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Solve the system of equations.

[tex]
\begin{array}{l}
y=2x \\
y=x^2-15
\end{array}
\]

A. [tex]\((3, 6)\)[/tex] and [tex]\((-5, -10)\)[/tex]

B. [tex]\((3, 6)\)[/tex] and [tex]\((5, 10)\)[/tex]

C. [tex]\((-3, -6)\)[/tex] and [tex]\((5, 10)\)[/tex]

D. [tex]\((-3, -6)\)[/tex] and [tex]\((-5, -10)\)[/tex]


Sagot :

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]