Solve x/5+1/3/frac{x{5}-1/3}=frac{frac{x}{3}+frac{1}{15}}{frac{x}{3}-frac{4}{5}}

$$\frac { \frac { x } { 5 } + \frac { 1 } { 3 } } { \frac { x } { 5 } - \frac { 1 } { 3 } } = \frac { \frac { x } { 3 } + \frac { 1 } { 15 } } { \frac { x } { 3 } - \frac { 4 } { 5 } }$$

Solve for x

$$x=5$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5 and 3 is 15. Multiply \frac{x}{5} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{5}{5}.

$$\frac{\frac{3x}{15}+\frac{5}{15}}{\frac{x}{5}-\frac{1}{3}}=\frac{\frac{x}{3}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

Since \frac{3x}{15} and \frac{5}{15} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3x+5}{15}}{\frac{x}{5}-\frac{1}{3}}=\frac{\frac{x}{3}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5 and 3 is 15. Multiply \frac{x}{5} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{5}{5}.

$$\frac{\frac{3x+5}{15}}{\frac{3x}{15}-\frac{5}{15}}=\frac{\frac{x}{3}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

Since \frac{3x}{15} and \frac{5}{15} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3x+5}{15}}{\frac{3x-5}{15}}=\frac{\frac{x}{3}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

Divide \frac{3x+5}{15} by \frac{3x-5}{15} by multiplying \frac{3x+5}{15} by the reciprocal of \frac{3x-5}{15}.

$$\frac{\left(3x+5\right)\times 15}{15\left(3x-5\right)}=\frac{\frac{x}{3}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

Cancel out 15 in both numerator and denominator.

$$\frac{3x+5}{3x-5}=\frac{\frac{x}{3}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 15 is 15. Multiply \frac{x}{3} times \frac{5}{5}.

$$\frac{3x+5}{3x-5}=\frac{\frac{5x}{15}+\frac{1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

Since \frac{5x}{15} and \frac{1}{15} have the same denominator, add them by adding their numerators.

$$\frac{3x+5}{3x-5}=\frac{\frac{5x+1}{15}}{\frac{x}{3}-\frac{4}{5}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 5 is 15. Multiply \frac{x}{3} times \frac{5}{5}. Multiply \frac{4}{5} times \frac{3}{3}.

$$\frac{3x+5}{3x-5}=\frac{\frac{5x+1}{15}}{\frac{5x}{15}-\frac{4\times 3}{15}}$$

Since \frac{5x}{15} and \frac{4\times 3}{15} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x+5}{3x-5}=\frac{\frac{5x+1}{15}}{\frac{5x-4\times 3}{15}}$$

Do the multiplications in 5x-4\times 3.

$$\frac{3x+5}{3x-5}=\frac{\frac{5x+1}{15}}{\frac{5x-12}{15}}$$

Divide \frac{5x+1}{15} by \frac{5x-12}{15} by multiplying \frac{5x+1}{15} by the reciprocal of \frac{5x-12}{15}.

$$\frac{3x+5}{3x-5}=\frac{\left(5x+1\right)\times 15}{15\left(5x-12\right)}$$

Cancel out 15 in both numerator and denominator.

$$\frac{3x+5}{3x-5}=\frac{5x+1}{5x-12}$$

Subtract \frac{5x+1}{5x-12} from both sides.

$$\frac{3x+5}{3x-5}-\frac{5x+1}{5x-12}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3x-5 and 5x-12 is \left(3x-5\right)\left(5x-12\right). Multiply \frac{3x+5}{3x-5} times \frac{5x-12}{5x-12}. Multiply \frac{5x+1}{5x-12} times \frac{3x-5}{3x-5}.

$$\frac{\left(3x+5\right)\left(5x-12\right)}{\left(3x-5\right)\left(5x-12\right)}-\frac{\left(5x+1\right)\left(3x-5\right)}{\left(3x-5\right)\left(5x-12\right)}=0$$

Since \frac{\left(3x+5\right)\left(5x-12\right)}{\left(3x-5\right)\left(5x-12\right)} and \frac{\left(5x+1\right)\left(3x-5\right)}{\left(3x-5\right)\left(5x-12\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3x+5\right)\left(5x-12\right)-\left(5x+1\right)\left(3x-5\right)}{\left(3x-5\right)\left(5x-12\right)}=0$$

Do the multiplications in \left(3x+5\right)\left(5x-12\right)-\left(5x+1\right)\left(3x-5\right).

$$\frac{15x^{2}-36x+25x-60-15x^{2}+25x-3x+5}{\left(3x-5\right)\left(5x-12\right)}=0$$

Combine like terms in 15x^{2}-36x+25x-60-15x^{2}+25x-3x+5.

$$\frac{11x-55}{\left(3x-5\right)\left(5x-12\right)}=0$$

Variable x cannot be equal to any of the values \frac{5}{3},\frac{12}{5} since division by zero is not defined. Multiply both sides of the equation by \left(3x-5\right)\left(5x-12\right).

$$11x-55=0$$

Add 55 to both sides. Anything plus zero gives itself.

$$11x=55$$

Divide both sides by 11.

$$x=\frac{55}{11}$$

Divide 55 by 11 to get 5.

$$x=5$$