Question

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

Answer 1

To solve this system using the substitution method, we will solve one equation for one variable and substitute it into the other equation.

We can solve the second equation for y:
y = -3x - 5

Now substitute this expression for y in the first equation:
x^2 - x - 40 = -3x - 5

Rearrange to get all terms on one side:
x^2 - x + 3x - 40 + 5 = 0

Combine like terms:
x^2 + 2x - 35 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. In this case, the quadratic factors nicely:
(x + 7)(x - 5) = 0

Setting each factor equal to zero gives us two possible solutions:
x + 7 = 0 -> x = -7
x - 5 = 0 -> x = 5

Now substitute these values of x back into one of the original equations to find the corresponding y-values.

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).

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