Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point) Responses (5,−20) and (−7,16) left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis (−7,26) left parenthesis negative 7 comma 26 right parenthesis (5,−10) left parenthesis 5 comma negative 10 right parenthesis (5,−10) and (−7,26)

1 answer

To solve the linear-quadratic system using the substitution method, we have the following equations:

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

We can substitute the expression for \( y \) from the linear equation into the quadratic equation:

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

Now, rearranging the equation gives us:

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

Combining like terms:

\[ x^2 + 2x - 35 = 0 \]

Next, we will factor the quadratic:

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

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

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

Now, we will find the corresponding \( y \) values for each \( x \) using the linear equation \( y = -3x - 5 \).

For \( x = -7 \):

\[ y = -3(-7) - 5 = 21 - 5 = 16 \]

So, one solution is \( (-7, 16) \).

For \( x = 5 \):

\[ y = -3(5) - 5 = -15 - 5 = -20 \]

So, the other solution is \( (5, -20) \).

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

The choices given were:

  • (5, -20) and (-7, 16)

So the answer is: (5, -20) and (-7, 16).