Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?
We can substitute the expression for y from the second equation into the first equation to solve the system.
Substitute -3x - 5 for y in the first equation:
-3x - 5 = x^2 - x - 40
Rearrange the equation to get it in standard form:
x^2 - 4x - 35 = 0
Factor the quadratic equation:
(x - 7)(x + 5) = 0
Setting each factor equal to zero:
x - 7 = 0 or x + 5 = 0
x = 7 or x = -5
Now substitute these x values back into the second equation to find the corresponding y values:
y = -3(7) - 5 = -21 - 5 = -26
y = -3(-5) - 5 = 15 - 5 = 10
Therefore, the solutions to the system are (7, -26) and (-5, 10).