System of equations

A system of equations is a set of equations which share the same variables. Below is an example of a system of equations.

\[\left\{\begin{array}{l}a^2 + b^2 = 25\\a+b=7\end{array}\right.\]

Solve 2 variable equations in less than 5 seconds!!!

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Solving Linear Systems

A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations.

Gaussian Elimination

Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example:


Find the ordered pair $(x,y)$ for which



We can eliminate $y$ by adding twice the second equation to the first:

$x - 12y= 2$
$+2($ $3x + 6y = 6)$
$\overline{7x + 0=14}$

Thus $x=2$. We can then plug in for $x$ in either of the equations: \begin{align*} (2)-12y &= 2 \\ y &= 0 \end{align*}

Thus, the solution to the system is $(2,0)$.


The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution.


Find the ordered pair $(x,y)$ for which



The first equation can be solved for $x$:

$x = 12y + 2.$

Plugging this into the second equation yields

$3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0.$

Thus $y=0$. Plugging this into either of the equations and solving for $x$ yields $x=2$.


The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect. We'll go back to our same example to illustrate this.


Find the ordered pair $(x,y)$ for which



We graph the two lines as follows:


From the graph, we can see that the solution to the system is $(2,0)$.

Advanced Methods

Matrices can also be used to solve systems of linear equations. In fact, they provide a way to make much broader statements about systems of linear equations.

There is a whole field of mathematics devoted to the study of linear equations called linear algebra.

Convenient Systems

Some systems can be solved by taking advantage of specific forms. Such systems can often seem tough to solve at first, however.


Consider the below system.


The key here is to take advantage of the symmetry. If we add up all 5 equations we will have a total of 4 of each variable on the LHS. On the RHS we will have $4+8+12+16+20 = 60$. Thus

$4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15.$

So then subtracting the first equation from this leaves $e$ on the LHS and $15-4=11$ on the RHS. Subtracting this equation from the second equation leaves $d$ on the LHS and $15-8=7$ on the RHS. And thus we continue on in this way to find that $(a,b,c,d,e)=(-5,-1,3,7,11).$

Clever Substitution

Consider the below system.

\[\left\{\begin{array}{l}x+y+3xy = 5\\ 2x+2y+4xy=8 \end{array}\right.\]

We can let $x+y=a$ and $b = xy$ to get the two-variable linear system below.

\[\left\{\begin{array}{l}a+3b = 5\\ 2a+4b=8 \end{array}\right.\]

Solving the system results in $b = 1$ and $a = 2$. Substituting that back results in $xy = 1$ and $x + y = 2$. We can do another substitution by letting $y = 2 - x$ and substituting to get $x(2-x) = 1$. Rearranging results in $0 = x^2 - 2x + 1$, so $x = 1$. Finally, by substituting $x$ back in, we get $y = 1$. Plugging $(1,1)$ back satisfies the system.




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