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 using the elimination method, we have the following two equations:

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

Since both equations are equal to \( y \), we can set them equal to each other:

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

Now, let's rearrange this equation to get all terms on one side:

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

Combining like terms, we get:

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

Next, we can factor this quadratic equation:

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

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

\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

Now that we have the \( x \)-values, we can substitute them back into either original equation to find the corresponding \( y \)-values. Let's use \( y = -7x - 2 \).

  1. For \( x = -3 \): \[ y = -7(-3) - 2 = 21 - 2 = 19 \] So, one solution is \( (-3, 19) \).

  2. For \( x = -4 \): \[ y = -7(-4) - 2 = 28 - 2 = 26 \] So, another solution is \( (-4, 26) \).

Therefore, the solutions to the system are:

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

The correct response is:

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