Are there any other methods to apply to solving simultaneous equations?
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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
edited yesterday
Rodrigo de Azevedo
13.2k41962
asked yesterday
user477343user477343
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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
edited yesterday
Rodrigo de Azevedo
13.2k41962
asked yesterday
user477343user477343
3,62031345
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5
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you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
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– Doug M
yesterday
2
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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.
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– Mefitico
yesterday
3
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I hope someone performs GMRES by hand on this system and reports the steps.
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– Rahul
yesterday
2
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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?
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– Teepeemm
yesterday
2
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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.
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– 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
edited yesterday
Rodrigo de Azevedo
13.2k41962
asked yesterday
user477343user477343
3,62031345
$endgroup$
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
asked yesterday
user477343user477343
3,62031345
edited yesterday
Rodrigo de Azevedo
13.2k41962
asked yesterday
user477343user477343
3,62031345
edited yesterday
Rodrigo de Azevedo
13.2k41962
edited yesterday
Rodrigo de Azevedo
13.2k41962
edited yesterday
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked yesterday
user477343user477343
3,62031345
asked yesterday
user477343user477343
3,62031345
asked yesterday
user477343user477343
3,62031345
3,62031345
5
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you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
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– Doug M
yesterday
2
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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.
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– Mefitico
yesterday
3
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I hope someone performs GMRES by hand on this system and reports the steps.
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– Rahul
yesterday
2
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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?
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– Teepeemm
yesterday
2
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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.
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– littleO
yesterday
|
show 8 more comments
5
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you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
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– 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.
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– Mefitico
yesterday
3
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I hope someone performs GMRES by hand on this system and reports the steps.
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– 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.
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– littleO
yesterday
5
5
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you can use the substitution $y = 18 - frac 32 x.$ Or, you could use Cramer's rule
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– 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.
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– Mefitico
yesterday
3
3
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I hope someone performs GMRES by hand on this system and reports the steps.
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– Rahul
yesterday
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I hope someone performs GMRES by hand on this system and reports the steps.
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– Rahul
yesterday
2
2
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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.
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– littleO
yesterday
|
show 8 more comments
11 Answers
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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.
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
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I have never seen that! What is it? :D
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– user477343
yesterday
1
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elementary operations!
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– Chinnapparaj R
yesterday
1
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I assume $R$ stands for Row.
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– user477343
yesterday
23
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It's also called Gaussian elimination.
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– YiFan
yesterday
3
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See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… .
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– Eric Towers
yesterday
add a comment |
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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 $y=dfrac{detDelta_y}{detDelta_0}$.
answered yesterday
Paras KhoslaParas Khosla
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Wow! Very useful! I have never heard of this method, before! $(+1)$
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– 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!"
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– 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...
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– mlk
yesterday
3
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@user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome.
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– 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.
answered yesterday
Yves DaoustYves Daoust
133k676231
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This is a great method. +1 :)
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– Paras Khosla
yesterday
add a comment |
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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.
answered yesterday
Elements in SpaceElements in Space
1,28211228
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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. $(+1)$
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– 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.)
answered yesterday
Yves DaoustYves Daoust
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+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.
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– Surb
yesterday
add a comment |
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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:
answered yesterday
Kelly LowderKelly Lowder
20515
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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
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– 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 $x=8$
answered yesterday
Fareed AFFareed AF
762112
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Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$
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– user477343
yesterday
add a comment |
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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 21 hours ago
Eric TowersEric Towers
33.8k22370
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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!
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
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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.
answered yesterday
PedroPedro
10.9k23475
$endgroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
answered 7 hours ago
BPPBPP
2,160927
$endgroup$
add a comment |
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11 Answers
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11 Answers
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$begingroup$
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.
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
$endgroup$
$begingroup$
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.
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
$endgroup$
$begingroup$
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.
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
$endgroup$
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.
