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

Question

Responses

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

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

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

(−3,19) and (−4,26)

Answer 1

To solve the system of equations

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

using the elimination method, we can set the two equations equal to each other since both equal \( y \):

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

Now, we will move all terms to one side of the equation:

\[ x^2 + 7x + 10 + 2 = 0 \]

Simplifying this, we get:

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

Next, we will factor the quadratic equation:

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

Setting each factor to zero gives us:

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

Now we will find the corresponding \( y \) values for each \( x \) by substituting back into one of the original equations. We'll use \( y = -7x - 2 \) for this purpose.

1. For \( x = -3 \): \[ y = -7(-3) - 2 = 21 - 2 = 19 \Rightarrow (-3, 19) \]

2. For \( x = -4 \): \[ y = -7(-4) - 2 = 28 - 2 = 26 \Rightarrow (-4, 26) \]

Thus, the solutions to the system of equations are:

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

Therefore, the correct response is:

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