Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point)
To solve the system using the substitution method, we can start by solving one of the equations for one variable and then substitute it into the other equation.
Let's solve the second equation for y:
y = -3x - 5
Now substitute this expression for y in the first equation:
-3x - 5 = x^2 - x - 40
Rearrange this equation to be in standard quadratic form:
x^2 - x + 3x - 5 + 40 = 0
x^2 + 2x + 35 = 0
Next, we can try factoring or using the quadratic formula to solve for x. However, this quadratic equation does not factor and the discriminant is negative, so we cannot find real roots.
Therefore, there are no solutions to this linear-quadratic system.