Solve y-p/3-0=x-2/-1-2

$$\frac { y - p } { 3 - 0 } = \frac { x - 2 } { - 1 - 2 }$$

Solve for p

$$p=x+y-2$$

Steps for Solving Linear Equation

Subtract 0 from 3 to get 3.

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

Subtract 2 from -1 to get -3.

$$\frac{y-p}{3}=\frac{x-2}{-3}$$

Multiply both numerator and denominator by -1.

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

Divide each term of y-p by 3 to get \frac{1}{3}y-\frac{1}{3}p.

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

Divide each term of -x+2 by 3 to get -\frac{1}{3}x+\frac{2}{3}.

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

Subtract \frac{1}{3}y from both sides.

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

The equation is in standard form.

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

Multiply both sides by -3.

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

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

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

Divide \frac{-x+2-y}{3} by -\frac{1}{3} by multiplying \frac{-x+2-y}{3} by the reciprocal of -\frac{1}{3}.

$$p=x+y-2$$

Solve for x

$$x=2+p-y$$

Steps for Solving Linear Equation

Subtract 0 from 3 to get 3.

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

Subtract 2 from -1 to get -3.

$$\frac{y-p}{3}=\frac{x-2}{-3}$$

Multiply both numerator and denominator by -1.

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

Divide each term of y-p by 3 to get \frac{1}{3}y-\frac{1}{3}p.

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

Divide each term of -x+2 by 3 to get -\frac{1}{3}x+\frac{2}{3}.

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

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

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

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

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

The equation is in standard form.

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

Multiply both sides by -3.

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

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

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

Divide \frac{-2+y-p}{3} by -\frac{1}{3} by multiplying \frac{-2+y-p}{3} by the reciprocal of -\frac{1}{3}.

$$x=2+p-y$$