Systems by Elimination

Grade 8 · algebra · 89 practice problems · read aloud

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Systems by Elimination

What is Elimination? 🤔

Elimination is a method to solve a system of linear equations (two or more lines). The goal is to add the equations together to eliminate one variable, making it easy to solve for the other. It's perfect when the variables are already lined up!

Step-by-Step Guide

  1. Line Up the Equations: Write them in Ax + By = C form, stacked.
  2. Match Coefficients: Multiply one or both equations so one variable has opposite coefficients.
  3. Add Equations: Add the equations vertically. One variable should cancel out!
  4. Solve for One Variable: Solve the new equation for the remaining variable.
  5. Substitute Back: Plug your answer into one of the original equations to find the other variable.
  6. Check Your Solution: Plug both values into the other equation to verify.

Visual Examples

Example 1: Simple Elimination

Solve: 2x + y = 7 and 3x - y = 8

  1. They are already lined up. The 'y' terms are +y and -y.
  2. Add the equations:
    (2x + y = 7)
    + (3x - y = 8)
    ------------------
    5x = 15
  3. Solve for x: x = 3
  4. Substitute x=3 into 2x + y = 7 → 2(3) + y = 7 → 6 + y = 7 → y = 1
  5. Solution: (3, 1)

Example 2: Multiplying First

Solve: 3x + 2y = 4 and 2x + y = 1

  1. Lined up. No coefficients are opposites.
  2. Multiply the second equation by -2: -4x - 2y = -2
  3. Now add:
    (3x + 2y = 4)
    + (-4x - 2y = -2)
    ------------------
    -x = 2
  4. Solve for x: x = -2
  5. Substitute x=-2 into 2x + y = 1 → 2(-2) + y = 1 → -4 + y = 1 → y = 5
  6. Solution: (-2, 5)

Common Mistakes ⚠️

Sign Errors: The most common mistake! When you multiply an equation by a negative number, every term must change sign.

Forgetting to Multiply the Constant: Remember, what you do to one side, you do to the ENTIRE other side. Multiply the constant term too.

Not Checking the Solution: Always plug your final answer into both original equations to catch calculation errors.

Tips & Tricks

Look for Opposites: First, check if any variables already have opposite coefficients (like +2y and -2y).

Pick the Easiest Variable: Choose to eliminate the variable that requires the least multiplication.

Shortcut for Substitution: After finding one variable, plug it into the easiest-looking original equation to minimize work.

How to Practice

  • Start Simple: Begin with systems where one variable is already set up to cancel.
  • Mix It Up: Practice problems where you need to multiply one or both equations.
  • Use Online Quizzes: Find interactive websites that give immediate feedback.
  • Write Your Own: Create a solved system, then scramble it and give it to a friend to solve!

Practice problems

6 of the 89, worked through step by step — try them before opening the answer.

1 2x + 3y = 16; 2x - y = 4

Hint: Notice that both equations have the same coefficient for one variable. Consider subtracting one equation from the other to eliminate that variable.

Show the answer

Answer: x = 3.5, y = 3

  1. Notice that both equations have the same "2x" term. We can subtract equation (2) from equation (1) to eliminate x. (2x + 3y) - (2x - y) = 16 - 4 2x + 3y - 2x + y = 12 The 2x terms cancel: 3y + y = 12 4y = 12
  2. Solve for y. y = 12 / 4 y = 3
  3. Substitute y = 3 into equation (2) to find x. 2x - y = 4 2x - 3 = 4 2x = 4 + 3 2x = 7 x = 7 / 2 x = 3.5
  4. Check in equation (1). 2(3.5) + 3(3) = 7 + 9 = 16, correct. Final answer: x = 3.5, y = 3

We are given the system of equations: (1) 2x + 3y = 16 (2) 2x - y = 4

2 2x + 3y = 12; 2x - y = 4

Hint: Look for coefficients that can be eliminated by adding or subtracting the equations. Consider what happens when you subtract one equation from the other.

Show the answer

Answer: x = 3, y = 2

  1. Solve one equation for one variable. From equation (2): 2x - y = 4 Add y to both sides: 2x = 4 + y Subtract 4 from both sides: 2x - 4 = y So y = 2x - 4
  2. Substitute y = 2x - 4 into equation (1). Equation (1): 2x + 3y = 12 Replace y with (2x - 4): 2x + 3(2x - 4) = 12
  3. Simplify and solve for x. 2x + 6x - 12 = 12 8x - 12 = 12 Add 12 to both sides: 8x = 24 Divide both sides by 8: x = 3
  4. Substitute x = 3 into y = 2x - 4. y = 2(3) - 4 y = 6 - 4 y = 2
  5. Check the solution in both original equations. Equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ Equation (2): 2(3) - 2 = 6 - 2 = 4 ✓ Final answer: x = 3, y = 2

We have the system of equations: (1) 2x + 3y = 12 (2) 2x - y = 4

3 2x + 3y = 16; 2x - y = 0

Hint: When both equations have the same coefficient for one variable, you can eliminate it by subtracting one equation from the other. This will leave you with an equation containing only one variable to solve.

