Solve 4xy^2-3x-3y^2=1

$$4x { y }^{ 2 } -3x-3 { y }^{ 2 } =1$$

Solve for x (complex solution)

$$x=\frac{3y^{2}+1}{4y^{2}-3}$$

$$y\neq -\frac{\sqrt{3}}{2}\text{ and }y\neq \frac{\sqrt{3}}{2}$$

Steps for Solving Linear Equation

Add 3y^{2} to both sides.

$$4xy^{2}-3x=1+3y^{2}$$

Combine all terms containing x.

$$\left(4y^{2}-3\right)x=1+3y^{2}$$

The equation is in standard form.

$$\left(4y^{2}-3\right)x=3y^{2}+1$$

Divide both sides by 4y^{2}-3.

$$\frac{\left(4y^{2}-3\right)x}{4y^{2}-3}=\frac{3y^{2}+1}{4y^{2}-3}$$

Dividing by 4y^{2}-3 undoes the multiplication by 4y^{2}-3.

$$x=\frac{3y^{2}+1}{4y^{2}-3}$$

Solve for x

$$x=\frac{3y^{2}+1}{4y^{2}-3}$$

$$|y|\neq \frac{\sqrt{3}}{2}$$

Steps for Solving Linear Equation

Add 3y^{2} to both sides.

$$4xy^{2}-3x=1+3y^{2}$$

Combine all terms containing x.

$$\left(4y^{2}-3\right)x=1+3y^{2}$$

The equation is in standard form.

$$\left(4y^{2}-3\right)x=3y^{2}+1$$

Divide both sides by 4y^{2}-3.

$$\frac{\left(4y^{2}-3\right)x}{4y^{2}-3}=\frac{3y^{2}+1}{4y^{2}-3}$$

Dividing by 4y^{2}-3 undoes the multiplication by 4y^{2}-3.

$$x=\frac{3y^{2}+1}{4y^{2}-3}$$

Solve for y (complex solution)

$$y=-i\left(4x-3\right)^{-\frac{1}{2}}\sqrt{-3x-1}$$

$$y=i\left(4x-3\right)^{-\frac{1}{2}}\sqrt{-3x-1}\text{, }x\neq \frac{3}{4}$$

Solve for y

$$y=\sqrt{\frac{3x+1}{4x-3}}$$

$$y=-\sqrt{\frac{3x+1}{4x-3}}\text{, }x\leq -\frac{1}{3}\text{ or }x>\frac{3}{4}$$