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Solve the system of equations:
[tex]\[
\begin{array}{l}
y = 3x \\
y = x^2 - 10
\end{array}
\][/tex]

A. [tex]\((-2, -6)\)[/tex] and [tex]\((-5, -15)\)[/tex]
B. [tex]\((-2, -6)\)[/tex] and [tex]\((5, 15)\)[/tex]
C. [tex]\((2, 6)\)[/tex] and [tex]\((5, 15)\)[/tex]
D. [tex]\((2, 6)\)[/tex] and [tex]\((-5, -15)\)[/tex]



Answer :

GinnyAnswer
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]

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