To solve the given system of equations:
[tex]\[
\begin{cases}
y = 3x \\
y = x^2 - 4
\end{cases}
\][/tex]
we follow these steps:
1. Substitute [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[
3x = x^2 - 4
\][/tex]
2. Rearrange the equation to form a standard quadratic equation:
[tex]\[
x^2 - 3x - 4 = 0
\][/tex]
3. Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[
x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 1 \cdot (-4)}}}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{{3 \pm \sqrt{{9 + 16}}}}{2}
\][/tex]
[tex]\[
x = \frac{{3 \pm \sqrt{25}}}{2}
\][/tex]
[tex]\[
x = \frac{{3 \pm 5}}{2}
\][/tex]
4. Calculate [tex]\( x \)[/tex]:
[tex]\[
x = \frac{{3 + 5}}{2} = \frac{8}{2} = 4
\][/tex]
[tex]\[
x = \frac{{3 - 5}}{2} = \frac{-2}{2} = -1
\][/tex]
5. Find the corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex] using [tex]\( y = 3x \)[/tex]:
For [tex]\( x = 4 \)[/tex]:
[tex]\[
y = 3 \cdot 4 = 12
\][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 3 \cdot (-1) = -3
\][/tex]
Thus, the solutions to the system of equations are [tex]\((-1, -3)\)[/tex] and [tex]\((4, 12)\)[/tex].
The correct answer is C. [tex]\((-1, -3)\)[/tex] and [tex]\((4, 12)\)[/tex].