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
edited yesterday
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
answered yesterday
Chinnapparaj RChinnapparaj R
6,2872929
6,2872929
$begingroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
answered yesterday
Paras KhoslaParas Khosla
3,183626
$endgroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
answered yesterday
Paras KhoslaParas Khosla
3,183626
$endgroup$
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 $y=dfrac{detDelta_y}{detDelta_0}$.
answered yesterday
Paras KhoslaParas Khosla
3,183626
$endgroup$
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
3,183626
answered yesterday
Paras KhoslaParas Khosla
3,183626
answered yesterday
Paras KhoslaParas Khosla
3,183626
answered yesterday
Paras KhoslaParas Khosla
3,183626
3,183626
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.
answered yesterday
Yves DaoustYves Daoust
133k676231
$endgroup$
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.
answered yesterday
Yves DaoustYves Daoust
133k676231
$endgroup$
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.
answered yesterday
Yves DaoustYves Daoust
133k676231
$endgroup$
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
answered yesterday
Yves DaoustYves Daoust
133k676231
answered yesterday
Yves DaoustYves Daoust
133k676231
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.
answered yesterday
Elements in SpaceElements in Space
1,28211228
$endgroup$
$begingroup$
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.
answered yesterday
Elements in SpaceElements in Space
1,28211228
$endgroup$
$begingroup$
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.
answered yesterday
Elements in SpaceElements in Space
1,28211228
$endgroup$
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
answered yesterday
Elements in SpaceElements in Space
1,28211228
answered yesterday
Elements in SpaceElements in Space
1,28211228
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.)
answered yesterday
Yves DaoustYves Daoust
133k676231
$endgroup$
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.)
answered yesterday
Yves DaoustYves Daoust
133k676231
$endgroup$
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.)
answered yesterday
Yves DaoustYves Daoust
133k676231
$endgroup$
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.)
answered yesterday
Yves DaoustYves Daoust
133k676231
edited yesterday
answered yesterday
Yves DaoustYves Daoust
133k676231
answered yesterday
Yves DaoustYves Daoust
133k676231
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:
answered yesterday
Kelly LowderKelly Lowder
20515
$endgroup$
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:
answered yesterday
Kelly LowderKelly Lowder
20515
$endgroup$
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:
answered yesterday
Kelly LowderKelly Lowder
20515
$endgroup$
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
answered yesterday
Kelly LowderKelly Lowder
20515
answered yesterday
Kelly LowderKelly Lowder
20515
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 $x=8$
answered yesterday
Fareed AFFareed AF
762112
$endgroup$
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 $x=8$
answered yesterday
Fareed AFFareed AF
762112
$endgroup$
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 $x=8$
answered yesterday
Fareed AFFareed AF
762112
$endgroup$
$$begin{align}3x+2y&=36 tag1\ 5x+4y&=64tag2end{align}$$
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 $x=8$
answered yesterday
Fareed AFFareed AF
762112
edited yesterday
answered yesterday
Fareed AFFareed AF
762112
answered yesterday
Fareed AFFareed AF
762112
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.)
answered 21 hours ago
Eric TowersEric Towers
33.8k22370
$endgroup$
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.)
answered 21 hours ago
Eric TowersEric Towers
33.8k22370
$endgroup$
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.)
answered 21 hours ago
Eric TowersEric Towers
33.8k22370
$endgroup$
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 21 hours ago
Eric TowersEric Towers
33.8k22370
answered 21 hours ago
Eric TowersEric Towers
33.8k22370
answered 21 hours ago
Eric TowersEric Towers
33.8k22370
answered 21 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!
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
$begingroup$
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!
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
$begingroup$
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!
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
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!
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
edited yesterday
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
answered yesterday
TheSimpliFireTheSimpliFire
13.2k62464
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.
answered yesterday
PedroPedro
10.9k23475
$endgroup$
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.
answered yesterday
PedroPedro
10.9k23475
$endgroup$
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.
answered yesterday
PedroPedro
10.9k23475
$endgroup$
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.
answered yesterday
PedroPedro
10.9k23475
edited yesterday
answered yesterday
PedroPedro
10.9k23475
answered yesterday
PedroPedro
10.9k23475
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
answered 7 hours ago
BPPBPP
2,160927
$endgroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
answered 7 hours ago
BPPBPP
2,160927
$endgroup$
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 :
$$textbf{X}=dfrac{atextbf{A}}{textbf{A}^2}+dfrac{textbf{A}_{perp}}{textbf{A}_{perp}cdottextbf{B}}left(b-dfrac{atextbf{A}cdottextbf{B}}{textbf{A}^2}right)$$
answered 7 hours ago
BPPBPP
2,160927
$endgroup$
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 7 hours ago
BPPBPP
2,160927
answered 7 hours ago
BPPBPP
2,160927
answered 7 hours ago
BPPBPP
2,160927
answered 7 hours ago
BPPBPP
2,160927
2,160927
add a comment |
add a comment |
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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