Question

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 into linear equation

Substitute $y = x^2 - 4$ into $x + y = 8$:
$x + (x^2 - 4) = 8$

Step2: Simplify to quadratic equation

Rearrange and combine like terms:
$x^2 + x - 12 = 0$

Step3: Factor the quadratic equation

Find factors of -12 that sum to 1:
$(x + 4)(x - 3) = 0$

Step4: Solve for x values

Set each factor equal to 0:
$x + 4 = 0 \implies x = -4$; $x - 3 = 0 \implies x = 3$

Step5: Find corresponding y values

For $x=3$: $y = 3^2 - 4 = 5$; For $x=-4$: $y = (-4)^2 - 4 = 12$

Answer:

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