Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point)

Question

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 solve the linear-quadratic system using the substitution method, we can start by setting the two equations equal to each other:

x^2 - x - 40 = -3x - 5

Rearranging the equation, we have:

x^2 - x + 3x - 40 + 5 = 0

Combining like terms, we get:

x^2 + 2x - 35 = 0

Factorizing the quadratic equation, we can write it as:

(x + 7)(x - 5) = 0

Setting each factor equal to zero, we have:

x + 7 = 0 or x - 5 = 0

Solving for x, we get:

x = -7 or x = 5

Substituting these values back into one of the original equations, we can find the corresponding y-values.

For x = -7, using the equation y = -3x - 5, we have:

y = -3(-7) - 5
= 21 - 5
= 16

Therefore, one solution is (-7, 16).

For x = 5, using the same equation, we have:

y = -3(5) - 5
= -15 - 5
= -20

Therefore, another solution is (5, -20).

So the correct solution is (5, -20) and (-7, 16), which matches with the option (5,−20) and (-7,26).