Question

question 1 (multiple choice worth 2 points) (06.03r lc) solve the system of equations. x + y = 8 y = x² - 4 (3, 5) and (-4, 12) (-3, 11) and (4, 4) (4, 4) and (5, 3) (12, -4) and (7, 1)

Explanation:

Step1: Substitute \( y \) from the second equation into the first.

We have \( x + (x^2 - 4) = 8 \).
Simplify to get \( x^2 + x - 4 - 8 = 0 \), so \( x^2 + x - 12 = 0 \).

Step2: Factor the quadratic equation.

Factor \( x^2 + x - 12 \) as \( (x + 4)(x - 3) = 0 \).

Step3: Solve for \( x \).

Set each factor to zero: \( x + 4 = 0 \) gives \( x = -4 \), and \( x - 3 = 0 \) gives \( x = 3 \).

Step4: Find corresponding \( y \) values.

For \( x = 3 \), substitute into \( x + y = 8 \): \( 3 + y = 8 \) gives \( y = 5 \).
For \( x = -4 \), substitute into \( x + y = 8 \): \( -4 + y = 8 \) gives \( y = 12 \).

Answer:

\((3, 5)\) and \((-4, 12)\)