# 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.

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

## 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:

#### Problem

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

$$\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.$$

#### Solution

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 $(2,0)$.

### Substitution

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.

#### Problem

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

$$\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.$$

#### Solution

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 $x=2$.

### Graphing

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.

#### Problem

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

$$\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.$$

#### Solution

We graph the two lines as follows:

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

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.

### Symmetry

Consider the below system.

$$\left\{\begin{array}{l}a+b+c+d=4\\a+b+c+e=8\\a+b+d+e=12\\a+c+d+e=16\\b+c+d+e=20\end{array}\right.$$

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.