Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40

Question

, 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)

Answer 1

To find the solutions to the system of equations \( y = x^2 - x - 40 \) and \( y = -3x - 5 \) using the substitution method, we can set the two equations equal to each other since they both equal \( y \).

1. Set the equations equal to each other:

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

2. Rearranging the equation gives:

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

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

3. Now, we will factor the quadratic:

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

4. Setting each factor to zero gives:

\[ x + 7 = 0 \implies x = -7 \]

\[ x - 5 = 0 \implies x = 5 \]

5. Now substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We can use \( y = -3x - 5 \):

• 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, another solution is \( (5, -20) \).

6. Thus, the solutions to the system are:

\[ (-7, 16) \text{ and } (5, -20) \]

Therefore, the correct responses are \( (5, -20) \) and \( (-7, 16) \).