Use the elimination method to solve the linear-quadratic system y=x2+10

Question

, 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 linear-quadratic system using the elimination method, we start with the equations given:

1. \( y = x^2 + 10 \) (This is the quadratic equation)
2. \( y = -7x - 2 \) (This is the linear equation)

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

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

Now, rearranging the equation to one side gives us:

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

Next, we need to factor the quadratic:

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

Setting each factor equal to zero gives:

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

Now we substitute these \( x \)-values back into the linear equation \( y = -7x - 2 \) to find the corresponding \( y \)-values.

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 other solution is \( (-4, 26) \).

Thus, the complete solution set for the system is:

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

The correct response from the options provided is:

(−3,19) and (−4,26).