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

$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$

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 $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ 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:

$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$


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

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    $\begingroup$ you can use the substitution $y = 18 - \frac 32 x.$ Or, you could use Cramer's rule $\endgroup$ – Doug M Apr 9 '19 at 5:32
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    $\begingroup$ 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 Apr 9 '19 at 12:28
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    $\begingroup$ I hope someone performs GMRES by hand on this system and reports the steps. $\endgroup$ – Rahul Apr 9 '19 at 17:02
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    $\begingroup$ 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 Apr 10 '19 at 0:02
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    $\begingroup$ 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 Apr 10 '19 at 0:25

13 Answers 13


Is this method allowed ?

$$\left[\begin{array}{rr|rr} 3 & 2 & 36 \\ 5 & 4 & 64 \end{array}\right] \sim \left[\begin{array}{rr|rr} 1 & \frac{2}{3} & 12 \\ 5 & 4 & 64 \end{array}\right] \sim \left[\begin{array}{rr|rr} 1 & \frac{2}{3} & 12 \\ 0 & \frac{2}{3} & 4 \end{array}\right] \sim \left[\begin{array}{rr|rr} 1 & 0 & 8 \\ 0 & \frac{2}{3} & 4 \end{array}\right] \sim \left[\begin{array}{rr|rr} 1 & 0 & 8 \\ 0 & 1 & 6 \end{array}\right] $$

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 $R_2 \to R_2\times \frac{3}{2}$

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

  • $\begingroup$ I have never seen that! What is it? :D $\endgroup$ – Mr Pie Apr 9 '19 at 6:07
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    $\begingroup$ elementary operations! $\endgroup$ – Chinnapparaj R Apr 9 '19 at 6:09
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    $\begingroup$ I assume $R$ stands for Row. $\endgroup$ – Mr Pie Apr 9 '19 at 6:28
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    $\begingroup$ It's also called Gaussian elimination. $\endgroup$ – YiFan Apr 9 '19 at 8:50
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    $\begingroup$ See also augmented matrix and, for typesetting, tex.stackexchange.com/questions/2233/… . $\endgroup$ – Eric Towers Apr 9 '19 at 14:52

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

Now computation is trivial as you have: $x=\dfrac{\det\Delta_x}{\det\Delta_0}$ and $y=\dfrac{\det\Delta_y}{\det\Delta_0}$.

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    $\begingroup$ Wow! Very useful! I have never heard of this method, before! $(+1)$ $\endgroup$ – Mr Pie Apr 9 '19 at 6:07
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    $\begingroup$ 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 Apr 9 '19 at 9:06
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    $\begingroup$ @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 Apr 9 '19 at 10:11
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    $\begingroup$ @user477343 asked for different ways to solve, not more efficient ways to solve. This is awesome. $\endgroup$ – user1717828 Apr 9 '19 at 12:09
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    $\begingroup$ @alephzero I'm not quite sure what you're talking about. Finding the determinant of a matrix can be done in roughly the same time as matrix multiplication, which stands at $O(n^2.38)$. $\endgroup$ – DreamConspiracy Apr 10 '19 at 18:18

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 $\begin{bmatrix} x\\ y\end{bmatrix}$ on the left hand side.

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

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

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

Now plug that back into f

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

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

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

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

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    $\begingroup$ So we just have $f\begin{bmatrix} 0 \\ 0\end{bmatrix}$ and then $f\bigg(f\begin{bmatrix} 0 \\ 0\end{bmatrix}\bigg)$ and by letting $f^k(\cdot ) = f(f(\ldots f(f(\cdot))\ldots )$ $k$ times, this overall goes to $$f^{100}\begin{bmatrix} 0 \\ 0\end{bmatrix}$$ and etc... hmm... it actually seems quite appealing to me, regardless of its low efficiency, as you say :P $\endgroup$ – Mr Pie Apr 10 '19 at 0:46
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    $\begingroup$ Note that this doesn't always work, $f$ needs to be a contraction. $\endgroup$ – flawr Apr 11 '19 at 12:10

By false position:

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


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


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

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    $\begingroup$ This is a great method. +1 :) $\endgroup$ – Paras Khosla Apr 9 '19 at 16:39
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    $\begingroup$ This is like a variation of the elimination method, but breaks things down better! Already upvoted :P $\endgroup$ – Mr Pie Apr 12 '19 at 0:15

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.)


