Lesson 3-4: Systems by Substitution

Core Concept

What is a System?

Two equations, two unknowns. We need values of x and y that satisfy BOTH equations at the same time. Graphically, it's the intersection point of two lines.

Equation 1: y = 2x + 1 Equation 2: y = -x + 7 Solution: the point where these lines cross

The Method

Substitution: Step by Step

Solve the system: $y = 2x + 1$ and $3x + y = 13$

1. Equation 1 already gives y: y = 2x + 1 2. Substitute into Equation 2: 3x + (2x + 1) = 13 3. Simplify: 5x + 1 = 13 4. 5x = 12 \to x = 12/5 5. Find y: y = 2(12/5) + 1 = 24/5 + 5/5 = 29/5 6. Solution: (12/5, 29/5)

Example 2

Solving for a Variable First

Solve: $2x + y = 10$ and $x + 3y = 15$

1. Isolate y in Eq.1: y = 10 - 2x 2. Substitute into Eq.2: x + 3(10 - 2x) = 15 3. x + 30 - 6x = 15 4. -5x = -15 \to x = 3 5. y = 10 - 2(3) = 4 6. Solution: (3, 4) 7. Check: 2(3)+4=10 ✓ and 3+3(4)=15 ✓

Watch Out

Common Mistakes

Forgetting to check your answer in BOTH equations
Substituting back into the same equation you used to isolate — use the other one
Sign errors when distributing a negative
Always verify: plug solution back into both original equations

Your Turn

Try It Yourself

Q: Solve using substitution: $y = x - 2$ and $2x + y = 7$ .

Show Answer

Substitute y = x-2 into Eq.2: 2x + (x-2) = 7 3x - 2 = 7 \to 3x = 9 \to x = 3 y = 3 - 2 = 1 Solution: (3, 1)

Key Takeaways

1-Minute Summary

Isolate one variable in one equation
Substitute the expression into the other equation
Solve for the remaining variable, then back-substitute
Always check your answer in both original equations

← Lesson 3: Graphing Next: Systems – Elimination →

Systems of Equations: Substitution