Systems of Equations: Elimination Method
What You’ll Learn
Section titled “What You’ll Learn”In this lesson you’ll learn the elimination method for solving systems of two linear equations, including how to make coefficients match and handle special cases.
The Concept
Section titled “The Concept”The elimination method (also called addition method) solves a system by adding or subtracting the equations to eliminate one variable, leaving an equation with only one variable.
Steps:
- Make the coefficients of one variable the same (or opposites) in both equations. Multiply one or both equations by numbers if needed
- Add (if opposites) or subtract (if same) the equations to cancel that variable
- Solve the resulting one-variable equation
- Substitute the value back into either original equation to find the second variable
- Write the solution as (x, y) and check in both equations
Example system:
Add the equations (y terms cancel):
4x = 12 → x = 3
Substitute x = 3 into second: 3 − 2y = 1 → −2y = −2 → y = 1
Solution: (3, 1)
Special cases:
- No solution: elimination gives contradiction (e.g., 0 = 5)
- Infinite solutions: elimination gives identity (0 = 0)
Worked Example
Section titled “Worked Example”Solve:
-
Add the equations (y terms cancel):
6x = 24 → x = 4
-
Substitute x = 4 into first: 4(4) + 3y = 25 → 16 + 3y = 25 → 3y = 9 → y = 3
Solution: (4, 3)
Check:
- 4(4) + 3(3) = 16 + 9 = 25 ✓
- 2(4) − 3(3) = 8 − 9 = −1 ✓
Both true.
Another (need to multiply):
Multiply first by 2: 10x + 4y = 36
Subtract second: 10x + 4y − (3x + 4y) = 36 − 26 → 7x = 10 → x = 10/7
Substitute: 5(10/7) + 2y = 18 → 50/7 + 2y = 18 → 2y = 76/7 → y = 38/7
Solution: (10/7, 38/7)
Real-World Application
Section titled “Real-World Application”Elimination is efficient for systems with aligned coefficients:
- Two investment plans: Plan A: 200 + 60m, Plan B: 300 + 40m (m = months). 200 + 60m = 300 + 40m → 20m = 100 → m = 5 months when equal
- Two costs: Fixed 50 dollars + 8 dollars per item vs. 30 dollars + 12 dollars per item. 50 + 8i = 30 + 12i → 20 = 4i → i = 5 items when costs match
- Mixture problems: 5 dollars per pound coffee + 9 dollars per pound to get 7 dollars per pound blend. Use elimination to find amounts.
This method is fast for comparison, break-even, or blending calculations.