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Question 4 of 10

Solve the system of equations:
[tex]\[
\begin{array}{l}
y = 2x \\
y = x^2 - 8
\end{array}
\][/tex]

A. [tex]\((-2, 4)\)[/tex] and [tex]\((4, 2)\)[/tex]
B. [tex]\((-4, 8)\)[/tex] and [tex]\((2, -4)\)[/tex]
C. [tex]\((-2, -4)\)[/tex] and [tex]\((4, 8)\)[/tex]
D. [tex]\((-4, -8)\)[/tex] and [tex]\((2, 4)\)[/tex]



Answer :

GinnyAnswer
To solve the system of equations, we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations:

[tex]\[ y = 2x \][/tex]
[tex]\[ y = x^2 - 8 \][/tex]

We start by substituting [tex]\( y \)[/tex] from the first equation into the second equation because both are equal to [tex]\( y \)[/tex].

[tex]\[ 2x = x^2 - 8 \][/tex]

Next, we rearrange this equation to form a standard quadratic equation:

[tex]\[ x^2 - 2x - 8 = 0 \][/tex]

To solve the quadratic equation, we can factorize:

[tex]\[ x^2 - 2x - 8 = (x - 4)(x + 2) = 0 \][/tex]

Setting each factor to zero gives us the roots of the equation:

[tex]\[ x - 4 = 0 \implies x = 4 \][/tex]
[tex]\[ x + 2 = 0 \implies x = -2 \][/tex]

Now, we substitute each value of [tex]\( x \)[/tex] back into the first equation [tex]\( y = 2x \)[/tex] to find the corresponding [tex]\( y \)[/tex] values.

For [tex]\( x = 4 \)[/tex]:

[tex]\[ y = 2(4) = 8 \][/tex]

So one solution is [tex]\((4, 8)\)[/tex].

For [tex]\( x = -2 \)[/tex]:

[tex]\[ y = 2(-2) = -4 \][/tex]

So another solution is [tex]\((-2, -4)\)[/tex].

Thus, the solutions to the system of equations are:

[tex]\[ (-2, -4) \text{ and } (4, 8) \][/tex]

From the given choices, the correct option is:

C. [tex]\((-2, -4)\)[/tex] and [tex]\((4, 8)\)[/tex]

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