Use a graphing calculator to locate the solutions of this system of equations:

[tex]\[
\begin{array}{l}
y = (x-2)^2 - 15 \\
-5x + y = -1
\end{array}
\][/tex]

One solution of the system of equations is:
A second solution of the system of equations is:



Answer :

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].