Solve Systems of Equations by Graphing
When we have more than one equation which shares the same solution,
we have a system of equations. In this section, we will focus on a system
of two linear equations which will have a solution given as an ordered pair
(x, y).
3x − y = 5
Example 1: Is (2, 1) the solution to the system {
x+y=3
Substitute 2 for x and 1 for y in each equation
3x – y = 5
3(2) – 1 = 5
6–1=5
5=5
True, (2, 1) is a solution for this equation
x+y=3
2+1=3
3=3
True, (2, 1) is a solution for this equation
Since (2, 1) satisfies both equations, it is the solution to the system
2x + 4y = 10
Example 2: Is (-1, 3) the solution to the system {
y = 3x − 6
Substitute -1 for x and 3 for y in each equation
2x + 4y = 10
2(-1) + 4(3) = 10
-2 + 12 = 10
10 = 10 True, (-1, 3) is a solution for this equation
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
y = 3x – 6
3 = 3(-1) – 6
3 = -3 – 6
3 ≠ -9
(-1, 3) is not a solution for this equation
Since (-1, 3) does not satisfy both equations, it is not the solution
One way to find the solution point for a system of equations is by
graphing. We graph the two lines on the same coordinate plane and
determine the point at which they intersect.
1
y= − x+3
2
Example 3: Solve by graphing: {
3
y= x−2
4
1
First graph y = − x + 3
2
Find two points:
x
0
2
y
1
y = − (0) + 3
2
y=0+3
y=3
The first point is (0, 3)
1
y = − (2) + 3
2
y = -1 + 3
y=2
The second point is (2, 2)
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
y
6
Graph the line:
4
2
6
4
2
2
4
6
x
2
4
6
Now graph y =
3
4
x−2
Find two points:
x
0
4
y=
3
4
y
(0) − 2
y=0–2
y = -2
y=
3
4
The first point is (0, -2)
(4) − 2
y=3–2
y=1
The second point is (4, 1)
y
Graph the line:
6
4
2
6
4
2
2
4
6
x
2
4
6
The solution is where the two lines intersect at (4, 1)
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 4: Solve by graphing: {
2x + 3y = 6
x = −3
First graph 2x + 3y = 6
Find two points:
x
0
y
0
2(0) + 3y = 6
0 + 3y = 6
3y = 6
y=2
2x + 3(0) = 6
2x + 0 = 6
2x = 6
x=3
The first point is (0, 2)
The second point is (3, 0)
y
Graph the line:
6
4
2
6
4
2
2
4
6
x
2
4
6
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Now graph x = -3 which a vertical line going through the x-axis at -3
y
6
4
2
6
4
2
2
4
6
x
2
4
6
The solution is where the two lines intersect at (-3, 4)
Notice that we can check our solution by substituting the point into
each equation:
2x + 3y = 6
2(-3) + 3(4) = 6
-6 + 12 = 6
6 = 6 True
x = -3
-3 = -3 True
Since (-3, 4) satisfies both equations, it is the solution to the system
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)