Question

Use the elimination method to solve the linear-quadratic system y=x2+10, y=−7x−2.(1 point)%0D%0AResponses%0D%0A%0D%0A(−3,19)%0D%0Aleft parenthesis negative 3 comma 19 right parenthesis%0D%0A%0D%0A(−3,19) and (−4,26)%0D%0Aleft parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis%0D%0A%0D%0A(−4,26)%0D%0Aleft parenthesis negative 4 comma 26 right parenthesis%0D%0A%0D%0A(3,−23) and (4,−30)

Answer 1

To solve the system of equations using the elimination method, we have:

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

First, we can set the two equations for \( y \) equal to each other:

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

Next, we rearrange the equation to bring all terms to one side:

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

This simplifies to:

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

Now, we will factor the quadratic:

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

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

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

Now that we have the \( x \)-values, we can substitute them back into one of the original equations to find the corresponding \( y \)-values. We'll use \( y = -7x - 2 \) for this purpose.

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 \( (-4, 26) \).

So the solutions to the system of equations are:

1. \( (-3, 19) \)
2. \( (-4, 26) \)

The correct response is:

**(−3,19) and (−4,26)**