Section 5.1: Solving Systems by Substitution
Substitution is a method for solving systems of linear equations by solving one equation for one variable and then substituting that expression into the other equation.
Example 1
Solve the system:
\( x + y = 5 \)
\( 2x - y = 1 \)
Step 1: Solve the first equation for \( y \): \( y = 5 - x \)
Step 2: Substitute into the second equation: \( 2x - (5 - x) = 1 \)
Step 3: Simplify: \( 2x - 5 + x = 1 \) → \( 3x - 5 = 1 \) → \( 3x = 6 \) → \( x = 2 \)
Step 4: Substitute back: \( y = 5 - 2 = 3 \)
Solution: \( (x,y) = (2,3) \)
Example 2
Solve:
\( 3x + 2y = 12 \)
\( x - y = 1 \)
Step 1: Solve the second equation for \( x \): \( x = y + 1 \)
Step 2: Substitute into the first equation: \( 3(y + 1) + 2y = 12 \)
Step 3: Simplify: \( 3y + 3 + 2y = 12 \) → \( 5y + 3 = 12 \) → \( 5y = 9 \) → \( y = \frac{9}{5} \)
Step 4: Substitute back: \( x = \frac{9}{5} + 1 = \frac{14}{5} \)
Solution: \( (x,y) = \left(\frac{14}{5}, \frac{9}{5}\right) \)
Practice Problems
- Solve \( x + 2y = 7 \) and \( 3x - y = 5 \).
- Solve \( 2x + 3y = 12 \) and \( x - 2y = 1 \).
- Solve \( y = 2x + 1 \) and \( x + y = 7 \).
- Solve \( 4x - y = 3 \) and \( 2x + 5y = 16 \).
- Solve \( x - y = 4 \) and \( 2x + y = 10 \).