Solve z=y^2+6xy-3y

$$z= { y }^{ 2 } +6xy-3y$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{y}{6}+\frac{z}{6y}+\frac{1}{2}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&z=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$y^{2}+6xy-3y=z$$

Subtract y^{2} from both sides.

$$6xy-3y=z-y^{2}$$

Add 3y to both sides.

$$6xy=z-y^{2}+3y$$

The equation is in standard form.

$$6yx=z+3y-y^{2}$$

Divide both sides by 6y.

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

Dividing by 6y undoes the multiplication by 6y.

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

Divide z-y^{2}+3y by 6y.

$$x=-\frac{y}{6}+\frac{z}{6y}+\frac{1}{2}$$

Solve for y

$$y=-\frac{\sqrt{36x^{2}-36x+4z+9}}{2}-3x+\frac{3}{2}$$

$$y=\frac{\sqrt{36x^{2}-36x+4z+9}}{2}-3x+\frac{3}{2}\text{, }z\geq -\frac{9\left(2x-1\right)^{2}}{4}$$