To solve the given system of equations:
[tex]\[
\begin{array}{l}
y = 2x \\
y = x^2 - 15
\end{array}
\][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
1. Substitute [tex]\( y = 2x \)[/tex] from the first equation into the second equation [tex]\( y = x^2 - 15 \)[/tex]:
[tex]\[
2x = x^2 - 15
\][/tex]
2. Rearrange this equation to form a quadratic equation:
[tex]\[
x^2 - 2x - 15 = 0
\][/tex]
3. Solve the quadratic equation by factoring:
To factor [tex]\( x^2 - 2x - 15 \)[/tex], we need two numbers that multiply to [tex]\(-15\)[/tex] and add up to [tex]\(-2\)[/tex]. These numbers are [tex]\(-5\)[/tex] and [tex]\(3\)[/tex]:
[tex]\[
(x - 5)(x + 3) = 0
\][/tex]
4. Set each factor to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
x - 5 = 0 \quad \text{or} \quad x + 3 = 0
\][/tex]
[tex]\[
x = 5 \quad \text{or} \quad x = -3
\][/tex]
5. Find the corresponding [tex]\( y \)[/tex] values using [tex]\( y = 2x \)[/tex]:
- For [tex]\( x = 5 \)[/tex]:
[tex]\[
y = 2(5) = 10
\][/tex]
- For [tex]\( x = -3 \)[/tex]:
[tex]\[
y = 2(-3) = -6
\][/tex]
6. List the solutions as ordered pairs:
- [tex]\( (x, y) = (5, 10) \)[/tex]
- [tex]\( (x, y) = (-3, -6) \)[/tex]
Therefore, the solutions to the system of equations are [tex]\( (5, 10) \)[/tex] and [tex]\( (-3, -6) \)[/tex].
The correct answer is:
C. [tex]\( (-3, -6) \)[/tex] and [tex]\( (5, 10) \)[/tex]
