To solve the system of equations:
[tex]\[
\begin{array}{l}
y = 3x \\
y = x^2 - 10
\end{array}
\][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
1. Substitute the expression for [tex]\(y\)[/tex] from the first equation into the second equation.
Given: [tex]\( y = 3x \)[/tex], substitute into [tex]\( y = x^2 - 10 \)[/tex].
[tex]\[ 3x = x^2 - 10 \][/tex]
2. Rearrange the equation to form a standard quadratic equation.
[tex]\[ x^2 - 3x - 10 = 0 \][/tex]
3. Solve the quadratic equation. The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our quadratic equation [tex]\( x^2 - 3x - 10 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -3 \)[/tex]
- [tex]\( c = -10 \)[/tex]
Substitute these values into the quadratic formula:
[tex]\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{9 + 40}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{49}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm 7}{2}
\][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-4}{2} = -2
\][/tex]
4. Find the corresponding [tex]\(y\)[/tex] values using [tex]\( y = 3x \)[/tex]:
- For [tex]\( x = 5 \)[/tex]:
[tex]\[
y = 3(5) = 15
\][/tex]
- For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = 3(-2) = -6
\][/tex]
5. List the solutions as pairs (x, y):
[tex]\[
(5, 15) \quad \text{and} \quad (-2, -6)
\][/tex]
Therefore, the solutions to the system of equations are:
B. [tex]\((-2,-6)\)[/tex] and [tex]\((5,15)\)[/tex]
