Question

Use the elimination method to solve the linear-quadratic system y=x^2+x−42, y=−x+6
.(1 point)
Responses

A) (−8,14) and (6,0)
left parenthesis negative 8 comma 14 right parenthesis and left parenthesis 6 comma 0 right parenthesis

B) (6,0)
left parenthesis 6 comma 0 right parenthesis

C) (8,−2) and (−6,12)
left parenthesis 8 comma negative 2 right parenthesis and left parenthesis negative 6 comma 12 right parenthesis

D) (−8,14)

Answer 1

To solve the system of equations \(y = x^2 + x - 42\) and \(y = -x + 6\) using the elimination method, we can set the two expressions for \(y\) equal to each other:

\[
x^2 + x - 42 = -x + 6
\]

Now, let's rearrange the equation:

1. Add \(x\) to both sides:

\[
x^2 + x + x - 42 = 6
\]

2. Combine like terms:

\[
x^2 + 2x - 42 = 6
\]

3. Subtract 6 from both sides:

\[
x^2 + 2x - 48 = 0
\]

Now we will factor the quadratic equation:

\[
(x + 8)(x - 6) = 0
\]

This gives us the solutions for \(x\):

\[
x + 8 = 0 \quad \Rightarrow \quad x = -8
\]

\[
x - 6 = 0 \quad \Rightarrow \quad x = 6
\]

Now we will find the corresponding \(y\) values for each \(x\):

1. For \(x = -8\):

\[
y = -(-8) + 6 = 8 + 6 = 14
\]

So one solution is \((-8, 14)\).

2. For \(x = 6\):

\[
y = -6 + 6 = 0
\]

So the other solution is \((6, 0)\).

Thus, the solutions to the system of equations are \((-8, 14)\) and \((6, 0)\).

This corresponds to option A):

A) \((-8, 14)\) and \((6, 0)\)