1. The Graphing Method
Topic 5.1.1

2. The Graphing Method 5.1.1 Topic California Standard:
9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets.
What it means for you:
You’ll solve systems of linear equations by graphing the lines and working out where they intersect.
Key Words:
system of linear equations
simultaneous equations

3. The Graphing Method 5.1.1 Topic
In Section 4.5 you graphed two inequalities to find the region of points that satisfied both inequalities.
Plotting two linear equations on a graph involves fewer steps, and it means you can show the joint solution to the equations graphically.

4. The Graphing Method 5.1.1 Topic Systems of Linear Equations
A system of linear equations consists of two or more linear equations in the same variables.
For example: 3x + 2y = 7 and x – 3y = –5 form a
system of linear equations in two variables — x and y.

5. The Graphing Method 5.1.1 Topic
The solution of a system of linear equations in two variables is a pair of values like x and y, or (x, y), that satisfies each of the equations in the system.
For example, x = 1, y = 2 or (1, 2) is the solution of the
system of equations 3x + 2y = 7 and x – 3y = –5 , since it
satisfies both equations:
3x + 2y = 7
3(1) + 2(2) = 7
3 + 4 = 7
x – 3y = –5
1 – 3(2) = –5
1 – 6 = –5

6. The Graphing Method 5.1.1 Topic
Equations in a system are often called simultaneous equations because any solution has to satisfy the equations simultaneously (at the same time).
The equations can’t be solved independently of one another.

7. The Graphing Method 5.1.1 Topic
Solving Systems of Equations by Graphing
A system of two linear equations can be solved graphically, by graphing both equations in the same coordinate plane.
Every point on the line of an equation is a solution of that equation.
The point at which the two lines cross lies on both lines and so is the solution of both equations.

8. The Graphing Method 5.1.1 Topic
The solution of a system of linear equations in two variables is the point of intersection (x, y) of their graphs.
Point of intersection

9. The Graphing Method 5.1.1 Topic
Example 1
Solve this system of equations by graphing:
2x – 3y = 7
Solution
–2x + y = –1
Step 1: Graph both equations in the same coordinate plane.
First find some points that lie on each of the lines.
Line of first equation:
Line of second equation:
2x – 3y = 7 x y –1 1
3y = 2x – 7 x y 2 –1 –3
–2x + y = –1
y = x – 2 3 7
y = 2x –1
The line goes through the points (2, –1) and (–1, –3).
The line goes through the points (0, –1) and (1, 1).
Solution follows…

10. The Graphing Method 5.1.1 Topic
Example 1
Solve this system of equations by graphing:
2x – 3y = 7
–2x + y = –1
Solution (continued)
y = 2x –1
Now you can draw the graph:
Step 2: Read off the coordinates of the point of intersection.
y = x – 2 3 7
(1, 1)
The point of intersection is (–1, –3).
(0, –1)
(2, –1)
(–1, –3)
(–1, –3)
Solution continues…

11. The Graphing Method 5.1.1 Topic
Example 1
Solve this system of equations by graphing:
2x – 3y = 7
–2x + y = –1
Solution (continued)
Step 3: Check whether your coordinates give true statements when they are substituted into each equation.
2x – 3y = 7 Þ 2(–1) – 3(–3) = Þ 7 = 7 — True statement
–2x + y = –1 Þ –2(–1) + (–3) = –1 Þ –1 = –1 — True statement
So x = –1, y = –3 is the solution of the system of equations.

12. The Graphing Method 5.1.1 Topic Guided Practice
Solve each system of equations in Exercises 1–2 by graphing on x- and y-axes spanning from –6 to 6.
1. y + x = 2 and y = x + 2 2 3
2. y + x = 3 and 3y – x = 5
(0, 2)
(1, 2) 2 3
y = x + 2
y + x = 2
y + x = 3
3y – x = 5
Solution follows…

13. The Graphing Method 5.1.1 Topic Guided Practice
Solve each system of equations in Exercises 3–4 by graphing on x- and y-axes spanning from –6 to 6.
4. y – x = 1 and y + x = –3 3 2 1
3. y = x – 3 and y + 2x = 3
(2, –1)
(–2, –2)
y + 2x = 3
y – x = 1 3 2
y = x – 3
y x = –3 1 2
Solution follows…

14. The Graphing Method 5.1.1 Topic Guided Practice
Solve each system of equations in Exercises 5–6 by graphing on x- and y-axes spanning from –6 to 6.
6. 2y – x = –6 and y + x = –3 1 2
5. y – x = 3 and y + x = –1
(–2, 1)
(0, –3)
y – x = 3
y + x = –1
y x = –3 1 2
2y – x = –6
Solution follows…

15. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 1–2 by graphing on x- and y-axes spanning from –6 to 6.
1. 2x + y = 7 and y = x + 1
2. x + y = 0 and y = –2x
(2, 3)
(0, 0)
y = x + 1
2x + y = 7
x + y = 0
y = –2x
Solution follows…

16. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 3–4 by graphing on x- and y-axes spanning from –6 to 6.
3. y = –3 and x – y = 2
4. x – y = 4 and x + 4y = –1
(–1, –3)
(3, –1)
x – y = 2
x + 4y = –1
x – y = 4
y = –3
Solution follows…

17. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 5–6 by graphing on x- and y-axes spanning from –6 to 6.
5. 2y + 4x = 4 and y = –x + 3
6. y = –x and y = 4x
(–1, 4)
(0, 0)
y = –x + 3
2y + 4x = 4
y = –x
y = 4x
Solution follows…

18. The Graphing Method 5.1.1 Topic Independent Practice
Determine the solution to the systems of equations graphed in Exercises 7 and 8. 7. 8.
y = 2x – 7
y = –x
x + y = –4
(3, –3)
(2, –3)
y = x – 4 1 3
Solution follows…

19. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 9–10 by graphing on x- and y-axes spanning from –6 to 6.
9. x – y = 6 and x + y = 0
10. y = 2x – 1 and x + y = 8
(3, –3)
(3, 5)
x + y = 0
y = 2x – 1
x + y = 8
x – y = 6
Solution follows…

20. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 11–12 by graphing on x- and y-axes spanning from –6 to 6.
11. 4x – 3y = 0 and 4x + y = 16
12. x – y = 0 and x + y = 8
(3, 4)
(4, 4)
4x – 3y = 0
4x + y = 16
x – y = 0
x + y = 8
Solution follows…

21. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 13–14 by graphing on x- and y-axes spanning from –6 to 6.
13. y = –x + 6 and x – y = –4
14. x – y = 1 and x + y = –3
(1, 5)
(–1, –2)
x – y = –4
y = –x + 6
x – y = 1
x + y = –3
Solution follows…

22. The Graphing Method 5.1.1 Topic Independent Practice
Solve each system of equations in Exercises 15–16 by graphing on x- and y-axes spanning from –6 to 6.
15. x + y = 1 and x – 2y = 1
16. 2x + y = –8 and 3x + y = –13
(1, 0)
(–5, 2)
x + y = 1
3x + y = –13
2x + y = –8
x – 2y = 1
Solution follows…

23. The Graphing Method 5.1.1 Topic Round Up
There’s something very satisfying about taking two long linear equations and coming up with just a one-coordinate-pair solution.
You should always substitute your solution back into the original equations, to check that you’ve got the correct answer.