Question

select the correct answer. solve the system of equations. $y = 4x + 5$ $y = x^2 + 12x + 20$ a. $(-3,-15)$ and $(-5,-7)$ b. $(-3,-17)$ and $(-5,-25)$ c. $(-3,-7)$ and $(-5,-15)$ d. $(3,17)$ and $(5,-15)$

Explanation:

Step1: Set the two equations equal

Since both equations are equal to \( y \), we can set them equal to each other: \( 4x + 5 = x^2 + 12x + 20 \)

Step2: Rearrange into standard quadratic form

Subtract \( 4x + 5 \) from both sides: \( 0 = x^2 + 8x + 15 \)

Step3: Factor the quadratic

Factor \( x^2 + 8x + 15 \): \( (x + 3)(x + 5) = 0 \)

Step4: Solve for \( x \)

Set each factor equal to zero: \( x + 3 = 0 \) gives \( x = -3 \); \( x + 5 = 0 \) gives \( x = -5 \)

Step5: Find corresponding \( y \) values

For \( x = -3 \), substitute into \( y = 4x + 5 \): \( y = 4(-3) + 5 = -12 + 5 = -7 \)
For \( x = -5 \), substitute into \( y = 4x + 5 \): \( y = 4(-5) + 5 = -20 + 5 = -15 \)

Answer:

C. \((-3,-7)\) and \((-5,-15)\)