Question

Use the elimination method to solve the linear-quadratic system y = x^2 + 10
, y =-7x - 2. (1 point)
• (-3, 19)
• (-3, 19) and (-4, 26)
O (3, -23) and (4, -30)
O (-4,26)

Answer 1

To solve the linear-quadratic system using the elimination method, we start by substituting the expression for \( y \) from the linear equation into the quadratic equation.

The equations we have are:

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

### Step 1: Set the equations equal to each other

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

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

### Step 2: Rearrange the equation

Move all terms to one side to set the equation to zero:

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

### Step 3: Factor the quadratic equation

Now we need to factor the quadratic equation:

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

### Step 4: Solve for \( x \)

Setting each factor equal to zero gives us:

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

### Step 5: Find the corresponding \( y \) values

Now we substitute these \( x \) values back into either of the original equations to find the corresponding \( y \) values. We'll use the linear equation \( y = -7x - 2 \):

1. For \( x = -3 \):

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

Thus, one solution is \( (-3, 19) \).

2. For \( x = -4 \):

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

Thus, the second solution is \( (-4, 26) \).

### Final solutions

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

From the options provided, the correct answer is:

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