# Solve systems of equations by graphing and identifying the point of intersection

We have solved problems like 3x – 4 = 11 by adding 4 to both sides and then dividing by 3 (solution is x = 5). We also have methods to solve equations with more than one variable in them. It turns out that to solve for more than one variable we will need the same number of equations as variables. For example, to solve for two variables such as x and y we will need two equations. When we have several equations we are using to solve, we call the equations a system of equations. When solving a system of equations we are looking for a solution that works in both equations. This solution is usually given as an ordered pair (x, y). The following example illustrates a solution working in both equations.

Example 1:

Show (2 ,1) is the solution to the system

3x – y = 5

x + y = 3

Solution:

Identify x and y from the ordered pair

(2, 1)

x = 2, y = 1

Plug these values into each equation

First equation

3(2) – (1) = 5

Evaluate

5 =5 (true)

Second equation, evaluate

(2) + (1) =3

3 = 3 (True)

As we found a true statement for both equations we know (2,1) is the solution to the system. It is in fact the only combination of numbers that works in both equations. In this lesson we will be working to find this point given the equations. It seems to follow that if we use points to describe the solution, we can use graphs to find the solutions.

If the graph of a line is a picture of all the solutions, we can graph two lines on the same coordinate plane to see the solutions of both equations. We are inter ested in the point that is a solution for both lines, this would be where the lines intersect! If we can find the intersection of the lines we have found the solution that works in both equations.

Example 2:

Find the intersection of $-\frac{1}{2}x+3$ $\frac{3}{4}x-2$

Solution:

To graph we identify slopes and y –intercepts

First: $-\frac{1}{2}x+3$ $m=-\frac{1}{2}$ and b = 3

Second: $\frac{3}{4}x-2$ $m=\frac{3}{4}$ and b = – 2

To graph each equation, we start at the y-intercept and use the slope $\frac{rise}{run}$ to get the next point and connect the dots.

Remember a negative slope is down hill!

Find the intersection point, (4,1)

(4,1) Our Solution Often our equations won’t be in slope-intercept form and we will have to solve both equations for y first so we can identify the slope and y-intercept.

Example 2:

Solve each equation for y

6x – 3y = – 9

2x + 2y = – 6

Solution:

Subtract x terms Put x terms first and Divide by coefficient of y  Identify slope and y –intercepts

First; $m=\frac{2}{1}$, b = 3

Second, $m=-\frac{1}{1}$, b = – 3

Now we can graph both lines on the same plane

To graph each equation, we start at the y-intercept and use the slope $\frac{rise}{run}$ to get the next point and connect the dots. Remember a negative slope is downhill!

Find the intersection point, ( -2, -1) So the solution is (-2,1).