Are there any other methods to apply to solving simultaneous equations? The 2019 Stack...
Why hard-Brexiteers don't insist on a hard border to prevent illegal immigration after Brexit?
Pokemon Turn Based battle (Python)
What is the accessibility of a package's `Private` context variables?
What did it mean to "align" a radio?
Why did Acorn's A3000 have red function keys?
What to do when moving next to a bird sanctuary with a loosely-domesticated cat?
How to notate time signature switching consistently every measure
What is the meaning of the verb "bear" in this context?
Did 3000BC Egyptians use meteoric iron weapons?
Have you ever entered Singapore using a different passport or name?
Lightning Grid - Columns and Rows?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
What tool would a Roman-age civilization have for the breaking of silver and other metals into dust?
"as much details as you can remember"
Is flight data recorder erased after every flight?
Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?
Identify boardgame from Big movie
Return to UK after being refused entry years previously
What is the motivation for a law requiring 2 parties to consent for recording a conversation
Does a dangling wire really electrocute me if I'm standing in water?
Why is the maximum length of OpenWrt’s root password 8 characters?
Is "plugging out" electronic devices an American expression?
Multiply Two Integer Polynomials
A poker game description that does not feel gimmicky
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
$begingroup$
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 $y$?
linear-algebra systems-of-equations
$endgroup$
|
show 8 more comments
$begingroup$
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 $y$?
linear-algebra systems-of-equations
$endgroup$
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
|
show 8 more comments
$begingroup$
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 $y$?
linear-algebra systems-of-equations
$endgroup$
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
linear-algebra systems-of-equations
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
|
show 8 more comments
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
|
show 8 more comments
11 Answers
11
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
$endgroup$
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
|
show 3 more comments
$begingroup$
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.
$endgroup$
1
$begingroup$
This is a great method. +1 :)
$endgroup$
– Paras Khosla
yesterday
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.)
$endgroup$
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
add a comment |
$begingroup$
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:
$endgroup$
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
add a comment |
$begingroup$
$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$
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 $x=8$
$endgroup$
1
$begingroup$
Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
$endgroup$
– user477343
yesterday
add a comment |
$begingroup$
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.)
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180580%2fare-there-any-other-methods-to-apply-to-solving-simultaneous-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
11 Answers
11
active
oldest
votes
11 Answers
11
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
$endgroup$
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
|
show 3 more comments
$begingroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
$endgroup$
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
|
show 3 more comments
$begingroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
$endgroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
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
|
show 3 more comments
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
|
show 3 more comments
$begingroup$
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.
$endgroup$
1
$begingroup$
This is a great method. +1 :)
$endgroup$
– Paras Khosla
yesterday
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
This is a great method. +1 :)
$endgroup$
– Paras Khosla
yesterday
add a comment |
$begingroup$
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.
$endgroup$
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.
answered yesterday
Yves DaoustYves Daoust
133k676231
133k676231
1
$begingroup$
This is a great method. +1 :)
$endgroup$
– Paras Khosla
yesterday
add a comment |
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.)
$endgroup$
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
add a comment |
$begingroup$
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.)
$endgroup$
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
add a comment |
$begingroup$
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.)
$endgroup$
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.)
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
add a comment |
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
add a comment |
$begingroup$
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:
$endgroup$
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
add a comment |
$begingroup$
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:
$endgroup$
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
add a comment |
$begingroup$
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:
$endgroup$
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:
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
add a comment |
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
add a comment |
$begingroup$
$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$
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 $x=8$
$endgroup$
1
$begingroup$
Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
$endgroup$
– user477343
yesterday
add a comment |
$begingroup$
$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$
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 $x=8$
$endgroup$
1
$begingroup$
Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
$endgroup$
– user477343
yesterday
add a comment |
$begingroup$
$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$
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 $x=8$
$endgroup$
$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$
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 $x=8$
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
add a comment |
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
add a comment |
$begingroup$
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.)
$endgroup$
add a comment |
$begingroup$
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.)
$endgroup$
add a comment |
$begingroup$
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.)
$endgroup$
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.)
answered 19 hours ago
Eric TowersEric Towers
33.8k22370
33.8k22370
add a comment |
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
$endgroup$
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!
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
PedroPedro
10.9k23475
10.9k23475
add a comment |
add a comment |
$begingroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
$endgroup$
add a comment |
$begingroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
$endgroup$
add a comment |
$begingroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
$endgroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
answered 5 hours ago
BPPBPP
2,160927
2,160927
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180580%2fare-there-any-other-methods-to-apply-to-solving-simultaneous-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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