Question

Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point)
Responses

(−3,19)
left parenthesis negative 3 comma 19 right parenthesis

(−3,19) and (−4,26)
left parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis

(3,−23) and (4,−30)
left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis

(−4,26)

Answer 1

To solve the system of equations \( y = x^2 + 10 \) and \( y = -7x - 2 \) using the elimination method, we can set the equations equal to each other because they both equal \( y \).

Thus, we can set up the equation:

\[
x^2 + 10 = -7x - 2
\]

Now, rearranging this equation to one side gives us:

\[
x^2 + 7x + 12 = 0
\]

Next, we can factor this quadratic equation:

\[
(x + 3)(x + 4) = 0
\]

Setting each factor to zero gives:

\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]

\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]

Now we will substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. Let's use \( y = -7x - 2 \).

1. For \( x = -3 \):

\[
y = -7(-3) - 2 = 21 - 2 = 19
\]

This gives us the point \( (-3, 19) \).

2. For \( x = -4 \):

\[
y = -7(-4) - 2 = 28 - 2 = 26
\]

This gives us the point \( (-4, 26) \).

Thus, the solutions to the system of equations are the points:

\[
(-3, 19) \quad \text{and} \quad (-4, 26)
\]

So the correct response is:

\[
\textbf{(−3,19) and (−4,26)}
\]