Question

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

(3,−23)

and (4,−30)
left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis
(−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
(−4,26)

Answer 1

To solve the system using the elimination method, we need to set both equations equal to each other.

The system of equations is:
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, let's rearrange this equation:

1. Move all terms to one side of the equation:
\[ x^2 + 7x + 12 = 0 \]

Next, we can factor the quadratic:

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

Setting each factor 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 these \( x \)-values by substituting back into one of the original equations. We will use \( y = -7x - 2 \).

**For \( x = -3 \):**

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

**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 is:

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