Solve 8y-8xy-8y^2=1

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

Solve for x

$$x=-y+1-\frac{1}{8y}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Subtract 8y from both sides.

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

Add 8y^{2} to both sides.

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

The equation is in standard form.

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

Divide both sides by -8y.

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

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

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

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

$$x=-y+1-\frac{1}{8y}$$

Solve for y (complex solution)

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

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

Solve for y

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

$$y=\frac{\sqrt{4x^{2}-8x+2}}{4}-\frac{x}{2}+\frac{1}{2}\text{, }x\geq \frac{\sqrt{2}}{2}+1\text{ or }x\leq -\frac{\sqrt{2}}{2}+1$$