Solve 6x^2+6x+6=3x^2+yx+8

$$6 x ^ { 2 } + 6 x + 6 = 3 x ^ { 2 } + y x + 8$$

Solve for y

$$y=3x+6-\frac{2}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+yx+8=6x^{2}+6x+6$$

Subtract 3x^{2} from both sides.

$$yx+8=6x^{2}+6x+6-3x^{2}$$

Combine 6x^{2} and -3x^{2} to get 3x^{2}.

$$yx+8=3x^{2}+6x+6$$

Subtract 8 from both sides.

$$yx=3x^{2}+6x+6-8$$

Subtract 8 from 6 to get -2.

$$yx=3x^{2}+6x-2$$

The equation is in standard form.

$$xy=3x^{2}+6x-2$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{3x^{2}+6x-2}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{3x^{2}+6x-2}{x}$$

Divide 3x^{2}+6x-2 by x.

$$y=3x+6-\frac{2}{x}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-12y+60}+y-6}{6}$$

$$x=\frac{-\sqrt{y^{2}-12y+60}+y-6}{6}$$