1. Solve Systems of Linear Equations by Elimination
Chapter 3.2

2. The Elimination Method
Another method for solving systems of linear equations is called the elimination method (or the addition method)
The elimination method is best used when both linear equations are in standard form, as shown below 𝐴𝑥+𝐵𝑦=𝐶 𝐷𝑥+𝐸𝑦=𝐹
In general, the elimination method allows you to solve a system of equations without having to work with fractions

3. The Elimination Method
These basic principles underlie the use of the elimination method
If equals are added to equals, the result will also be equal
Equations can be multiplied on both sides by any non-zero number
The first principle can be illustrated as follows
5+4=9
7−1=6
Both of the above equations are true statements
We can add the first equation to the second and the result is also a true statement

4. The Elimination Method
5+4=9
7−1=6
12+3=15
This means that if the equations of the system below are assumed to be true and are added together, the resulting equation is also true
4𝑥−5𝑦=7 3𝑥+2𝑦=1
7𝑥−3𝑦=8

5. The Elimination Method
We can also change an true equation by multiplying all terms on both sides of the equal sign by a non-zero number
4−12=−8
The equation above is a true statement
If we multiply it by, say, −3, the result is again a true statement
−3 4 + −3 −12 =−3(−8)
−12+36=24

6. The Elimination Method
The key to solving by elimination is to have either the x-terms or the y-terms of each equation cancel to zero
The following example shows how this works
𝑥+3𝑦=14 −𝑥−4𝑦=−18
What equation results when the top equation is added to the bottom equation (combining like terms, of course)?

7. The Elimination Method
𝑥+3𝑦=14 −𝑥−4𝑦=−18 𝑥+3𝑦=14 −𝑥−4𝑦=−18 −𝑦=−4 𝑦=4 The y-coordinate of our solution is 4 To find the x-coordinate, substitute the y-coordinate into either equation and solve for x

8. The Elimination Method
𝑥+3𝑦=14 −𝑥−4𝑦=−18 𝑥+3 4 =14 𝑥+12=14 𝑥=2 The solution is (2,4)

9. Guided Practice
Solve the system by elimination.
𝑥−9𝑦=−24 −3𝑥+9𝑦=18

10. Guided Practice Solve the system by elimination. 𝑥−9𝑦=−24 −3𝑥+9𝑦=18
−2𝑥=−6
𝑥=3
Find y: 3−9𝑦=−24
−9𝑦=−27
𝑦=3
The solution is (3,3)

11. Multiply by Negative 1 Not all systems have terms that are opposites
In some cases, we must multiply one of the equations by an integer that will transform it (but keep it true) so that one term will cancel with the same term from the second equation
Example
3𝑥+𝑦=−5 3𝑥−4𝑦=5
Adding the two equations will not eliminate either the x-terms or the y-terms
However, the x-terms would cancel if one of them was negative

12. Multiply by Negative 1 3𝑥+𝑦=−5 3𝑥−4𝑦=5
How can we change one of the equations so that the x-terms will add to zero?

13. Multiply by Negative 1 3𝑥+𝑦=−5 3𝑥−4𝑦=5
How can we change one of the equations so that the x-terms will add to zero?
We can multiply, say, the second equation by −1
3𝑥+𝑦=−5
−1 3𝑥 − −1 4𝑦 =−1(5)
−3𝑥+4𝑦=−5
5𝑦=−10
𝑦=−2

14. Multiply by Negative 1 3𝑥+𝑦=−5 3𝑥−4𝑦=5
Now find x by substituting −2 for y
3𝑥−2=−5
3𝑥=−3
𝑥=−1
The solution is (−1,−3)

15. Guided Practice
Solve the system by elimination.
−2𝑥−4𝑦=18 −2𝑥−7𝑦=21

16. Guided Practice Solve the system by elimination. −2𝑥−4𝑦=18 −2𝑥−7𝑦=21
−1 −2𝑥 −1 −4𝑦 =−1 18
−2𝑥−7𝑦=21
2𝑥+4𝑦=−18
−3𝑦=3
𝑦=−1
Substitute: −2𝑥−4 −1 =−2𝑥+4=18
−2𝑥=14
𝑥=−7

17. Multiply One Equation
Some systems may have an equation where either the x-term or the y- term (or both) have coefficient 1
In this case, we multiply the equation by the number that will cancel the same term in the second equation
Example
2𝑥+3𝑦=−5 4𝑥−𝑦=11

18. Multiply One Equation 2𝑥+3𝑦=−5 4𝑥−𝑦=11
By what number can we multiply the second equation so that the y- terms will cancel to zero when both equations are added?

