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

(−4,26)
left parenthesis negative 4 comma 26 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
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Answer 1

To solve the linear-quadratic system using the elimination method, we start with the two equations:

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

Since both equations are equal to \( y \), we can set them equal to each other:

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

Now, rearranging this equation by bringing all terms to one side:

\[
x^2 + 7x + 10 + 2 = 0
\]
\[
x^2 + 7x + 12 = 0
\]

Next, we will factor this quadratic equation:

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

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

\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]

Now we have two \( x \)-values: \( x = -3 \) and \( x = -4 \). We will find the corresponding \( y \)-values by substituting these \( x \)-values back into either original equation. We'll use the second equation \( y = -7x - 2 \).

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

\[
y = -7(-3) - 2 = 21 - 2 = 19
\]
Thus, one solution is \( (-3, 19) \).

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

\[
y = -7(-4) - 2 = 28 - 2 = 26
\]
Thus, the other solution is \( (-4, 26) \).

Therefore, the solutions to the system are:

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

So the correct response based on the options provided is:

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