Question

find the ordered pair solutions for the system of equations.
$\begin{cases}y = x^{2}-2x - 6\\y = 4x + 10end{cases}$
$(?, )$ and $( , )$
enter the smallest x first.

Explanation:

Step1: Set the two equations equal

Since both expressions equal $y$, we set $x^{2}-2x - 6=4x + 10$.

Step2: Rearrange to standard quadratic form

Move all terms to one - side: $x^{2}-2x-4x - 6 - 10=0$, which simplifies to $x^{2}-6x - 16=0$.

Step3: Factor the quadratic equation

We factor $x^{2}-6x - 16$ as $(x - 8)(x+2)=0$.

Step4: Solve for x

Set each factor equal to zero:
If $x - 8=0$, then $x = 8$; if $x + 2=0$, then $x=-2$.

Step5: Find corresponding y - values

When $x=-2$, substitute into $y = 4x + 10$, so $y=4\times(-2)+10=-8 + 10=2$.
When $x = 8$, substitute into $y = 4x + 10$, so $y=4\times8+10=32 + 10=42$.

Answer:

$(-2,2)$ and $(8,42)$