Use the elimination method to solve the linear-quadratic system y=x2+10

Question

, y=−7x−2
.(1 point)
Responses

(−4,26)
left parenthesis negative 4 comma 26 right parenthesis

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

(3,−23)
and (4,−30)
left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis

(−3,19)

Answer 1

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

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

We can set the equations equal to each other since both equal \( y \):

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

Now, we'll rearrange the equation to form a standard quadratic equation:

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

Next, we can factor the quadratic:

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

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

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

Now, we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We can use the second equation \( y = -7x - 2 \).

For \( x = -3 \):

\[ y = -7(-3) - 2 = 21 - 2 = 19 \] So we have one point: \( (-3, 19) \).

For \( x = -4 \):

\[ y = -7(-4) - 2 = 28 - 2 = 26 \] So we have another point: \( (-4, 26) \).

Thus, the solution set for the system of equations is:

1. \( (-3, 19) \)
2. \( (-4, 26) \)

So the correct response is:

(−3,19) and (−4,26)