y-64}) = {y-6, x-8} $$
        and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this problem. However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.





        Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$. A sequence of reductions (similar to the Euclidean algorithm for GCDs): begin{align*}
        (5,4,-64) - (3,2,-36) &= (2,2,-28) \
        (3,2,-36) - (2,2,-28) &= (1,0,-8) tag{1} \
        (2,2,-28) - 2(1,0,-8) &= (0,2,-12) tag{2} \
        end{align*}

        From (1), we have $x=8$. From (2), $2y = 12$, so $y = 6$. (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components. One implementation of the LLL algorithm, terminates with the shorter vectors ${(-1,2,4), (-2,2,4)}$, but we would continue to manipulate these to get the desired zeroes.)






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
          $$ mathrm{GB}({3x+2y-36, 5x+4y-64}) = {y-6, x-8} $$
          and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this problem. However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.





          Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$. A sequence of reductions (similar to the Euclidean algorithm for GCDs): begin{align*}
          (5,4,-64) - (3,2,-36) &= (2,2,-28) \
          (3,2,-36) - (2,2,-28) &= (1,0,-8) tag{1} \
          (2,2,-28) - 2(1,0,-8) &= (0,2,-12) tag{2} \
          end{align*}

          From (1), we have $x=8$. From (2), $2y = 12$, so $y = 6$. (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components. One implementation of the LLL algorithm, terminates with the shorter vectors ${(-1,2,4), (-2,2,4)}$, but we would continue to manipulate these to get the desired zeroes.)






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
            $$ mathrm{GB}({3x+2y-36, 5x+4y-64}) = {y-6, x-8} $$
            and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this problem. However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.





            Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$. A sequence of reductions (similar to the Euclidean algorithm for GCDs): begin{align*}
            (5,4,-64) - (3,2,-36) &= (2,2,-28) \
            (3,2,-36) - (2,2,-28) &= (1,0,-8) tag{1} \
            (2,2,-28) - 2(1,0,-8) &= (0,2,-12) tag{2} \
            end{align*}

            From (1), we have $x=8$. From (2), $2y = 12$, so $y = 6$. (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components. One implementation of the LLL algorithm, terminates with the shorter vectors ${(-1,2,4), (-2,2,4)}$, but we would continue to manipulate these to get the desired zeroes.)






            share|cite|improve this answer









            $endgroup$



            Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
            $$ mathrm{GB}({3x+2y-36, 5x+4y-64}) = {y-6, x-8} $$
            and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this problem. However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.





            Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$. A sequence of reductions (similar to the Euclidean algorithm for GCDs): begin{align*}
            (5,4,-64) - (3,2,-36) &= (2,2,-28) \
            (3,2,-36) - (2,2,-28) &= (1,0,-8) tag{1} \
            (2,2,-28) - 2(1,0,-8) &= (0,2,-12) tag{2} \
            end{align*}

            From (1), we have $x=8$. From (2), $2y = 12$, so $y = 6$. (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components. One implementation of the LLL algorithm, terminates with the shorter vectors ${(-1,2,4), (-2,2,4)}$, but we would continue to manipulate these to get the desired zeroes.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 19 hours ago









            Eric TowersEric Towers

            33.8k22370




            33.8k22370























                4












                $begingroup$

                Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.





                Method $1$: (multiplicity of $y$)




                Let $y=kx$ for some $kinBbb R$. Then $$3x+2y=36implies x(2k+3)=36implies x=frac{36}{2k+3}\5x+4y=64implies x(4k+5)=64implies x=frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)implies (144-128)k=(192-180)implies k=frac34.$$ Now $$x=frac{64}{4k+5}=frac{64}{4cdotfrac34+5}=8implies y=kx=frac34cdot8=6.quadsquare$$






                Method $2$: (use this if you really like quadratic equations :P)




                How about we try squaring the equations? We get $$3x+2y=36implies 9x^2+12xy+4y^2=1296\5x+4y=64implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!







                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  Ok. Thank you for clarifying!
                  $endgroup$
                  – user477343
                  yesterday
















                4












                $begingroup$

                Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.





                Method $1$: (multiplicity of $y$)




                Let $y=kx$ for some $kinBbb R$. Then $$3x+2y=36implies x(2k+3)=36implies x=frac{36}{2k+3}\5x+4y=64implies x(4k+5)=64implies x=frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)implies (144-128)k=(192-180)implies k=frac34.$$ Now $$x=frac{64}{4k+5}=frac{64}{4cdotfrac34+5}=8implies y=kx=frac34cdot8=6.quadsquare$$






                Method $2$: (use this if you really like quadratic equations :P)




                How about we try squaring the equations? We get $$3x+2y=36implies 9x^2+12xy+4y^2=1296\5x+4y=64implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!







