Solve x-1/2-1=y-3/-5-3

$$\frac { x - 1 } { 2 - 1 } = \frac { y - 3 } { - 5 - 3 }$$

Solve for x

$$x=\frac{11-y}{8}$$

Solution Steps

Subtract 1 from 2 to get 1.

$$\frac{x-1}{1}=\frac{y-3}{-5-3}$$

Anything divided by one gives itself.

$$x-1=\frac{y-3}{-5-3}$$

Subtract 3 from -5 to get -8.

$$x-1=\frac{y-3}{-8}$$

Multiply both numerator and denominator by -1.

$$x-1=\frac{-y+3}{8}$$

Divide each term of -y+3 by 8 to get -\frac{1}{8}y+\frac{3}{8}.

$$x-1=-\frac{1}{8}y+\frac{3}{8}$$

Add 1 to both sides.

$$x=-\frac{1}{8}y+\frac{3}{8}+1$$

Add \frac{3}{8} and 1 to get \frac{11}{8}.

$$x=-\frac{1}{8}y+\frac{11}{8}$$

Solve for y

$$y=11-8x$$

Steps for Solving Linear Equation

Subtract 1 from 2 to get 1.

$$\frac{x-1}{1}=\frac{y-3}{-5-3}$$

Anything divided by one gives itself.

$$x-1=\frac{y-3}{-5-3}$$

Subtract 3 from -5 to get -8.

$$x-1=\frac{y-3}{-8}$$

Multiply both numerator and denominator by -1.

$$x-1=\frac{-y+3}{8}$$

Divide each term of -y+3 by 8 to get -\frac{1}{8}y+\frac{3}{8}.

$$x-1=-\frac{1}{8}y+\frac{3}{8}$$

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

$$-\frac{1}{8}y+\frac{3}{8}=x-1$$

Subtract \frac{3}{8} from both sides.

$$-\frac{1}{8}y=x-1-\frac{3}{8}$$

Subtract \frac{3}{8} from -1 to get -\frac{11}{8}.

$$-\frac{1}{8}y=x-\frac{11}{8}$$

Multiply both sides by -8.

$$\frac{-\frac{1}{8}y}{-\frac{1}{8}}=\frac{x-\frac{11}{8}}{-\frac{1}{8}}$$

Dividing by -\frac{1}{8} undoes the multiplication by -\frac{1}{8}.

$$y=\frac{x-\frac{11}{8}}{-\frac{1}{8}}$$

Divide x-\frac{11}{8} by -\frac{1}{8} by multiplying x-\frac{11}{8} by the reciprocal of -\frac{1}{8}.

$$y=11-8x$$