Solve [(x^2+y^2)-(x-y)^2+2y(x-y)]div4y=1

$$[ ( x ^ { 2 } + y ^ { 2 } ) - ( x - y ) ^ { 2 } + 2 y ( x - y ) ] \div 4 y = 1$$

Solve for x

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

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-y\right)^{2}.

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

To find the opposite of x^{2}-2xy+y^{2}, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

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

Combine y^{2} and -y^{2} to get 0.

$$\left(2xy+2y\left(x-y\right)\right)y=4$$

Use the distributive property to multiply 2y by x-y.

$$\left(2xy+2yx-2y^{2}\right)y=4$$

Combine 2xy and 2yx to get 4xy.

$$\left(4xy-2y^{2}\right)y=4$$

Use the distributive property to multiply 4xy-2y^{2} by y.

$$4xy^{2}-2y^{3}=4$$

Add 2y^{3} to both sides.

$$4xy^{2}=4+2y^{3}$$

The equation is in standard form.

$$4y^{2}x=2y^{3}+4$$

Divide both sides by 4y^{2}.

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

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

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

Divide 4+2y^{3} by 4y^{2}.

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