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

(−4,26)
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

(−3,19) and (−4,26)

Answer 1

To solve the system of equations using the elimination method, we can first set the two equations equal to each other since they both equal \( y \).

The given equations are:

1. \( y = x^2 + 10 \)
2. \( y = -7x - 2 \)

We can set these equal to each other:

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

Now, let’s rearrange this to form a standard quadratic equation:

\[
x^2 + 7x + 10 + 2 = 0
\]

This simplifies to:

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

Next, we can factor this quadratic equation:

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

Setting each factor equal to zero gives us the solutions for \( x \):

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x + 4 = 0 \) → \( x = -4 \)

Now we need to find the corresponding \( y \) values for each \( x \) using either of the original equations. We will use \( y = -7x - 2 \) for this purpose.

For \( x = -3 \):

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

So one solution is \( (-3, 19) \).

For \( x = -4 \):

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

So the second solution is \( (-4, 26) \).

The solutions to the system of equations are:

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

Thus, the correct response is:

**(−3,19) and (−4,26)**.