Solve +4y^2-8xy-2x+1=0

$$+ 4 y ^ { 2 } - 8 x y - 2 x + 1 = 0$$

Solve for x

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

$$y\neq -\frac{1}{4}$$

Steps for Solving Linear Equation

Subtract 4y^{2} from both sides. Anything subtracted from zero gives its negation.

$$-8xy-2x+1=-4y^{2}$$

Subtract 1 from both sides.

$$-8xy-2x=-4y^{2}-1$$

Combine all terms containing x.

$$\left(-8y-2\right)x=-4y^{2}-1$$

Divide both sides by -8y-2.

$$\frac{\left(-8y-2\right)x}{-8y-2}=\frac{-4y^{2}-1}{-8y-2}$$

Dividing by -8y-2 undoes the multiplication by -8y-2.

$$x=\frac{-4y^{2}-1}{-8y-2}$$

Divide -4y^{2}-1 by -8y-2.

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

Solve for y (complex solution)

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

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

Solve for y

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

$$y=-\frac{\sqrt{4x^{2}+2x-1}}{2}+x\text{, }x\geq \frac{\sqrt{5}-1}{4}\text{ or }x\leq \frac{-\sqrt{5}-1}{4}$$