Solve 1^2/1*3=x+1/y+2

$$\frac{ { 1 }^{ 2 } }{ 1 \times 3 } = \frac{ x+1 }{ y+2 }$$

Solve for x

$$x=\frac{y-1}{3}$$

$$y\neq -2$$

Solution Steps

Multiply both sides of the equation by y+2.

$$\left(\frac{1}{3}y+\frac{2}{3}\right)\times 1^{2}=x+1$$

Calculate 1 to the power of 2 and get 1.

$$\left(\frac{1}{3}y+\frac{2}{3}\right)\times 1=x+1$$

Use the distributive property to multiply \frac{1}{3}y+\frac{2}{3} by 1.

$$\frac{1}{3}y+\frac{2}{3}=x+1$$

Swap sides so that all variable terms are on the left hand side.

$$x+1=\frac{1}{3}y+\frac{2}{3}$$

Subtract 1 from both sides.

$$x=\frac{1}{3}y+\frac{2}{3}-1$$

Subtract 1 from \frac{2}{3} to get -\frac{1}{3}.

$$x=\frac{1}{3}y-\frac{1}{3}$$

Solve for y

$$y=3x+1$$

$$x\neq -1$$

Steps for Solving Linear Equation

Variable y cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by y+2.

$$\left(\frac{1}{3}y+\frac{2}{3}\right)\times 1^{2}=x+1$$

Calculate 1 to the power of 2 and get 1.

$$\left(\frac{1}{3}y+\frac{2}{3}\right)\times 1=x+1$$

Use the distributive property to multiply \frac{1}{3}y+\frac{2}{3} by 1.

$$\frac{1}{3}y+\frac{2}{3}=x+1$$

Subtract \frac{2}{3} from both sides.

$$\frac{1}{3}y=x+1-\frac{2}{3}$$

Subtract \frac{2}{3} from 1 to get \frac{1}{3}.

$$\frac{1}{3}y=x+\frac{1}{3}$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}y}{\frac{1}{3}}=\frac{x+\frac{1}{3}}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$y=\frac{x+\frac{1}{3}}{\frac{1}{3}}$$

Divide x+\frac{1}{3} by \frac{1}{3} by multiplying x+\frac{1}{3} by the reciprocal of \frac{1}{3}.

$$y=3x+1$$

Variable y cannot be equal to -2.

$$y=3x+1\text{, }y\neq -2$$