Core Concept

The Idea Behind Elimination

If one equation has +3y and the other has −3y, adding them makes the y-terms cancel — eliminating that variable. Then you solve for the remaining variable.

x + 3y = 11 ... (1)
x − 3y = 1 ... (2)
─────────────
2x = 12 (add equations)
x = 6
Then: 6 + 3y = 11 → y = 5/3... (wait, let's use (1): 6+3y=11 → 3y=5 → y=5/3)

Tip: always double-check by substituting back.

Example 1

Direct Elimination

Solve: 2x + y = 8 and 3x − y = 7.

1. Notice: +y and −y — they cancel when added
2. Add equations: (2x+y) + (3x−y) = 8+7
3. 5x = 15 → x = 3
4. Substitute x=3 into Eq.1: 2(3)+y=8 → y=2
5. Solution: (3, 2)
6. Check Eq.2: 3(3)−2 = 9−2 = 7 ✓

Example 2

Multiplying to Create Opposite Coefficients

Solve: 3x + 2y = 12 and x + y = 5.

1. No variables cancel directly. Multiply Eq.2 by −2:
2. −2(x + y) = −2(5) → −2x − 2y = −10
3. Now add to Eq.1:
4. 3x + 2y = 12
5. −2x − 2y = −10
6. ──────────────
7. x = 2
8. Sub into Eq.2: 2 + y = 5 → y = 3
9. Solution: (2, 3) ✓

Special Cases

No Solution or Infinite Solutions

Sometimes when you add the equations, both variables cancel:

No Solution
x + y = 3
x + y = 7
─────
0 = −4 (impossible!)

Parallel lines — never intersect.

Infinite Solutions
2x + 4y = 8
x + 2y = 4
─────
0 = 0 (always true!)

Same line — infinite intersections.

Watch Out

Common Mistakes

Your Turn

Try It Yourself

Q: Solve using elimination: 4x + 3y = 25 and 2x − 3y = 5.

Show Answer
+3y and −3y cancel when added:
6x = 30 → x = 5
4(5) + 3y = 25 → 3y = 5 → y = 5/3
Solution: (5, 5/3)

Key Takeaways

1-Minute Summary

← Lesson 4: Substitution Take the Unit 3 Quiz →