To find the solutions to the linear-quadratic system using the substitution method, we will substitute the value of y in the first equation with the corresponding equation in the second equation.
Given equations:
1) y = x^2 - x - 40
2) y = -3x - 5
Substituting y in equation 1 with -3x - 5 from equation 2:
-3x - 5 = x^2 - x - 40
Rearrange this equation to get a quadratic equation:
x^2 - x + 3x - 5 + 40 = 0
x^2 + 2x + 35 = 0
This quadratic equation does not factorize nicely, so we will use the quadratic formula to find the values of x:
x = (-2 ± sqrt(2^2 - 4(1)(35))) / 2(1)
x = (-2 ± sqrt(-136)) / 2
x = (-2 ± 2i√(34)) / 2
x = -1 ± i√(34)
Therefore, the solutions to the system of equations are:
(-7, 26) and (5, -10)