Section 2.7: Solving Systems Algebraically

Algebraic methods allow solving systems of linear equations without graphing. The two main methods are:

Substitution: Solve one equation for a variable, then substitute into the other equation.
Elimination: Add or subtract equations to eliminate one variable, then solve for the other.

Example 1: Substitution

Solve the system: \[ \begin{cases} y = 2x + 3 \\ x + y = 7 \end{cases} \]

Step 1: Substitute \( y = 2x + 3 \) into \( x + y = 7 \): \( x + (2x + 3) = 7 \) → \( 3x + 3 = 7 \) → \( 3x = 4 \) → \( x = \frac{4}{3} \)

Step 2: Find \( y = 2(4/3) + 3 = 8/3 + 3 = 17/3 \)

Solution: \( (x, y) = \left(\frac{4}{3}, \frac{17}{3}\right) \)

Solve the system: \[ \begin{cases} 2x + 3y = 12 \\ 4x - 3y = 6 \end{cases} \]

Step 1: Add the two equations to eliminate \( y \): \( (2x + 3y) + (4x - 3y) = 12 + 6 \) → \( 6x = 18 \) → \( x = 3 \)

Step 2: Substitute \( x = 3 \) into first equation: \( 2(3) + 3y = 12 \) → \( 6 + 3y = 12 \) → \( 3y = 6 \) → \( y = 2 \)

Solution: \( (x, y) = (3, 2) \)

Solve by substitution: \( y = x + 4 \), \( 2x + y = 10 \)
Solve by elimination: \( 3x + 2y = 12 \), \( 5x - 2y = 8 \)
Solve by substitution: \( y = -x + 3 \), \( x + 2y = 5 \)
Solve by elimination: \( 4x + 3y = 20 \), \( 2x - 3y = 4 \)
Determine if the system has one, none, or infinitely many solutions: \( y = 2x + 1 \), \( 2y - 4x = 2 \)