Solve y-2/3-2=x-4/1-4

$$\frac { y - 2 } { 3 - 2 } = \frac { x - 4 } { 1 - 4 }$$

Solve for x

$$x=10-3y$$

Steps for Solving Linear Equation

Subtract 2 from 3 to get 1.

$$\frac{y-2}{1}=\frac{x-4}{1-4}$$

Anything divided by one gives itself.

$$y-2=\frac{x-4}{1-4}$$

Subtract 4 from 1 to get -3.

$$y-2=\frac{x-4}{-3}$$

Multiply both numerator and denominator by -1.

$$y-2=\frac{-x+4}{3}$$

Divide each term of -x+4 by 3 to get -\frac{1}{3}x+\frac{4}{3}.

$$y-2=-\frac{1}{3}x+\frac{4}{3}$$

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

$$-\frac{1}{3}x+\frac{4}{3}=y-2$$

Subtract \frac{4}{3} from both sides.

$$-\frac{1}{3}x=y-2-\frac{4}{3}$$

Subtract \frac{4}{3} from -2 to get -\frac{10}{3}.

$$-\frac{1}{3}x=y-\frac{10}{3}$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}x}{-\frac{1}{3}}=\frac{y-\frac{10}{3}}{-\frac{1}{3}}$$

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

$$x=\frac{y-\frac{10}{3}}{-\frac{1}{3}}$$

Divide y-\frac{10}{3} by -\frac{1}{3} by multiplying y-\frac{10}{3} by the reciprocal of -\frac{1}{3}.

$$x=10-3y$$

Solve for y

$$y=\frac{10-x}{3}$$

Solution Steps

Subtract 2 from 3 to get 1.

$$\frac{y-2}{1}=\frac{x-4}{1-4}$$

Anything divided by one gives itself.

$$y-2=\frac{x-4}{1-4}$$

Subtract 4 from 1 to get -3.

$$y-2=\frac{x-4}{-3}$$

Multiply both numerator and denominator by -1.

$$y-2=\frac{-x+4}{3}$$

Divide each term of -x+4 by 3 to get -\frac{1}{3}x+\frac{4}{3}.

$$y-2=-\frac{1}{3}x+\frac{4}{3}$$

Add 2 to both sides.

$$y=-\frac{1}{3}x+\frac{4}{3}+2$$

Add \frac{4}{3} and 2 to get \frac{10}{3}.

$$y=-\frac{1}{3}x+\frac{10}{3}$$