Question

solve the system of equations:
y = 2x + 1
y = x² + 2x - 8

a. (-4, 0) and (2, 0)
b. (0, 1) and (2, 5)
c. (-3, -5) and (3, 7)
d. (-3, 5) and (3, 2)

Explanation:

Step1: Set the two equations equal

Since both equal \( y \), set \( 2x + 1 = x^2 + 2x - 8 \).
\( 2x + 1 = x^2 + 2x - 8 \)

Step2: Simplify the equation

Subtract \( 2x \) from both sides: \( 1 = x^2 - 8 \).
Then add 8 to both sides: \( x^2 = 9 \).

Step3: Solve for \( x \)

Take square roots: \( x = \pm 3 \), so \( x = 3 \) or \( x = -3 \).

Step4: Find corresponding \( y \) values

For \( x = 3 \): Use \( y = 2x + 1 \), so \( y = 2(3) + 1 = 7 \).
For \( x = -3 \): Use \( y = 2x + 1 \), so \( y = 2(-3) + 1 = -5 \).

Answer:

C. \((-3, -5)\) and \((3, 7)\)