Show the answer

Answer: x = 2, y = 4

  1. Solve equation (2) for y. From equation (2): 2x - y = 0 Add y to both sides: 2x = y So y = 2x.
  2. Substitute y = 2x into equation (1). Equation (1): 2x + 3y = 16 Replace y with 2x: 2x + 3*(2x) = 16 Multiply: 2x + 6x = 16 Combine like terms: 8x = 16
  3. Solve for x. Divide both sides by 8: x = 16/8 x = 2.
  4. Solve for y using y = 2x. y = 2*(2) = 4.
  5. Check the solution in both original equations. Equation (1): 2*(2) + 3*(4) = 4 + 12 = 16 ✓ Equation (2): 2*(2) - 4 = 4 - 4 = 0 ✓ Final answer: x = 2, y = 4

We are given the system of equations: (1) 2x + 3y = 16 (2) 2x - y = 0

4 3x + 2y = 16; 3x - 2y = 8

Hint: When the coefficients of one variable are opposites, adding the equations eliminates that variable, allowing you to solve for the other variable first.

Show the answer

Answer: x = 4, y = 2

  1. Notice that both equations have the same coefficient for x (3x). We can eliminate y by adding the two equations together. Add equation (1) and equation (2): (3x + 2y) + (3x - 2y) = 16 + 8 3x + 2y + 3x - 2y = 24 6x + 0y = 24 6x = 24
  2. Solve for x. Divide both sides by 6: x = 24 / 6 x = 4
  3. Substitute x = 4 into equation (1) to find y. 3(4) + 2y = 16 12 + 2y = 16
  4. Isolate 2y. 2y = 16 - 12 2y = 4
  5. Solve for y. y = 4 / 2 y = 2
  6. Check in equation (2) to verify. 3(4) - 2(2) = 12 - 4 = 8, which matches. Final answer: x = 4, y = 2

We are given the system of equations: (1) 3x + 2y = 16 (2) 3x - 2y = 8

5 3x + 2y = 16, 2x - 2y = 4

Hint: Look for coefficients that can be eliminated by adding the equations together. Check if any variable terms will cancel out when you combine both equations.

Show the answer

Answer: x = 4, y = 2

  1. Notice that the second equation has -2y and the first has +2y. If we add the two equations directly, the y terms will cancel. Add equation (1) and equation (2): (3x + 2y) + (2x - 2y) = 16 + 4 3x + 2x + 2y - 2y = 20 5x + 0 = 20 5x = 20
  2. Solve for x. Divide both sides by 5: x = 20 / 5 x = 4
  3. Substitute x = 4 into equation (1) to find y. 3(4) + 2y = 16 12 + 2y = 16
  4. Isolate 2y. 2y = 16 - 12 2y = 4
  5. Solve for y. y = 4 / 2 y = 2
  6. Check in equation (2) to verify. 2(4) - 2(2) = 8 - 4 = 4, which matches. Final answer: x = 4, y = 2

We are given the system of equations: (1) 3x + 2y = 16 (2) 2x - 2y = 4

6 2x + 3y = 12; 4x - 3y = 6

Hint: Look for coefficients that are opposites to eliminate one variable when adding the equations together.

Show the answer

Answer: x = 3, y = 2

  1. Choose a method to solve** Notice that the coefficients of \( y \) are \( +3 \) in the first equation and \( -3 \) in the second equation. If we add the two equations directly, the \( y \) terms will cancel out. --- **
  2. Add the equations** Add equation (1) and equation (2): (2x + 3y) + (4x - 3y) = 12 + 6 2x + 4x + 3y - 3y = 18 6x + 0 = 18 6x = 18 --- **
  3. Solve for x** Divide both sides by 6: x = 18 / 6 x = 3 --- **
  4. Substitute x = 3 into equation (1)** 2(3) + 3y = 12 6 + 3y = 12 --- **
  5. Solve for y** Subtract 6 from both sides: 3y = 12 - 6 3y = 6 Divide both sides by 3: y = 6 / 3 y = 2 --- **
  6. Check the solution** Substitute x = 3, y = 2 into equation (2): 4(3) - 3(2) = 12 - 6 = 6, which matches equation (2). Both equations are satisfied. --- **Final Answer:** x = 3, y = 2

We are given the system of equations: (1) 2x + 3y = 12 (2) 4x - 3y = 6 --- **

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