Sure, let's solve the system of equations step by step:
[tex]\[
\begin{array}{l}
1. ~ y = x^2 + 8x - 5 \\
2. ~ y = 8x - 4
\end{array}
\][/tex]
Given these two equations, we can set them equal to each other because they both equal [tex]\( y \)[/tex]:
[tex]\[
x^2 + 8x - 5 = 8x - 4
\][/tex]
Next, we subtract [tex]\( 8x \)[/tex] from both sides of the equation to simplify:
[tex]\[
x^2 + 8x - 5 - 8x = 8x - 4 - 8x \implies x^2 - 5 = -4
\][/tex]
Then, we add 4 to both sides of the equation:
[tex]\[
x^2 - 5 + 4 = -4 + 4 \implies x^2 - 1 = 0
\][/tex]
This results in a standard quadratic equation, which we can solve by factoring:
[tex]\[
x^2 - 1 = 0 \implies (x - 1)(x + 1) = 0
\][/tex]
Setting each factor to zero gives us the possible values for [tex]\( x \)[/tex]:
[tex]\[
x - 1 = 0 \implies x = 1
\][/tex]
[tex]\[
x + 1 = 0 \implies x = -1
\][/tex]
Now we have two values for [tex]\( x \)[/tex]. We need to find the corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex], using the equation [tex]\( y = 8x - 4 \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 8(1) - 4 = 8 - 4 = 4
\][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 8(-1) - 4 = -8 - 4 = -12
\][/tex]
Thus, the solutions to the system of equations are:
[tex]\[
\boxed{(-1, -12)} \text{ and } \boxed{(1, 4)}
\][/tex]
So the values of [tex]\( x \)[/tex] that satisfy the system of equations are [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], with the corresponding [tex]\( y \)[/tex] values being [tex]\( y = 4 \)[/tex] and [tex]\( y = -12 \)[/tex], respectively.
