Question

question 46 points 2 consider the system of equations $y + x + 6 = x^2$ and $y - 2x + 2 = 0$. solve the system of equations algebraically. \\(\circ\\) (-1, -4) and (4, 6) \\(\circ\\) (-1, 4) and (6, 4) \\(\circ\\) (1, -4) and (4, -6) \\(\circ\\) (-1, 4) and (-4, 6)

Explanation:

Step1: Isolate y from linear equation

From $y - 2x + 2 = 0$, rearrange to get:
$y = 2x - 2$

Step2: Substitute y into quadratic equation

Substitute $y = 2x - 2$ into $y + x + 6 = x^2$:
$(2x - 2) + x + 6 = x^2$
Simplify left side:
$3x + 4 = x^2$
Rearrange to standard quadratic form:
$x^2 - 3x - 4 = 0$

Step3: Factor the quadratic equation

Factor $x^2 - 3x - 4$:
$(x - 4)(x + 1) = 0$
Solve for x:
$x - 4 = 0 \implies x = 4$
$x + 1 = 0 \implies x = -1$

Step4: Find corresponding y-values

For $x = 4$:
$y = 2(4) - 2 = 6$
For $x = -1$:
$y = 2(-1) - 2 = -4$

Answer:

(-1, -4) and (4, 6)