7.3 Solving Systems of Linear Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression. The third method of solving systems of linear equations is called elimination. When we solve a system by substitution, we start with two equations and two variables and reduce it to one equation with one variable. We will do the same with elimination, but use a different method.

Solve a System of Equations by Elimination

Elimination is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality.

Table 7.3.1
Addition Property of Equality
For any expressions a, b, c, and d,

if               a = b
and           c = d
then    a + c = b + d

We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal. When we use elimination, we add or subtract two equations with the goal of eliminating one of the variables. Once there is only one variable left, the equation can be solved algebraically.

Table 7.3.2
To solve a system of equations by elimination:
Step 1: Write both equations in standard form (Ax + By = C). This makes it easier to use this method. If any coefficients are fractions, clear them.

Step 2: Make the coefficients of one variable opposite to each other (for example, if one is positive, make the other one negative).

  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.

Step 3: Add the equations resulting from Step 2 to eliminate one variable.

Step 4: Solve for the remaining variable.

Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.

Step 6: Write the solution as an ordered pair.

Step 7: Check that the ordered pair is a solution to both original equations.

Example 1

Solve the system by elimination.
2x + y = 7
x - 2y = 6

Table 7.3.3
Steps Solution
Write both equations in standard form. In this case, both are already written in standard form. 2x + y = 7
x - 2y = 6
Make the coefficients of one variable opposites. In this case, we will make both equations have  a 2y term. 2x + y = 7
2(2x + y) = 2(7)
4x + 2y = 14
Add the equations resulting from Step 2 to eliminate one variable. 4x + 2y = 14
x - 2y = 6
5x = 20
Solve for the remaining variable. x = 4
Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. x - 2y = 6
4 - 2y = 6
-2y = 2
y = -1
Write the solution as an ordered pair. (4,-1)
Check that the ordered pair is a solution to both original equations. 2x + y = 7
2(4) + (-1) = 7
7 = 7
4 - 2(-1) = 6
6 = 6
Practice 1

Solve the systems by elimination.

a)

x + y = 10
x - y = 12

 

 

 

 

b)

9x + 3y = 15
2x - 3y = 7

 

 

 

 

c)

4x + y = -5
-2x - 2y = -2

 

 

 

 

d)

3x - 2y = -2
5x - 6y = 10

 

 

 

 

e)

7x + 8y = 4
3x - 5y = 27

 

 

 

 

f)

4x - 3y = 9
7x + 2y = -6

 

 

 

 

Homework 

  1. Solve the systems by elimination.

    a)

    -3x + y = -9
    x - 2y = -12

     

     

     

     

    b)

    3x - y = -1
    4x + 2y = -6

     

     

     

     

    c)

    x + y = -8
    x - y = -6

     

     

     

     

    d)

    -7x + 6y = -10
    x - 6y = 22

     

     

     

     

    e)

    5x + 2y = 1
    -5x - 4y = -7

     

     

     

     

    f)

    3x - 4y = -11
    x - 2y = -5

     

     

     

     

    g)

    6x - 5y = -75
    -x - 2y = -13

     

     

     

     

    h)

    2x - 5y = 7
    3x - y = 17

     

     

     

     

    i)

    7x + y = -4
    13x + 3y = 4

     

     

     

     

    j)

    3x - 5y = -9
    5x + 2y = 16

     

     

     

     

    k)

    4x + 7y = 14
    -2x + 3y = 32

     

     

     

     

    l)

    3x + 8y = -3
    2x + 5y = -3

     

     

     

     

    m)

    3x + 8y = 67
    5x + 3y = 60

     

     

     

     

    n)

    2x + y = 3
    6x + 3y = 9

     

     

     

     

    o)

    -3x - y = 8
    6x + 2y = 16

     

     

     

     

  2. In the following exercises, state whether it would be more convenient to solve the system of equations by substitution or elimination.

    a)

    8x - 15y = -32
    6x + 3y = -5

     

    b)

    4x - 2y = -6
    x = 4y -6

     

    c)

    y = 4x + 9
    5x - 2y = -21

     

    d)

    9x - 4y = 24
    3x + 5y = -14

     

    e)

    y = 7x - 5
    y = -7x + 5

     

    f)

    12x - 5y = -42
    3x + 7y = -15

     

  3. Norris can row 3 miles upstream against the current in 1 hour, the same amount of time it takes him to row 5 miles downstream, with the current.
    r - c = 3
    r + c = 5

    Solve the system
    a) for r, his rowing speed in still water
    b) for c, the speed of the river current

     

     

     

  4. Rosemary wants to make 10 pounds of trail mix using nuts and raisins, and she wants the total cost of the trail mix to be $54. Nuts cost $6 per pound and raisins cost $3 per pound.
    n + r = 10
    6n + 3r = 54

    Solve the system to find
    a) the number of pounds of nuts
    b) the number of pounds of raisins she should use

     

     

     

     

  5. Solve the system of equations using substitution.
    3x + 4y = 22
    5x - 2y = 14

    a) x = \dfrac{3}{4}, y = \dfrac{16}{3}
    b) x = \dfrac{50}{13}, y = \dfrac{34}{13}
    c) x = 2, y = 4
    d) x = 3, y = 2
  6. Which of the following is true about the elimination method for solving systems of linear equations?

    a) The elimination method is used to eliminate one variable by multiplying both equations by constants and then adding or subtracting the equations.
    b) The elimination method always requires the equations to be in standard form.
    c) The elimination method can only be used if both equations have the same number of variables.
    d) The elimination method only works if the system has exactly one solution.

Answers

1.

a) (6, 9) b) \left(-\dfrac{4}{5},-\dfrac{7}{5}\right) c) (–7, –1) d) (–2, –4) e) (–1, 3)
f) (–1, 2) g) (–5, 9) h) (6, 1) i) (–2, 10) j) (2, 3)
k) (–7, 6) l) (–9, 3) m) (9, 5) n) Infinitely many solutions o) No solutions

2.

a) Elimination b) Substitution c) Substitution
d) Elimination e) Substitution f) Elimination
3. r = 4 and c = 1 4. n = 8 and r = 2 5. b 6. a

Attribution

Unless otherwise indicated, material on this page has been adapted from the following resource(s):

Berg, T. (2020). Intermediate algebra. BCcampus. https://collection.bccampus.ca/textbooks/intermediate-algebra-bccampus-412/, licensed under CC BY-NC-SA 4.0

Marecek, L., Anthony-Smith, M., & Honeycutt, M. (2020). Elementary algebra 2e. OpenStax. https://collection.bccampus.ca/textbooks/elementary-algebra-2e-openstax-106/, licensed under CC BY 4.0

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