Solve 4y-3/5n-4=5/3x+20/3

$$4 y - \frac { 3 } { 5 } n - 4 = \frac { 5 } { 3 } x + \frac { 20 } { 3 }$$

Solve for n

$$n=\frac{20y}{3}-\frac{25x}{9}-\frac{160}{9}$$

Steps for Solving Linear Equation

Subtract 4y from both sides.

$$-\frac{3}{5}n-4=\frac{5}{3}x+\frac{20}{3}-4y$$

Add 4 to both sides.

$$-\frac{3}{5}n=\frac{5}{3}x+\frac{20}{3}-4y+4$$

Add \frac{20}{3} and 4 to get \frac{32}{3}.

$$-\frac{3}{5}n=\frac{5}{3}x+\frac{32}{3}-4y$$

The equation is in standard form.

$$-\frac{3}{5}n=\frac{5x}{3}-4y+\frac{32}{3}$$

Divide both sides of the equation by -\frac{3}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{5}n}{-\frac{3}{5}}=\frac{\frac{5x}{3}-4y+\frac{32}{3}}{-\frac{3}{5}}$$

Dividing by -\frac{3}{5} undoes the multiplication by -\frac{3}{5}.

$$n=\frac{\frac{5x}{3}-4y+\frac{32}{3}}{-\frac{3}{5}}$$

Divide \frac{5x}{3}+\frac{32}{3}-4y by -\frac{3}{5} by multiplying \frac{5x}{3}+\frac{32}{3}-4y by the reciprocal of -\frac{3}{5}.

$$n=\frac{20y}{3}-\frac{25x}{9}-\frac{160}{9}$$

Solve for x

$$x=\frac{12y}{5}-\frac{9n}{25}-\frac{32}{5}$$

Steps for Solving Linear Equation

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

$$\frac{5}{3}x+\frac{20}{3}=4y-\frac{3}{5}n-4$$

Subtract \frac{20}{3} from both sides.

$$\frac{5}{3}x=4y-\frac{3}{5}n-4-\frac{20}{3}$$

Subtract \frac{20}{3} from -4 to get -\frac{32}{3}.

$$\frac{5}{3}x=4y-\frac{3}{5}n-\frac{32}{3}$$

The equation is in standard form.

$$\frac{5}{3}x=-\frac{3n}{5}+4y-\frac{32}{3}$$

Divide both sides of the equation by \frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{3}x}{\frac{5}{3}}=\frac{-\frac{3n}{5}+4y-\frac{32}{3}}{\frac{5}{3}}$$

Dividing by \frac{5}{3} undoes the multiplication by \frac{5}{3}.

$$x=\frac{-\frac{3n}{5}+4y-\frac{32}{3}}{\frac{5}{3}}$$

Divide 4y-\frac{3n}{5}-\frac{32}{3} by \frac{5}{3} by multiplying 4y-\frac{3n}{5}-\frac{32}{3} by the reciprocal of \frac{5}{3}.

$$x=\frac{12y}{5}-\frac{9n}{25}-\frac{32}{5}$$