## Introduction

A **system of equations** or **inequalties** is several equations or inequalities that are solved simultaneously.

## Solving Systems of Equations

There are several methods you can use to **solve systems of equations problems**. To solve these problems, you can **graph**, or use the **substitution or elimination method**.

### Solving Systems of Equations by Graphing

The solution to this system of equations is where the **two lines intersect**. Graph each equation then look for the point of intersection on the graph. The point of intersection is where the two equations are **equal**.

$\begin{align} y_{1}&=3x +4\\ y_{2}&=\frac{1}{3}x + 2\\ \end{align}$

The solution for this system is point $(-0.75, 1.75)$.

### Solving Systems of Equations by Substitution

To **solve by substitution**, manipulate the problem to have just one of the variables then follow steps to solve.

$\begin{align} 3y &=6x + 3\\ 2y &=8x + 4 \end{align}$

To solve this system by substitution, follow these steps. Solve one equation for x or y-value. For this problem, use the first equation and solve for the y-value.

$\begin{align} 3y &=6x + 3\\ \frac{3}{3}y&=\frac{6x+3}{3}\\ y&=2x+1 \end{align}$

Substitute the new value into the other equation – for this problem, the second equation.

$\begin{align} 2y &=8x + 4\\ 2(2x+1)&=8x+4\\ 4x+2&=8x+4\\ 4x+2 -2&=8x+4 -2\\ 4x &=8x+2\\ 4x -8x &=8x -8x +2\\ -4x &= 2\\ \frac{-4}{-4}x&=\frac{2}{-4}\\ x&=\frac{-1}{2} \end{align}$

Next, to solve for y-value, substitute the value for x into one of the equations.

$\begin{align} 3y&=6x + 3\\ 3y&=(6\times\frac{-1}{2}) + 3\\ 3y&=-3 +3\\ 3y&=0\\ y&=0 \end{align}$

The solution to this system is $(\frac{-1}{2}, 0)$

### Solving Systems of Equations by Elimination

To **solve by elimination**, **add or subtract** to eliminate one variable then solve.

$\begin{align} 2y &= 3x + 2\\ y&=x+4 \end{align}$

Manipulate one equation so you can eliminate one of the variables by adding or subtracting. For this problem, we can multiply the second equation by 2 then subtract it from the first equation.

$\begin{align} 2y &= 3x + 2\\ -2(y&=x+4) \end{align}$

$\begin{align} 2y= 3x + 2\\ \underline{-2y=-2x+ -8}\\ 0=x + - 6\\ x=6 \end{align}$

Next, substitute the x-value into one of the equations to calculate the y-value.

$\begin{align} 2y &= 3(6) + 2\\ 2y&=18 +2\\ 2y&=20\\ y&= 10 \end{align}$

The solution to this system is $(6, 10)$

Although **all systems of equations problems can be solved by graphing, substitution, or elimination**, select the method that is easiest to use.

### Word Problems

**Word problems** often model real word situations. Set up a system of equations to solve this problem.

The school ensemble sold tickets for two performances. The equations 25x + 50y = 625 and 40x + 65y = 850 represent the value of the tickets sold for each of the two performances. If x represents the cost of a student ticket and y is the cost of an adult ticket, what is the cost of each ticket?

First set up each equation in the **slope-intercept form** then **graph**. The **intersection** of the two line is the solution for the system.

$\begin{align} 25x + 50y &= 625 \\ 50y &= -25x + 625\\ y &= -0.50x + 12.5 \end{align}$

$\begin{align} 40x + 64y &= 840 \\ 64y &= -40x + 840\\ y &= -0.625x + 13.125 \end{align}$

The solution is point (5, 10).

## Solving Systems of Inequalties by Graphing

**The only way to solve a systems of inequalities is by graphing.** Graph each inequality, then determine the area that **satisfies all of the inequalities in the system**. To check, pick a point in the solution area and verify that it works for all the inequalities in the system.

$\begin{align} y&\geq2x +2\\ y&\geq2x -2 \end{align}$

Verify the solution for both inequalities using the point (0, 5).

$\begin{align} 5&\geq2\times 0 +2\\ 5&\geq 2 \end{align}$

$\begin{align} 5&\geq2\times 0 -2\\ 5&\geq -2 \end{align}$

The solution works for both inequalities.