Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point)

Responses

(−3,19)
left parenthesis negative 3 comma 19 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

(−4,26)

1 answer

To solve the system of equations \( y = x^2 + 10 \) and \( y = -7x - 2 \) using the elimination method, we can set the equations equal to each other because they both equal \( y \).

Thus, we can set up the equation:

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

Now, rearranging this equation to one side gives us:

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

Next, we can factor this quadratic equation:

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

Setting each factor to zero gives:

\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

Now we will substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. Let's use \( y = -7x - 2 \).

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

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

This gives us the point \( (-3, 19) \).

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

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

This gives us the point \( (-4, 26) \).

Thus, the solutions to the system of equations are the points:

\[ (-3, 19) \quad \text{and} \quad (-4, 26) \]

So the correct response is:

\[ \textbf{(−3,19) and (−4,26)} \]