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Use the substitution method to find the solutions to the linear-quadratic system y=x^2−x−40 , y=−3x−5 ?(1 point) Responses (−7,26) left parenthesis negative 7 comma 26 right parenthesis (5,−10) left parenthesis 5 comma negative 10 right parenthesis (5,−20) and (−7,16) left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis (5,−10) and (−7,26)

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To solve the linear-quadratic system given by the equations:

  1. \( y = x^2 - x - 40 \)
  2. \( y = -3x - 5 \)

you can use the substitution method.

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

\[ x^2 - x - 40 = -3x - 5 \]

Now, we rearrange the equation to set it to zero:

\[ x^2 - x + 3x - 40 + 5 = 0 \] \[ x^2 + 2x - 35 = 0 \]

Next, we can factor the quadratic equation:

\[ (x + 7)(x - 5) = 0 \]

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

  1. \( x + 7 = 0 \) \(\Rightarrow x = -7\)
  2. \( x - 5 = 0 \) \(\Rightarrow x = 5\)

Now, we substitute these values back into one of the original equations to find the corresponding values of \( y \). We'll use the linear equation \( y = -3x - 5 \).

For \( x = -7 \):

\[ y = -3(-7) - 5 \] \[ y = 21 - 5 \] \[ y = 16 \] Thus, one solution is \((-7, 16)\).

For \( x = 5 \):

\[ y = -3(5) - 5 \] \[ y = -15 - 5 \] \[ y = -20 \] Thus, the other solution is \((5, -20)\).

Therefore, the solutions to the system are \((-7, 16)\) and \((5, -20)\).

Looking at your answer choices, it seems the correct response is:

\((-7, 16)\) and \((5, -20)\).