How to Solve: Systems of Equations (Elimination)

How to Solve: Systems of Equations (Elimination)

Introduction

"Master elimination, and you’ll solve two equations in under 60 seconds—no graphing, no guesswork. This is how you ace word problems on speed, cost, and mixtures in your exam!

What You Need To Know First

1. Linear equations in two variables – You must recognize equations like 2x + 3y = 6 and know they represent straight lines.
2. Like terms – You must be able to identify and combine terms with the same variable (e.g., 3x and -5x).
3. Basic arithmetic operations – You must be comfortable adding, subtracting, and multiplying integers and decimals.

Key Vocabulary

Formulas To Know

Step-by-Step Method

Step 1: Write both equations in standard form. - Standard form: Ax + By = C (no fractions, no decimals). - Example: y = 2x – 3 → -2x + y = -3.

Step 2: Align like terms. - Write equations one above the other, variables in the same order. - Example: 3x + 2y = 8 5x – 2y = 4

Step 3: Check for opposite coefficients. - If one variable already has opposite coefficients (e.g., +2y and -2y), skip to Step 5. - If not, multiply one or both equations to create opposites.

Step 4: Multiply to create opposite coefficients. - Choose the variable to eliminate (usually the one with smaller coefficients). - Multiply the first equation by the coefficient of the variable in the second equation. - Multiply the second equation by the coefficient of the variable in the first equation. - Keep the signs opposite! - Example: 3x + 2y = 8 (multiply by 3) 2x – 3y = -5 (multiply by 2) → 9x + 6y = 24 → 4x – 6y = -10

Step 5: Add or subtract the equations. - If coefficients are opposites, add the equations. - If coefficients are the same, subtract the equations. - This eliminates one variable.

Step 6: Solve for the remaining variable. - Isolate the variable using inverse operations.

Step 7: Substitute back to find the other variable. - Plug the value from Step 6 into either original equation. - Solve for the second variable.

Step 8: Write the solution as an ordered pair (x, y). - Always check by plugging into both original equations.

Step 9: Verify the solution. - Substitute (x, y) into both equations. - Both must be true (e.g., 5 = 5).

Worked Example Using Steps

Problem: 2x + 3y = 7 4x – y = 3

Step 1: Already in standard form.

Step 2: 2x + 3y = 7 4x – y = 3

Step 3: No opposite coefficients. Choose y to eliminate.

Step 4: Multiply the second equation by 3 to make y coefficients opposites. 4x – y = 3 → 12x – 3y = 9

Step 5: Add the equations: (2x + 3y) + (12x – 3y) = 7 + 9 14x = 16

Step 6: Solve for x: x = 16/14 = 8/7

Step 7: Substitute x = 8/7 into 4x – y = 3: 4(8/7) – y = 3 32/7 – y = 3 -y = 3 – 32/7 -y = 21/7 – 32/7 -y = -11/7 y = 11/7

Step 8: Solution: (8/7, 11/7)

Step 9: Verify: - 2(8/7) + 3(11/7) = 16/7 + 33/7 = 49/7 = 7 ✔️ - 4(8/7) – 11/7 = 32/7 – 11/7 = 21/7 = 3 ✔️

Worked Examples

Example 1 – Basic (No Multiplication Needed)

Problem: x + y = 5 x – y = 1

Step 1: Already in standard form.

Step 2: x + y = 5 x – y = 1

Step 3: y has opposite coefficients (+y and -y).

Step 5: Add the equations: (x + y) + (x – y) = 5 + 1 2x = 6

Step 6: x = 3

Step 7: Substitute x = 3 into x + y = 5: 3 + y = 5 y = 2

Step 8: Solution: (3, 2)

What we did and why: We added the equations to eliminate y because its coefficients were opposites. This gave us x directly.

Example 2 – Medium (Multiplication Required)

Problem: 3x + 2y = 8 2x – 3y = -5

Step 1: Already in standard form.

Step 2: 3x + 2y = 8 2x – 3y = -5

Step 3: No opposite coefficients. Choose y to eliminate.

Step 4: Multiply first equation by 3, second by 2: 3(3x + 2y) = 3(8) → 9x + 6y = 24 2(2x – 3y) = 2(-5) → 4x – 6y = -10

Step 5: Add the equations: (9x + 6y) + (4x – 6y) = 24 + (-10) 13x = 14

Step 6: x = 14/13

Step 7: Substitute x = 14/13 into 3x + 2y = 8: 3(14/13) + 2y = 8 42/13 + 2y = 8 2y = 8 – 42/13 2y = 104/13 – 42/13 2y = 62/13 y = 31/13

Step 8: Solution: (14/13, 31/13)

What we did and why: We multiplied both equations to create opposite y coefficients (+6y and -6y), then added to eliminate y.

Example 3 – Exam Style (Disguised Problem)

Problem: A school sells tickets for a play. Adult tickets cost $8, student tickets cost $5. On Friday, 120 tickets were sold for $720. How many adult and student tickets were sold?

Step 1: Define variables. - Let a = number of adult tickets. - Let s = number of student tickets.

Step 2: Write equations. - Total tickets: a + s = 120 - Total money: 8a + 5s = 720

Step 3: Align equations. a + s = 120 8a + 5s = 720

Step 4: Multiply first equation by 5 to eliminate s. 5(a + s) = 5(120) → 5a + 5s = 600

Step 5: Subtract the new equation from the second equation: (8a + 5s) – (5a + 5s) = 720 – 600 3a = 120

Step 6: a = 40

Step 7: Substitute a = 40 into a + s = 120: 40 + s = 120 s = 80

Step 8: Solution: 40 adult tickets, 80 student tickets.

What we did and why: We used elimination to remove s by making its coefficients equal, then subtracted to solve for a. This is a classic word problem setup.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, listen up—this is your 60-second crash course for elimination. First, write both equations in Ax + By = C form. Next, pick a variable to eliminate—usually the one with smaller numbers. Multiply one or both equations to make coefficients opposites. Add or subtract the equations to kill one variable. Solve for the remaining variable, then plug it back in to find the other. Always, always check your answer in both original equations. Watch out for fractions, sign errors, and word problems where you have to define variables first. You’ve got this—go practice three problems right now, and elimination will be your fastest way to full marks!