Question

Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point)
Responses

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

(−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

(−4,26)

Answer 1

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

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

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

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

Now, let's rearrange the equation to one side:

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

Next, we can factor this quadratic equation:

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

From the factored form, we can find the values of \( x \):

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

Now we will substitute these values of \( x \) back into one of the original equations to solve for \( y \). We'll use Equation 2 for this purpose:

1. For \( x = -3 \):
\[
y = -7(-3) - 2 = 21 - 2 = 19
\]
This gives us the point \( (-3, 19) \).

2. For \( x = -4 \):
\[
y = -7(-4) - 2 = 28 - 2 = 26
\]
This gives us the point \( (-4, 26) \).

So the solution to the system of equations is:

\((-3, 19)\) and \((-4, 26)\).

The correct response is: **(−3,19) and (−4,26)**.