                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  Ok. Thank you for clarifying!
                  $endgroup$
                  – user477343
                  yesterday














                4












                4








                4





                $begingroup$

                Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.





                Method $1$: (multiplicity of $y$)




                Let $y=kx$ for some $kinBbb R$. Then $$3x+2y=36implies x(2k+3)=36implies x=frac{36}{2k+3}\5x+4y=64implies x(4k+5)=64implies x=frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)implies (144-128)k=(192-180)implies k=frac34.$$ Now $$x=frac{64}{4k+5}=frac{64}{4cdotfrac34+5}=8implies y=kx=frac34cdot8=6.quadsquare$$






                Method $2$: (use this if you really like quadratic equations :P)




                How about we try squaring the equations? We get $$3x+2y=36implies 9x^2+12xy+4y^2=1296\5x+4y=64implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!







                share|cite|improve this answer











                $endgroup$



                Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.





                Method $1$: (multiplicity of $y$)




                Let $y=kx$ for some $kinBbb R$. Then $$3x+2y=36implies x(2k+3)=36implies x=frac{36}{2k+3}\5x+4y=64implies x(4k+5)=64implies x=frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)implies (144-128)k=(192-180)implies k=frac34.$$ Now $$x=frac{64}{4k+5}=frac{64}{4cdotfrac34+5}=8implies y=kx=frac34cdot8=6.quadsquare$$






                Method $2$: (use this if you really like quadratic equations :P)




                How about we try squaring the equations? We get $$3x+2y=36implies 9x^2+12xy+4y^2=1296\5x+4y=64implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!








                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered yesterday









                TheSimpliFireTheSimpliFire

                13.2k62464




                13.2k62464












                • $begingroup$
                  In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  Ok. Thank you for clarifying!
                  $endgroup$
                  – user477343
                  yesterday


















                • $begingroup$
                  In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
                  $endgroup$
                  – user477343
                  yesterday








                • 1




                  $begingroup$
                  No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
                  $endgroup$
                  – TheSimpliFire
                  yesterday










                • $begingroup$
                  Ok. Thank you for clarifying!
                  $endgroup$
                  – user477343
                  yesterday
















                $begingroup$
                In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
                $endgroup$
                – user477343
                yesterday






                $begingroup$
                In your first method, why do you substitute $k=frac34$ in the second equation $5x+4y=64$ as opposed to the first equation $3x+2y=36$? Also, hello! :D
                $endgroup$
                – user477343
                yesterday






                1




                1




                $begingroup$
                Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
                $endgroup$
                – TheSimpliFire
                yesterday




                $begingroup$
                Because for $3x+2y=36$, we get $2k$ in the denominator, but $2k=3/2$ leaves us with a fraction. If we use the other equation, we get $4k=3$ which is neater.
                $endgroup$
                – TheSimpliFire
                yesterday












                $begingroup$
                So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
                $endgroup$
                – user477343
                yesterday






                $begingroup$
                So, it doesn't really matter which one we substitute it in; but it is good to have some intuition when deciding! Thanks for your answer :P $(+1)$
                $endgroup$
                – user477343
                yesterday






                1




                1




                $begingroup$
                No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
                $endgroup$
                – TheSimpliFire
                yesterday




                $begingroup$
                No, at an intersection point between two lines, most of their properties at that point are the same (apart from gradient, of course)
                $endgroup$
                – TheSimpliFire
                yesterday












                $begingroup$
                Ok. Thank you for clarifying!
                $endgroup$
                – user477343
                yesterday




                $begingroup$
                Ok. Thank you for clarifying!
                $endgroup$
                – user477343
                yesterday











                2












                $begingroup$

                It is clear that:





                • $x=10$, $y=3$ is an integer solution of $(1)$.


                • $x=12$, $y=1$ is an integer solution of $(2)$.


                Then, from the theory of Linear Diophantine equations:




                • Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.

                • Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.


                Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that



                $$10+2t=x_0=12+4tqquadtext{and}qquad 3-3t=y_0=1-5t.$$



                Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
                $$x_0=12+4(-1)=8,; y_0=1-5(-1)=6.$$



                Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  It is clear that:





                  • $x=10$, $y=3$ is an integer solution of $(1)$.


