QuestionUse the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point)Responses(5,−10) and (−7,26)left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis(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)
To solve the linear-quadratic system using the substitution method, we need to solve one equation for one variable and substitute it into the other equation.
We can start by solving the second equation for y:
y = -3x - 5
Next, we substitute this expression for y in the first equation:
x^2 - x - 40 = -3x - 5
Now, we can simplify this equation and solve for x:
x^2 - x + 3x - 35 = 0
x^2 +2x - 35 = 0
This quadratic equation can be factored as:
(x + 7)(x - 5) = 0
Setting each factor equal to zero, we find x = -7 and x = 5.
Substituting these values back into y = -3x - 5, we can find the corresponding y-values:
For x = -7, y = -3(-7) - 5 = 21 - 5 = 16
For x = 5, y = -3(5) - 5 = -15 - 5 = -20
So, the solutions to the linear-quadratic system are: (5, -20) and (-7, 16).