Quadratic-liner system of equations, MATH B!!!?

Question

i have no idea how to do this any help please:

xy=12

y= 2x+2

Answer 1

12 / x = 2 x + 2
12 = 2 x ² + 2 x
x ² + x - 6 = 0
( x + 3 ) ( x - 2 ) = 0
x = - 3 , x = 2
y = - 4 , y = 6

Solution is (- 3 , - 4) , (2 , 6)

Answer 2

I'm assuming you want to solve for the variables.
Substitutes y =2x + 2 into xy = 12

x(2x + 2) = 12
2x^2 + 2x = 12
2x^2 + 2x -12 = 0
2(x^2 + x -6) = 0
2(x +3)(x-2) = 0

So x+3 = 0 and x-2 = 0
Therefore x = -3 and x = 2

Plug both values into xy=12 to solve for y
(-3)y = 12
y = -4

and

(2)y = 12
y = 6

So your intersection points are at (-3, -4) and (2, 6)

Answer 3

substitute y = 2x +2 = 2(x+1) to the other equation
xy = 12
x 2(x+1) = 12
x (x+1) = 6
x^2 + x = 6
x^2 + x - 6 = 0
(x+3)(x-2) = 0
x = - 3 . . . . . and .. . x = 2