Algebra 101 - Unit 13

Lesson 77: Solving Linear Systems Algebraically

Linear systems can also be solved algebraically using substitution or elimination. These methods allow precise solutions without relying on graphs.

Method 1: Substitution

1. Solve one equation for one variable in terms of the other.
2. Substitute this expression into the other equation.
3. Solve for the remaining variable.
4. Substitute back to find the other variable.

Example 1: Solve by substitution:

x + y = 5
2x − y = 1

Solve first equation for y: y = 5 − x
Substitute into second: 2x − (5 − x) = 1 → 2x − 5 + x = 1 → 3x = 6 → x = 2
Then y = 5 − 2 = 3 → Solution: (2, 3)

Method 2: Elimination

1. Multiply equations if necessary so that adding or subtracting eliminates one variable.
2. Add or subtract equations to eliminate one variable.
3. Solve for the remaining variable.
4. Substitute back to find the other variable.

Example 2: Solve by elimination:

3x + 2y = 12
x − y = 1

Multiply second equation by 2: 2x − 2y = 2
Add to first: (3x + 2y) + (2x − 2y) = 12 + 2 → 5x = 14 → x = 14/5
Substitute: 14/5 − y = 1 → y = 14/5 − 1 = 9/5 → Solution: (14/5, 9/5)

Practice Problems

1. Solve by substitution: x + 2y = 7, 3x − y = 8
2. Solve by elimination: 2x + 3y = 12, 4x − y = 6
3. Solve using substitution: 3x − y = 5, x + y = 4
4. Solve using elimination: 5x + 2y = 14, 3x − 2y = 4
5. Determine if the system has one solution, none, or infinitely many: x − y = 2, 2x − 2y = 4