Solve y=xy/1+x+y=

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

Solve for y (complex solution)

$$\left\{\begin{matrix}y=0\text{, }&x\neq -1\\y\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{1+x}{1+x}.

$$y=\frac{xy}{1+x}+\frac{y\left(1+x\right)}{1+x}$$

Since \frac{xy}{1+x} and \frac{y\left(1+x\right)}{1+x} have the same denominator, add them by adding their numerators.

$$y=\frac{xy+y\left(1+x\right)}{1+x}$$

Do the multiplications in xy+y\left(1+x\right).

$$y=\frac{xy+y+xy}{1+x}$$

Combine like terms in xy+y+xy.

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

Subtract \frac{2xy+y}{1+x} from both sides.

$$y-\frac{2xy+y}{1+x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{1+x}{1+x}.

$$\frac{y\left(1+x\right)}{1+x}-\frac{2xy+y}{1+x}=0$$

Since \frac{y\left(1+x\right)}{1+x} and \frac{2xy+y}{1+x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{y\left(1+x\right)-\left(2xy+y\right)}{1+x}=0$$

Do the multiplications in y\left(1+x\right)-\left(2xy+y\right).

$$\frac{y+xy-2yx-y}{1+x}=0$$

Combine like terms in y+xy-2yx-y.

$$\frac{-xy}{1+x}=0$$

Multiply both sides of the equation by x+1.

$$-xy=0$$

The equation is in standard form.

$$\left(-x\right)y=0$$

Divide 0 by -x.

$$y=0$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\neq -1\text{, }&y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$y\left(x+1\right)=xy+\left(x+1\right)y$$

Use the distributive property to multiply y by x+1.

$$yx+y=xy+\left(x+1\right)y$$

Use the distributive property to multiply x+1 by y.

$$yx+y=xy+xy+y$$

Combine xy and xy to get 2xy.

$$yx+y=2xy+y$$

Subtract 2xy from both sides.

$$yx+y-2xy=y$$

Combine yx and -2xy to get -yx.

$$-yx+y=y$$

Subtract y from both sides.

$$-yx=y-y$$

Combine y and -y to get 0.

$$-yx=0$$

The equation is in standard form.

$$\left(-y\right)x=0$$

Divide 0 by -y.

$$x=0$$

Solve for y

$$\left\{\begin{matrix}y=0\text{, }&x\neq -1\\y\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{1+x}{1+x}.

$$y=\frac{xy}{1+x}+\frac{y\left(1+x\right)}{1+x}$$

Since \frac{xy}{1+x} and \frac{y\left(1+x\right)}{1+x} have the same denominator, add them by adding their numerators.

$$y=\frac{xy+y\left(1+x\right)}{1+x}$$

Do the multiplications in xy+y\left(1+x\right).

$$y=\frac{xy+y+xy}{1+x}$$

Combine like terms in xy+y+xy.

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

Subtract \frac{2xy+y}{1+x} from both sides.

$$y-\frac{2xy+y}{1+x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{1+x}{1+x}.

$$\frac{y\left(1+x\right)}{1+x}-\frac{2xy+y}{1+x}=0$$

Since \frac{y\left(1+x\right)}{1+x} and \frac{2xy+y}{1+x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{y\left(1+x\right)-\left(2xy+y\right)}{1+x}=0$$

Do the multiplications in y\left(1+x\right)-\left(2xy+y\right).

$$\frac{y+xy-2yx-y}{1+x}=0$$

Combine like terms in y+xy-2yx-y.

$$\frac{-xy}{1+x}=0$$

Multiply both sides of the equation by x+1.

$$-xy=0$$

The equation is in standard form.

$$\left(-x\right)y=0$$

Divide 0 by -x.

$$y=0$$