Question

Use the elimination method to solve the linear-quadratic system y = x^2 + 10, y = -7x -2

A. (-3,19)
B. (-4,26)
C. (3,-23) and (4,-30)
D. (-3,19) and (-4,26)

Answer 1

To solve the system of equations using the elimination method, we start with the given equations:

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

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

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

Now, we'll rearrange this equation to one side:

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

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 substitute these values back into either original equation to find the corresponding values of \( y \).

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

Using Equation 2:

\[
y = -7(-3) - 2 = 21 - 2 = 19
\]

So one point is \( (-3, 19) \).

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

Using Equation 2 again:

\[
y = -7(-4) - 2 = 28 - 2 = 26
\]

So another point is \( (-4, 26) \).

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

Therefore, the correct answer is:

**D. (-3,19) and (-4,26)**.