Solve 2xsqrt{y}-6sqrt{x}=1

$$2x \sqrt{ y } -6 \sqrt{ x } =1$$

Solve for y

$$y=\frac{\left(\frac{6}{\sqrt{x}}+\frac{1}{x}\right)^{2}}{4}$$

$$x>0$$

Steps for Solving Radical Equations

Subtract -6\sqrt{x} from both sides of the equation.

$$2x\sqrt{y}-6\sqrt{x}-\left(-6\sqrt{x}\right)=1-\left(-6\sqrt{x}\right)$$

Subtracting -6\sqrt{x} from itself leaves 0.

$$2x\sqrt{y}=1-\left(-6\sqrt{x}\right)$$

Subtract -6\sqrt{x} from 1.

$$2x\sqrt{y}=6\sqrt{x}+1$$

Divide both sides by 2x.

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

Dividing by 2x undoes the multiplication by 2x.

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

Divide 1+6\sqrt{x} by 2x.

$$\sqrt{y}=\frac{3}{\sqrt{x}}+\frac{1}{2x}$$

Square both sides of the equation.

$$y=\frac{\left(6\sqrt{x}+1\right)^{2}}{4x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\left(\sqrt{2\sqrt{y}+9}+3\right)^{2}}{4y}\text{, }&arg(y^{-\frac{1}{2}}\left(\sqrt{2\sqrt{y}+9}+3\right))<\pi \text{ and }y\neq 0\\x=\frac{\left(\sqrt{2\sqrt{y}+9}-3\right)^{2}}{4y}\text{, }&arg(y^{-\frac{1}{2}}\times \frac{\sqrt{2\sqrt{y}+9}-3}{2})\geq \pi \text{ and }y\neq 0\end{matrix}\right.$$

Solve for y (complex solution)

$$y=\frac{\left(6x^{-\frac{1}{2}}+\frac{1}{x}\right)^{2}}{4}$$

$$x\neq 0\text{ and }arg(-3x^{-\frac{1}{2}}-\frac{1}{2x})\geq \pi$$

Solve for x

$$x=\frac{\left(\sqrt{2\sqrt{y}+9}+3\right)^{2}}{4y}$$

$$y>0$$