Question

question 15 of 25
solve the system of equations:
y = 3x
y = x² - 4
a. (-1, -3) and (4, 12)
b. (-2, -6) and (2, 6)
c. (-2, 0) and (2, 0)
d. (-1, 3) and (4, 5)

Explanation:

Step1: Substitute \( y = 3x \) into \( y = x^2 - 4 \)

Since both equations equal \( y \), we can set them equal to each other: \( 3x = x^2 - 4 \)

Step2: Rearrange into quadratic equation

Rearrange the equation to standard quadratic form \( ax^2 + bx + c = 0 \): \( x^2 - 3x - 4 = 0 \)

Step3: Factor the quadratic equation

Factor \( x^2 - 3x - 4 \): \( (x - 4)(x + 1) = 0 \)

Step4: Solve for \( x \)

Set each factor equal to zero: \( x - 4 = 0 \) or \( x + 1 = 0 \), so \( x = 4 \) or \( x = -1 \)

Step5: Find corresponding \( y \) values

• For \( x = 4 \), \( y = 3(4) = 12 \)
• For \( x = -1 \), \( y = 3(-1) = -3 \)

Answer:

A. \((-1, -3)\) and \((4, 12)\)