Certainly! Let's solve the system of equations step-by-step.
We have the following system of equations:
[tex]\[
\left\{
\begin{array}{c}
y = -6x - 16 \\
y = x^2 - 7
\end{array}
\right.
\][/tex]
To solve this system, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations. Here is the step-by-step solution:
1. Set the equations equal to each other:
Since both equations are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[
-6x - 16 = x^2 - 7
\][/tex]
2. Rearrange the equation:
Bring all terms to one side of the equation to set it to zero:
[tex]\[
x^2 + 6x + 9 = 0
\][/tex]
3. Factor the quadratic equation:
Notice that [tex]\( x^2 + 6x + 9 \)[/tex] is a perfect square trinomial. We can factor it as follows:
[tex]\[
(x + 3)^2 = 0
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
Set each factor equal to zero:
[tex]\[
x + 3 = 0
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x = -3
\][/tex]
5. Solve for [tex]\( y \)[/tex]:
Use the value of [tex]\( x \)[/tex] in either of the original equations to find [tex]\( y \)[/tex]. Let's use the first equation [tex]\( y = -6x - 16 \)[/tex]:
[tex]\[
y = -6(-3) - 16
\][/tex]
Simplify:
[tex]\[
y = 18 - 16
\][/tex]
Thus:
[tex]\[
y = 2
\][/tex]
6. Write the solution:
We have found that [tex]\( x = -3 \)[/tex] and [tex]\( y = 2 \)[/tex]. Therefore, the solution to the system of equations is:
[tex]\[
(-3, 2)
\][/tex]
So, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations are [tex]\( x = -3 \)[/tex] and [tex]\( y = 2 \)[/tex].