                  • $x=12$, $y=1$ is an integer solution of $(2)$.


                  Then, from the theory of Linear Diophantine equations:




                  • Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.

                  • Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.


                  Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that



                  $$10+2t=x_0=12+4tqquadtext{and}qquad 3-3t=y_0=1-5t.$$



                  Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
                  $$x_0=12+4(-1)=8,; y_0=1-5(-1)=6.$$



                  Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    It is clear that:





                    • $x=10$, $y=3$ is an integer solution of $(1)$.


                    • $x=12$, $y=1$ is an integer solution of $(2)$.


                    Then, from the theory of Linear Diophantine equations:




                    • Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.

                    • Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.


                    Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that



                    $$10+2t=x_0=12+4tqquadtext{and}qquad 3-3t=y_0=1-5t.$$



                    Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
                    $$x_0=12+4(-1)=8,; y_0=1-5(-1)=6.$$



                    Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.






                    share|cite|improve this answer











                    $endgroup$



                    It is clear that:





                    • $x=10$, $y=3$ is an integer solution of $(1)$.


                    • $x=12$, $y=1$ is an integer solution of $(2)$.


                    Then, from the theory of Linear Diophantine equations:




                    • Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.

                    • Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.


                    Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that



                    $$10+2t=x_0=12+4tqquadtext{and}qquad 3-3t=y_0=1-5t.$$



                    Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
                    $$x_0=12+4(-1)=8,; y_0=1-5(-1)=6.$$



                    Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited yesterday

























                    answered yesterday









                    PedroPedro

                    10.9k23475




                    10.9k23475























                        0












                        $begingroup$

                        Consider the three vectors $textbf{A}=(3,2)$, $textbf{B}=(5,4)$ and $textbf{X}=(x,y)$. Your system could be written as $$textbf{A}cdottextbf{X}=a\textbf{B}cdottextbf{X}=b$$ where $a=36$, $b=64$ and $textbf{A}_{perp}=(-2,3)$ is orthogonal to $textbf{A}$. The first equation gives us $textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+lambdatextbf{A}_{perp}$. Now to find $lambda$ we use the second equation, we get $lambda=dfrac{b}{textbf{A}_{perp}cdottextbf{B}}-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2timestextbf{A}_{perp}cdottextbf{B}}$. Et voilà :
                        $$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Consider the three vectors $textbf{A}=(3,2)$, $textbf{B}=(5,4)$ and $textbf{X}=(x,y)$. Your system could be written as $$textbf{A}cdottextbf{X}=a\textbf{B}cdottextbf{X}=b$$ where $a=36$, $b=64$ and $textbf{A}_{perp}=(-2,3)$ is orthogonal to $textbf{A}$. The first equation gives us $textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+lambdatextbf{A}_{perp}$. Now to find $lambda$ we use the second equation, we get $lambda=dfrac{b}{textbf{A}_{perp}cdottextbf{B}}-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2timestextbf{A}_{perp}cdottextbf{B}}$. Et voilà :
                          $$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$






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                          $endgroup$
















                            0












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                            0





                            $begingroup$

                            Consider the three vectors $textbf{A}=(3,2)$, $textbf{B}=(5,4)$ and $textbf{X}=(x,y)$. Your system could be written as $$textbf{A}cdottextbf{X}=a\textbf{B}cdottextbf{X}=b$$ where $a=36$, $b=64$ and $textbf{A}_{perp}=(-2,3)$ is orthogonal to $textbf{A}$. The first equation gives us $textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+lambdatextbf{A}_{perp}$. Now to find $lambda$ we use the second equation, we get $lambda=dfrac{b}{textbf{A}_{perp}cdottextbf{B}}-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2timestextbf{A}_{perp}cdottextbf{B}}$. Et voilà :
                            $$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$






                            share|cite|improve this answer









                            $endgroup$



                            Consider the three vectors $textbf{A}=(3,2)$, $textbf{B}=(5,4)$ and $textbf{X}=(x,y)$. Your system could be written as $$textbf{A}cdottextbf{X}=a\textbf{B}cdottextbf{X}=b$$ where $a=36$, $b=64$ and $textbf{A}_{perp}=(-2,3)$ is orthogonal to $textbf{A}$. The first equation gives us $textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+lambdatextbf{A}_{perp}$. Now to find $lambda$ we use the second equation, we get $lambda=dfrac{b}{textbf{A}_{perp}cdottextbf{B}}-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2timestextbf{A}_{perp}cdottextbf{B}}$. Et voilà :
                            $$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 5 hours ago









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