19. Multiply One Equation 2𝑥+3𝑦=−5 4𝑥−𝑦=11
Multiply the second equation by 3:
2𝑥+3𝑦=−5
3 4𝑥 −3 𝑦 =3 11
12𝑥−3𝑦=33
14𝑥=28
𝑥=2
Find y: 4 2 −𝑦=11, 8−𝑦=11, −𝑦=3, 𝑦=−3
The solution is (2,−3)

20. Guided Practice
Solve the system by elimination.
−4𝑥+6𝑦=−26 9𝑥+12𝑦=−18

21. Guided Practice Solve the system by elimination. −4𝑥+6𝑦=−26 9𝑥+12𝑦=−18
−2 −4𝑥 −2 6𝑦 =−2 −26
9𝑥+12𝑦=−18
8𝑥−12𝑦=52
17𝑥=34
𝑥=2
Find y: 𝑦=−18, 18+12𝑦=−18, 12𝑦=−36, 𝑦=−3
The solution is (2,−3)

22. Multiply Both Equations
For many systems of equations, we must multiply both equations by numbers that will transform one or the other term into opposites
Example
3𝑥+4𝑦=7 5𝑥−3𝑦=2
Note that the y-terms are opposite in sign
If we multiply the first equation by 3 and the second equation by 4, then the y-terms will become 12𝑦 and −12𝑦, respectively
We can then solve for x

23. Multiply Both Equations
3𝑥+4𝑦=7 5𝑥−3𝑦=2
3 3𝑥 +3 4𝑦 =3 7
4 5𝑥 −4 3𝑦 =4 2
9𝑥+12𝑦=21
20𝑥−12𝑦=8
29𝑥=29
𝑥=1
Find y: 𝑦=7, 3+4𝑦=7, 4𝑦=4, 𝑦=1
The solution is (1,1)

24. Guided Practice
Solve the system by elimination.
4𝑥+4𝑦=−8 9𝑥+3𝑦=6

25. Guided Practice Solve the system by elimination. 4𝑥+4𝑦=−8 9𝑥+3𝑦=6
−3 4𝑥 −3 4𝑦 =−3 −8
4 9𝑥 +4 3𝑦 =4(6)
−12𝑥−12𝑦=24
36𝑥+12𝑦=24
24𝑥=48
𝑥=2
Find y: 𝑦=−8, 8+4𝑦=−8, 4𝑦=−16, 𝑦=−4
The solution is (2,−4)

26. Substitution or Elimination?
Some systems can be more easily solved by substitution than by elimination
Example
𝑥+3𝑦=3 2𝑥+5𝑦=4
We can easily solve the first equation for x and substitute this for x in the second equation

27. Substitution or Elimination?
𝑥+3𝑦=3 2𝑥+5𝑦=4 𝑥=−3𝑦+3 Substitute 2 −3𝑦+3 +5𝑦=4 −6𝑦+6+5𝑦=4 −𝑦+6=4 −𝑦=−2 𝑦=2 Find x: 𝑥=−3 2 +3=−6+3=−3 The solution is (−3,2)

28. Guided Practice
Solve the system.
−7𝑥+4𝑦=−3 𝑥+4𝑦=5

29. Guided Practice Solve the system. −7𝑥+4𝑦=−3 𝑥+4𝑦=5
Solve for x: 𝑥=−4𝑦+5
Substitute: −7 −4𝑦+5 +4𝑦=−3
28𝑦−35+4𝑦=−3
32𝑦−35=−3
32𝑦=32
𝑦=1
Find x: 𝑥=−4 1 +5=−4+5=1
The solution is (1,1)

30. You Must Choose!
You must learn to choose the method that allows you to easily solve a system of equations
You can choose between the elimination method or the substitution method
Think a bit about a problem before deciding on the method!

31. Exercise 3.2b
Handout