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

plot of simultaneous equations

  • $\begingroup$ 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. $(+1)$ $\endgroup$ – Mr Pie Apr 10 '19 at 0:45

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 $2\times2$ systems, the different variants make little difference in this respect. Things become more interesting for $n\times n$ 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 $3\times3$ 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:


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

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

Solve for $y$,

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

Solve for $x$,

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

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 $2\times2$ 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 $100\times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)

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    $\begingroup$ +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 Apr 9 '19 at 19:55

$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$

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 $x=8$

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    $\begingroup$ Pure algebra! I personally prefer the second method. Thanks for that! $(+1)$ $\endgroup$ – Mr Pie Apr 9 '19 at 7:55

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 $k\in\Bbb R$. Then $$3x+2y=36\implies x(2k+3)=36\implies x=\frac{36}{2k+3}\\5x+4y=64\implies x(4k+5)=64\implies x=\frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)\implies (144-128)k=(192-180)\implies k=\frac34.$$ Now $$x=\frac{64}{4k+5}=\frac{64}{4\cdot\frac34+5}=8\implies y=kx=\frac34\cdot8=6.\quad\square$$

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

How about we try squaring the equations? We get $$3x+2y=36\implies 9x^2+12xy+4y^2=1296\\5x+4y=64\implies 25x^2+40xy+16y^2=4096$$ 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 $2x+2y=28\implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!

  • $\begingroup$ 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 $\endgroup$ – Mr Pie Apr 9 '19 at 8:39
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    $\begingroup$ 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 Apr 9 '19 at 8:41
  • $\begingroup$ 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 $(+1)$ $\endgroup$ – Mr Pie Apr 9 '19 at 9:02
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    $\begingroup$ 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 Apr 9 '19 at 9:06
  • $\begingroup$ Ok. Thank you for clarifying! $\endgroup$ – Mr Pie Apr 10 '19 at 0:43

As another iterative method I suggest the Jacobi Method. A sufficient criterion for its convergence is that the matrix must be diagonally dominant. Which this one in our system is not:

$\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}=\begin{bmatrix}36 \\ 64\end{bmatrix}$

We can however fix this by replacing e.g. $y' := \frac{1}{1.3} y$. Then the system is

$\underbrace{\begin{bmatrix} 3 & 2.6 \\ 5 & 5.2\end{bmatrix}}_{=:A}\begin{bmatrix} x \\ y'\end{bmatrix}=\begin{bmatrix}36 \\ 64\end{bmatrix}$

and $A$ is diagonally dominant. Then we can decompose $A = L + D + U$ into $L,U,D$ where $L,U$ are the strict upper and lower triangular parts and $D$ is the diagonal of $A$ and the iteration

$$\vec x_{i+1} = - D^{-1}((L+R)\vec x_i + b)$$

will converge to the solution $(x,y')$. Note that $D^{-1}$ is particularly easy to compute as you just have to invert the entries. So in theis case the iteration is

$$\vec x_{i+1} = -\begin{bmatrix} 1/3 & 0 \\ 0 & 1/5.2 \end{bmatrix}\left(\begin{bmatrix} 0 & 2.6 \\ 5 & 0 \end{bmatrix} \vec x_i + b\right)$$

So you can actually view this as a fixed point iteration of the function $f(\vec x) = -D^{-1}((L+R)\vec x + b)$ which is guaranteed to be a contraction in the case of diagonal dominance of $A$. It is actually quite slow and doesn't any practical application for directly solving systems of linear equations but it (or variations of it) is quite often used as a preconditioner.


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

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

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.


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

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    $\begingroup$ What if the dimension of the vectors is $n>3$, can one define a vector product as we have in three dimensions? $\endgroup$ – Chip May 6 '19 at 1:51

$$3x+2y=36\tag1$$ $$5x+4y=64\rightarrow \rightarrow \rightarrow \rightarrow 3x+2y+3x+2y-x=64$$ $$36+36-x=64$$ $$x=8$$

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    $\begingroup$ Holy cow, that was a clever substitution! I am so doing that next time! $(+1)$ :D $\endgroup$ – Mr Pie Jul 11 '19 at 22:35

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