How to Solve: Systems of Equations (Substitution)

How to Solve: Systems of Equations (Substitution)

Introduction

"Master substitution, and you’ll crack word problems on budgets, sports stats, and even chemistry mixtures—plus ace the 2-3 systems questions on your exam!

What You Need To Know First

1. Solving linear equations – Isolate a variable (e.g., x or y) in one step.
2. Substituting values – Replace a variable with a number or expression (e.g., y = 2x + 3).
3. Checking solutions – Plug values back into both original equations to verify.

Key Vocabulary

Formulas To Know

No new formulas—just the substitution method steps (below). Given on exam sheet: The standard form of a linear equation: Ax + By = C.

Step-by-Step Method

Goal: Find the (x, y) pair that satisfies both equations.

1. Pick an equation where one variable is already isolated (e.g., y = 3x + 2) OR isolate a variable yourself.
2. Substitute the expression from Step 1 into the other equation (replace y or x).
3. Solve the new equation for the remaining variable.
4. Back-substitute the value from Step 3 into either original equation to find the other variable.
5. Write the solution as an ordered pair: (x, y).
6. Check by plugging (x, y) into both original equations.

Worked Example (Using Steps Above)

System: 1. y = 2x + 1 2. 3x + y = 9

Step 1: y is already isolated in Equation 1. Step 2: Substitute y = 2x + 1 into Equation 2: 3x + (2x + 1) = 9 Step 3: Solve for x: 5x + 1 = 9 5x = 8 x = 8/5 (or 1.6) Step 4: Back-substitute x = 8/5 into Equation 1: y = 2(8/5) + 1 = 16/5 + 5/5 = 21/5 (or 4.2) Step 5: Solution: (8/5, 21/5) Step 6: Check: - Equation 1: 21/5 = 2(8/5) + 1 → 21/5 = 16/5 + 5/5 ✔️ - Equation 2: 3(8/5) + 21/5 = 24/5 + 21/5 = 45/5 = 9 ✔️

Worked Examples

Example 1 – Basic

System: 1. x + y = 5 2. y = 2x – 1

Step 1: y is isolated in Equation 2. Step 2: Substitute y = 2x – 1 into Equation 1: x + (2x – 1) = 5 Step 3: Solve for x: 3x – 1 = 5 3x = 6 x = 2 Step 4: Back-substitute x = 2 into Equation 2: y = 2(2) – 1 = 3 Step 5: Solution: (2, 3) Step 6: Check: - Equation 1: 2 + 3 = 5 ✔️ - Equation 2: 3 = 2(2) – 1 ✔️

What we did and why: We used the isolated y to replace it in the first equation, solving for x first. Always back-substitute to find the other variable.

Example 2 – Medium (Isolate a Variable First)

System: 1. 2x + 3y = 12 2. x – y = 1

Step 1: Isolate x in Equation 2: x = y + 1 Step 2: Substitute x = y + 1 into Equation 1: 2(y + 1) + 3y = 12 Step 3: Solve for y: 2y + 2 + 3y = 12 5y + 2 = 12 5y = 10 y = 2 Step 4: Back-substitute y = 2 into x = y + 1: x = 2 + 1 = 3 Step 5: Solution: (3, 2) Step 6: Check: - Equation 1: 2(3) + 3(2) = 6 + 6 = 12 ✔️ - Equation 2: 3 – 2 = 1 ✔️

What we did and why: We isolated x first because it had a coefficient of 1 (easier to solve). Always pick the variable with the simplest coefficient to isolate.

Example 3 – Exam Style (Disguised Variables)

Problem: A café sells small and large coffees. Small coffees cost $3, large cost $5. Yesterday, they sold 12 coffees total and made $50. How many small (s) and large (l) coffees were sold?

Step 1: Write the system: 1. s + l = 12 (total coffees) 2. 3s + 5l = 50 (total revenue)

Step 2: Isolate s in Equation 1: s = 12 – l Step 3: Substitute s = 12 – l into Equation 2: 3(12 – l) + 5l = 50 Step 4: Solve for l: 36 – 3l + 5l = 50 36 + 2l = 50 2l = 14 l = 7 Step 5: Back-substitute l = 7 into s = 12 – l: s = 12 – 7 = 5 Step 6: Solution: 5 small, 7 large Step 7: Check: - Total coffees: 5 + 7 = 12 ✔️ - Revenue: 3(5) + 5(7) = 15 + 35 = 50 ✔️

What we did and why: We translated the word problem into equations, isolated a variable, and solved step-by-step. Always define variables clearly (s = small, l = large).

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s your last-minute substitution cheat sheet: 1. Pick an equation where a variable is alone (like y = 2x + 1). If not, isolate one yourself. 2. Substitute that expression into the other equation. Replace y with 2x + 1 (or x with whatever you isolated). 3. Solve for the remaining variable. This gives you x or y. 4. Back-substitute to find the other variable. Plug your answer into either original equation. 5. Write the solution as (x, y) and check it in both equations. Pro tip: If you get 5 = 3, there’s no solution. If you get 0 = 0, there are infinite solutions. Now go crush that exam!"