Solve y-(-3)/-5-3=x-3/-4-3

$$\frac { y - ( - 3 ) } { - 5 - 3 } = \frac { x - 3 } { - 4 - 3 }$$

Solve for x

$$x=\frac{7y+45}{8}$$

Steps for Solving Linear Equation

The opposite of -3 is 3.

$$\frac{y+3}{-5-3}=\frac{x-3}{-4-3}$$

Subtract 3 from -5 to get -8.

$$\frac{y+3}{-8}=\frac{x-3}{-4-3}$$

Multiply both numerator and denominator by -1.

$$\frac{-y-3}{8}=\frac{x-3}{-4-3}$$

Subtract 3 from -4 to get -7.

$$\frac{-y-3}{8}=\frac{x-3}{-7}$$

Multiply both numerator and denominator by -1.

$$\frac{-y-3}{8}=\frac{-x+3}{7}$$

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

$$-\frac{1}{8}y-\frac{3}{8}=\frac{-x+3}{7}$$

Divide each term of -x+3 by 7 to get -\frac{1}{7}x+\frac{3}{7}.

$$-\frac{1}{8}y-\frac{3}{8}=-\frac{1}{7}x+\frac{3}{7}$$

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

$$-\frac{1}{7}x+\frac{3}{7}=-\frac{1}{8}y-\frac{3}{8}$$

Subtract \frac{3}{7} from both sides.

$$-\frac{1}{7}x=-\frac{1}{8}y-\frac{3}{8}-\frac{3}{7}$$

Subtract \frac{3}{7} from -\frac{3}{8} to get -\frac{45}{56}.

$$-\frac{1}{7}x=-\frac{1}{8}y-\frac{45}{56}$$

The equation is in standard form.

$$-\frac{1}{7}x=-\frac{y}{8}-\frac{45}{56}$$

Multiply both sides by -7.

$$\frac{-\frac{1}{7}x}{-\frac{1}{7}}=\frac{-\frac{y}{8}-\frac{45}{56}}{-\frac{1}{7}}$$

Dividing by -\frac{1}{7} undoes the multiplication by -\frac{1}{7}.

$$x=\frac{-\frac{y}{8}-\frac{45}{56}}{-\frac{1}{7}}$$

Divide -\frac{y}{8}-\frac{45}{56} by -\frac{1}{7} by multiplying -\frac{y}{8}-\frac{45}{56} by the reciprocal of -\frac{1}{7}.

$$x=\frac{7y+45}{8}$$

Solve for y

$$y=\frac{8x-45}{7}$$

Steps for Solving Linear Equation

The opposite of -3 is 3.

$$\frac{y+3}{-5-3}=\frac{x-3}{-4-3}$$

Subtract 3 from -5 to get -8.

$$\frac{y+3}{-8}=\frac{x-3}{-4-3}$$

Multiply both numerator and denominator by -1.

$$\frac{-y-3}{8}=\frac{x-3}{-4-3}$$

Subtract 3 from -4 to get -7.

$$\frac{-y-3}{8}=\frac{x-3}{-7}$$

Multiply both numerator and denominator by -1.

$$\frac{-y-3}{8}=\frac{-x+3}{7}$$

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

$$-\frac{1}{8}y-\frac{3}{8}=\frac{-x+3}{7}$$

Divide each term of -x+3 by 7 to get -\frac{1}{7}x+\frac{3}{7}.

$$-\frac{1}{8}y-\frac{3}{8}=-\frac{1}{7}x+\frac{3}{7}$$

Add \frac{3}{8} to both sides.

$$-\frac{1}{8}y=-\frac{1}{7}x+\frac{3}{7}+\frac{3}{8}$$

Add \frac{3}{7} and \frac{3}{8} to get \frac{45}{56}.

$$-\frac{1}{8}y=-\frac{1}{7}x+\frac{45}{56}$$

The equation is in standard form.

$$-\frac{1}{8}y=-\frac{x}{7}+\frac{45}{56}$$

Multiply both sides by -8.

$$\frac{-\frac{1}{8}y}{-\frac{1}{8}}=\frac{-\frac{x}{7}+\frac{45}{56}}{-\frac{1}{8}}$$

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

$$y=\frac{-\frac{x}{7}+\frac{45}{56}}{-\frac{1}{8}}$$

Divide -\frac{x}{7}+\frac{45}{56} by -\frac{1}{8} by multiplying -\frac{x}{7}+\frac{45}{56} by the reciprocal of -\frac{1}{8}.

$$y=\frac{8x-45}{7}$$