Question

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

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

(−4,26)

Answer 1

To solve the linear-quadratic system using the elimination method, we have the following equations:

1) \( y = x^2 + 10 \) (quadratic equation)

2) \( y = -7x - 2 \) (linear equation)

To use the elimination method, we can set the two expressions for \( y \) equal to each other:

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

Now, let's rearrange the equation to set it to zero:

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

Now, we factor the quadratic:

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

Setting each factor equal to zero gives us the possible \( x \) values:

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

Next, we substitute these \( x \) values back into the linear equation to find the corresponding \( y \) values.

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

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

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

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

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

So, another solution is \( (-4, 26) \).

Thus, the solutions to the system are:

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

The correct response is:

**(−3,19) and (−4,26)**.