Question

9. Find the solutions of the system algebraically.
y = x^2 − 2x − 6
y= 4x + 10

A. (2, -2) and (8, 42)
B. (-2, 2) and (-8, -42)
C. (-2, 2) and (8, 42)
D. no solution

Answer 1

To find the solutions for the system algebraically, we can set the two equations equal to each other:

x^2 − 2x − 6 = 4x + 10

Rearranging the terms, we get:

x^2 - 6x - 16 = 0

Now, we need to solve this quadratic equation. We can factor it or use the quadratic formula:

x = [6 ± sqrt((-6)^2 - 4(1)(-16))] / 2(1)
x = [6 ± sqrt(36 + 64)] / 2
x = [6 ± sqrt(100)] / 2
x = [6 ± 10] / 2

So, x = (6 + 10) / 2 or x = (6 - 10) / 2
x = 16 / 2 or x = -4 / 2
x = 8 or x = -2

Now, we can substitute these values of x back into one of the original equations to find the corresponding values of y:

For x = 8:
y = (8)^2 − 2(8) − 6
y = 64 - 16 - 6
y = 42

For x = -2:
y = (-2)^2 − 2(-2) − 6
y = 4 + 4 - 6
y = 2

So, the solutions are (8, 42) and (-2, 2), which matches option C: (-2, 2) and (8, 42).