Use the elimination method to solve the linear-quadratic system y=x2+x−42

Question

, y=−x+6
.(1 point)
Responses

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

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

(−8,14)
left parenthesis negative 8 comma 14 right parenthesis

(−8,14)
and (6,0)

Answer 1

To solve the system of equations using the elimination method, we have the following equations:

1. \( y = x^2 + x - 42 \)
2. \( y = -x + 6 \)

First, we will set them equal to each other because both equations equal \( y \):

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

Now, let's move all terms to one side of the equation:

\[ x^2 + x + x - 42 - 6 = 0 \]

This simplifies to:

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

Next, we can factor the quadratic:

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

Setting each factor to zero gives us the \( x \) values:

1. \( x + 8 = 0 \) → \( x = -8 \)
2. \( x - 6 = 0 \) → \( x = 6 \)

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

For \( x = -8 \): Substituting into the second equation \( y = -x + 6 \): \[ y = -(-8) + 6 = 8 + 6 = 14 \] So one solution is \( (-8, 14) \).

For \( x = 6 \): Substituting into the second equation \( y = -x + 6 \): \[ y = -6 + 6 = 0 \] So the other solution is \( (6, 0) \).

Thus, the solutions to the system are:

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

Therefore, the answer is:

(−8,14) and (6,0).