# System of equations

A system of equations is a set of equations which share the same variables. An example of a system of equations is $2a - 3b$ $= 4$ $3a - 2b$ $= 3$

## 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 $x - 12y$ $= 2$ $3x + 6y$ $= 6$

#### 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: $(2)-12y = 2 \Rightarrow y = 0$.

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 $x - 12y$ $= 2$ $3x + 6y$ $= 6$

#### 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 $x - 12y$ $= 2$ $3x + 6y$ $= 6$

#### Solution

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

Convenient systems usually seem very tough to solve at first. Often times a clever insight will make things much easier.

### Problem

Solve the following system: $a+$ $b+$ $c+$ $d$ $=4$ $a+$ $b+$ $c+$ $e$ $=8$ $a+$ $b+$ $d+$ $e$ $=12$ $a+$ $c+$ $d+$ $e$ $=16$ $b+$ $c+$ $d+$ $e$ $=20$

### Solution

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 $(a,b,c,d,e)=(-5,-1,3,7,11).$

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