Education.com
Try
Brainzy
Try
Plus

Systems of Equations—Solving with Elimination Practice Questions

based on 4 ratings
By
Updated on Oct 3, 2011

To review these concepts, go to Systems of Equations—Solving with Elimination Study Guide.

Systems of Equations—Solving with Elimination Practice Questions

Problems

Practice 1

Solve each system of equations.

  1. 8y + 3x = –2, –8y – 14 = 5x
  2. x – 2y = 20, 4x + 7y = 5
  3. –4x + 3y = –4, 6x – 5y = 2

Practice 2

Find the solution to this system of equations:

      x + y + 3z = 1
      2y + 9x + 4z = –1
      –8z – 4x = 3y

Solutions

Practice 1

1. Choose a variable to eliminate. The first equation contains an 8y term and the second equation contains a –8y term. Add these equations to eliminate y:
Solve for x:
3x – 14 = 5x – 2
–14 = 2x – 2
–12 = 2x
–6 = x
Substitute –6 for x in either equation and solve for y:
8y + 3x = –2
8y + 3(–6) = –2
8y – 18 = –2
8y = 16
y = 2
The solution to this system of equations is x = –6, y = 2.  
2. Choose a variable to eliminate. Adding the two equations will not eliminate a variable, and subtracting one equation from the other will not eliminate a variable, either. To eliminate y, we would have to multiply both equations by constants before adding. To eliminate x, we can multiply just the first equation by –4 before adding:
–4(x – 2y) = –4(20)
–4x + 8y = –80
The first equation now contains a –4x term and the second equation contains a 4x term. Add these equations to eliminate x:
Divide by 15 to solve for y:
15y = –75
y = –5
Substitute –5 for y in either equation and solve for x:
x – 2y = 20
x – 2(–5) = 20
x + 10 = 20
x = 10
The solution to this system of equations is x = 10, y = –5.
3. Choose a variable to eliminate. Adding the two equations will not eliminate a variable, and subtracting one equation from the other will not eliminate a variable, either. To eliminate x, we can multiply the first equation by 3 and multiply the second equation by 2 before adding:
3(–4x + 3y) = 3(–4)    2(6x – 5y) = 2(2)
–12x + 9y = –12         12x – 10y = 4
The first equation now contains a –12x term and the second equation contains a 12x term. Add these equations to eliminate x:
Divide by –1 to solve for y:
y = –8
y = 8
Substitute 8 for y in either equation and solve for x:
–4x + 3y = –4
–4x + 3(8) = –4
–4x + 24 = –4
–4x = –28
x = 7
The solution to this system of equations is x = 7, y = 8.
View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
Ask
150 Characters allowed