To solve the system of equations:
[tex]\[
\begin{array}{l}
y = (x-2)^2 - 15 \\
-5x + y = -1
\end{array}
\][/tex]
Let's go through the step-by-step solution:
1. Rewrite the second equation for [tex]\( y \)[/tex]:
The second equation is [tex]\(-5x + y = -1\)[/tex].
Add [tex]\( 5x \)[/tex] to both sides of the equation to isolate [tex]\( y \)[/tex]:
[tex]\[
y = 5x - 1
\][/tex]
2. Substitute [tex]\( y = 5x - 1 \)[/tex] into the first equation:
The first equation is [tex]\( y = (x-2)^2 - 15 \)[/tex]. Using our expression for [tex]\( y \)[/tex] from the second equation, we get:
[tex]\[
5x - 1 = (x-2)^2 - 15
\][/tex]
3. Solve the resulting equation for [tex]\( x \)[/tex]:
[tex]\[
5x - 1 = (x-2)^2 - 15
\][/tex]
Simplify the equation:
[tex]\[
5x - 1 = x^2 - 4x + 4 - 15
\][/tex]
[tex]\[
5x - 1 = x^2 - 4x - 11
\][/tex]
Bring all terms to one side to set the equation to zero:
[tex]\[
x^2 - 4x - 11 - 5x + 1 = 0
\][/tex]
[tex]\[
x^2 - 9x - 10 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
(x + 1)(x - 10) = 0
\][/tex]
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[
x = -1 \quad \text{and} \quad x = 10
\][/tex]
4. Calculate the corresponding [tex]\( y \)[/tex] values:
Substitute [tex]\( x = -1 \)[/tex] and [tex]\( x = 10 \)[/tex] back into either of the [tex]\( y \)[/tex] equations, preferably the simpler one [tex]\( y = 5x - 1 \)[/tex]:
For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 5(-1) - 1 = -5 - 1 = -6
\][/tex]
For [tex]\( x = 10 \)[/tex]:
[tex]\[
y = 5(10) - 1 = 50 - 1 = 49
\][/tex]
5. Complete solutions:
The solutions to the system of equations are:
- When [tex]\( x = -1 \)[/tex], [tex]\( y = -6 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 49 \)[/tex]
Therefore, one solution of the system of equations is [tex]\((-1, -6)\)[/tex], and a second solution is [tex]\((10, 49)\)[/tex].
