Use the substitution method to determine whether the linear-quadratic system y=x^2−2x−6, y=−5x−9 has 0, 1, or 2 solutions.
To determine the number of solutions, we will substitute the second equation into the first equation.
Substituting y = -5x - 9 into the equation y = x^2 - 2x - 6, we get:
-5x - 9 = x^2 - 2x - 6
Rearranging this equation, we have:
x^2 - 2x + 5x + 3 = 0
x^2 + 3x + 3 = 0
Now, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 3, and c = 3.
x = (-3 ± √(3^2 - 4(1)(3))) / (2(1))
x = (-3 ± √(9 - 12)) / 2
x = (-3 ± √(-3)) / 2
Since we have a negative value underneath the square root, this equation has no real solutions. Therefore, the linear-quadratic system y = x^2 - 2x - 6, y = -5x - 9 has 0 solutions.