1. Equations in Two variables
Lesson 3-5
Systems of linear
Equations in Two variables

2. A system of two linear equations consists of two equations that can be written in the form:
A solution of a system of linear equations is an ordered pair (x, y) that satisfies each equation.
There are several methods of solving systems of linear equations
- Substitution Method
- Linear combination Method
- Graphing Method

3. Linear combination
Method

4. x – y = 1 3x + y = 11 Example 1 Solution
Solve by the linear combination method
x – y = 1
3x + y = 11
Solution
Step1: rewrite both equations in the form Ax + By = C
The two equations are already in the form Ax + By = C, so we go to the step2
Step2: Adjust the coefficients of the variables so that the x-terms or the
y-terms will cancel out.
Note that the coefficients of the y-terms are already adjusted and will cancel out
when the two equations are added. so we go to step3

5. x – y = 1 3x + y = 11 4x = 12 Thus the Solution is (3, 2)
Step3: add the equations and solve
x – y = 1
3x + y = 11
4x = 12
Divide both sides of the equation by 4
Step4: Back-substitute and find the other variable.
Thus the Solution is (3, 2)

6. 3x – 4y = 2 x – 2y = 0 3x – 4y = 2 -3x +6y = 0 Example 2 Solution
Solve by the linear combination method
3x – 4y = 2
x – 2y = 0
Solution
Adjust the coefficients of the variables so that the x-terms or the y-terms
will cancel out.
Multiply the second equation by (-3), This will set up the x-terms to cancel.
3x – 4y = 2
-3x +6y = 0

7. 3x – 4y = 2 -3x +6y = 0 2y = 2 Thus the Solution is (1, 2)
add the equations
3x – 4y = 2
-3x +6y = 0
2y = 2
Divide both sides of the equation by 2
Back-substitute and find the other variable.
Thus the Solution is (1, 2)

8. 2x – 4y = -6 5x + 3y = 11 Example 3 Solution
Solve by the linear combination method
2x – 4y = -6
5x + 3y = 11
Solution
Adjust the coefficients of the variables so that the x-terms or the y-terms
will cancel out.
Neither variable is the obvious choice for cancellation. However I can multiply to
convert the x-terms to 10x or the y-terms to 12y. Since I'm lazy and 10 is smaller
than 12, I'll multiply to cancel the x-terms. I will multiply the first equation by (-5)
and the second row by (2); then I'll add down and solve

9. 2x – 4y = -6 5x + 3y = 11 -10x +20y = 30 10x + 6y = 22 26y = 52
add the equations and solve
2x – 4y = -6
5x + 3y = 11
Multiply by (-5)
Multiply by (2)
-10x +20y = 30
10x + 6y = 22
26y = 52
Divide both sides of the equation by 26

10. Back-substitute and find the other variable.
Thus the Solution is (1, 2)

11. Home Work (1)
(8, 10, 12)
Page 129

12. Written Exercises .. page 129
Solve each system 8)
5x + 6y +8= 0
3x – 2y +16= 0

13. 10)
8x – 3y= 3
3x – 2y + 5= 0

14. 12)
3p + 2q= -2
9p – q= -6

15. Substitution
method

16. 3x – 4y = 2 x – 2y = 0 Example Solution
Solve by the substitution method
3x – 4y = 2
x – 2y = 0
Solution
Step1: Rewrite either equation for one variable in terms of the other.
To avoid having fractions in the substitution process, let’s choose the 2nd equation and add (2y) to both sides of the equation. 2y
Step2: Substitute into the other equation and solve

17. Thus the Solution is (1, 2) simplify Simplify like terms
Divide both sides of the equation by 2
Step3: Back-substitute the value found into the other equation
Thus the Solution is (1, 2)

18. x – y = 1 3x + y = 11 Your Turn Solve by the substitution method
Step1: Rewrite either equation for one variable in terms of the other.
Step2: Substitute into the other equation and solve
Step3: Back-substitute the value found into the other equation

19. Home Work (2)
(18, 20, 24, 26, 28)
Page 129

20. Written Exercises .. page 129
Solve each system
18)
3x – 2y = 6
5x + 3y + 9= 0

21. 20)
6x = 4y + 5
6y = 9x – 5

22. 24)
2x + y = 2 – x
x + 2y = 2 + y

23. 26)
x + y = 4(y + 2)
x – y = 2(y + 4)

24. 28)
2(y – x) = 5 + 2x
2(y + x) = 5 – 2y

25. Graphing
method

26. x + 2y = 4 -x + y = -1 x + 2y = 4 -x + y = -1 2y = -x + 4 y = x – 1
Example
Solve by Graphing
x + 2y = 4
-x + y = -1
Solution
Step1: Put both equations in slope-intercept form: y = mx + b
x + 2y = 4
-x + y = -1 -x x
2y = -x + 4
y = x – 1 2

27. Step2: Graph both equations on the same coordinate plane.
Graph b: on the y-axis
Use m: rise then run and graph a second point b
Draw a line: it should pass through the two points.
Step3: Estimate the coordinates of the point where the lines intersect.

28. infinitely many solutions
CONCEPT
SUMMARY
system of linear equations y x y x y x
Lines intersect
one solution
Lines are parallel
no solution
Lines coincide
infinitely many solutions
Consistent system
dependent system
Inconsistent system

29. Home Work (3)
(14, 16, 30, 32)
Page 129

30. Written Exercises .. page 129
Graph both equations in the same coordinate system, then estimate the solution.
14)
2x + y = -2
2x – 3y = 15
Step1: Put both equations in slope-intercept form y = mx + b

31. Step2: Graph both equations on the same coordinate plane.
Graph b: on the y-axis
Use m: rise then run and graph a second point
Draw a line: it should pass through the two points.
Step3: Estimate the coordinates of the point where the lines intersect.

32. 16)
3x + 5y = 15
x – y = 4
Step1: Put both equations in slope-intercept form y = mx + b

33. Step2: Graph both equations on the same coordinate plane.
Graph b: on the y-axis
Use m: rise then run and graph a second point
Draw a line: it should pass through the two points.
Step3: Estimate the coordinates of the point where the lines intersect.

34. 30) 3x = 4y + 8 3y = 4x + 8 3x = 4y + 8 3y = 4x + 8 3x – 8 = 4y 3 4
Write each system in slope-intercept form. By Comparing the slopes and the y-intercepts, determine whether the equations are consistent or inconsistent.
30)
3x = 4y + 8
3y = 4x + 8
write both equations in slope-intercept form y = mx + b
3x = 4y + 8
3y = 4x + 8 -8 3
3x – 8 = 4y 4
The Slopes are not equal so the two lines will intersect at one point, thus
the system is Consistent.

35. 32)
3x – 6y = 9
4x – 3y = 12
Put both equations in slope-intercept form y = mx + b