y = x2 - 6 is Equation 1
8x + y = -18 is Equation 2
And we want to know where the line crosses the parabola
Since we already have one equation as
y = x2 - 6
We can easily try a substitution.
Substitute x2 - 6 into Equation 2 and solve for x
8x + x2 - 6 = -18
Add 18 to both sides of the equation
8x + x2 - 6 + 18 = 0
Combine like terms
8x + x2 + 12 = 0
We have a quadratic
x2 + 8x + 12 =0
This can be factored
(x + 6)(x +2)=0
x = -6
x = -2
Now we can substitute these values back into y = x2 - 6 and solve for y
y = (-6)2 - 6 = 36 - 6 = 30 and we have (-6, 30)
For the next value of x
y = x2 - 6
y = (-2)2 - 6 = 4 - 6 = -2 we have (-2, -2)
We can check both these values by plugging back into Equation 2
8x + y = -18
8(-6) + 30 = -18
-48 + 30 = -18
-18 = -18
For the next pair
8x + y = -18
8(-2) + (-2) = -18
(-16) + (- 2) = -18
-16 - 2 = -18
-18 = -18
So the line crosses the parabola at two points (-6, 30) and (-2, -2)
You might want to check that on graphing calculator or computer
I hope you find this useful and if you have any questions please send me an email.