Solve 1-x/y=x+y/2

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

Solve for x

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

$$y\neq -2\text{ and }y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2y, the least common multiple of y,2.

$$2\left(1-x\right)=y\left(x+y\right)$$

Use the distributive property to multiply 2 by 1-x.

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

Use the distributive property to multiply y by x+y.

$$2-2x=yx+y^{2}$$

Subtract yx from both sides.

$$2-2x-yx=y^{2}$$

Subtract 2 from both sides.

$$-2x-yx=y^{2}-2$$

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by -y-2.

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

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

$$x=\frac{y^{2}-2}{-y-2}$$

Divide y^{2}-2 by -y-2.

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

Solve for y (complex solution)

$$\left\{\begin{matrix}\\y=\frac{-\sqrt{x^{2}-8x+8}-x}{2}\text{, }&\text{unconditionally}\\y=\frac{\sqrt{x^{2}-8x+8}-x}{2}\text{, }&x\neq 1\end{matrix}\right.$$

Solve for y

$$\left\{\begin{matrix}y=\frac{\sqrt{x^{2}-8x+8}-x}{2}\text{, }&x\geq 2\sqrt{2}+4\text{ or }\left(x\neq 1\text{ and }x\leq 4-2\sqrt{2}\right)\\y=\frac{-\sqrt{x^{2}-8x+8}-x}{2}\text{, }&x\geq 2\sqrt{2}+4\text{ or }x\leq 4-2\sqrt{2}\end{matrix}\right.$$