Solve 18x^2-3yx+y=0

$$18 { x }^{ 2 } -3yx+y = 0$$

Solve for y

$$y=-\frac{18x^{2}}{1-3x}$$

$$x\neq \frac{1}{3}$$

Steps for Solving Linear Equation

Subtract 18x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-3yx+y=-18x^{2}$$

Combine all terms containing y.

$$\left(-3x+1\right)y=-18x^{2}$$

The equation is in standard form.

$$\left(1-3x\right)y=-18x^{2}$$

Divide both sides by -3x+1.

$$\frac{\left(1-3x\right)y}{1-3x}=-\frac{18x^{2}}{1-3x}$$

Dividing by -3x+1 undoes the multiplication by -3x+1.

$$y=-\frac{18x^{2}}{1-3x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y\left(y-8\right)}+y}{12}$$

$$x=\frac{-\sqrt{y\left(y-8\right)}+y}{12}$$

Solve for x

$$x=\frac{\sqrt{y\left(y-8\right)}+y}{12}$$

$$x=\frac{-\sqrt{y\left(y-8\right)}+y}{12}\text{, }y\geq 8\text{ or }y\leq 0$$