Question

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

(−3,19)
(
−
3
,
19
)
left parenthesis negative 3 comma 19 right parenthesis

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

(3,−23)
(
3
,
−
23
)
and (4,−30)
(
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:

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

we will use the elimination method by setting the two equations equal to each other.

First, set the right-hand sides equal:

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

Now, let's rearrange this equation:

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

Next, we can factor this quadratic equation:

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

Setting each factor equal to zero gives us the solutions for \( x \):

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

Now we can find the corresponding \( y \) values by substituting these \( x \) values back into either equation. We'll use \( y = -7x - 2 \).

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

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

**For \( x = -4 \):**

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

Thus the solutions to the system are \( (-3, 19) \) and \( (-4, 26) \).

The correct response is:
\[
(-3, 19) \text{ and } (-4, 26)
\]