Solution - Absolute value equations

$$\left(x+\frac{3}{5}\right)-x=\left(x+\frac{-7}{3}\right)-x$$

Group like terms:

$$\left(x-x\right)+\frac{3}{5}=\left(x+\frac{-7}{3}\right)-x$$

Simplify the arithmetic:

$$\frac{3}{5}=\left(x+\frac{-7}{3}\right)-x$$

Group like terms:

$$\frac{3}{5}=\left(x-x\right)+\frac{-7}{3}$$

Simplify the arithmetic:

$$\frac{3}{5}=\frac{-7}{3}$$

The statement is false:

$$\frac{3}{5}=\frac{-7}{3}$$

$$\left(x+\frac{3}{5}\right)=-x+\frac{7}{3}$$

Add $$$$ to both sides:

$$\left(x+\frac{3}{5}\right)+x=\left(-x+\frac{7}{3}\right)+x$$

Group like terms:

$$\left(x+x\right)+\frac{3}{5}=\left(-x+\frac{7}{3}\right)+x$$

Simplify the arithmetic:

$$2x+\frac{3}{5}=\left(-x+\frac{7}{3}\right)+x$$

Group like terms:

$$2x+\frac{3}{5}=\left(-x+x\right)+\frac{7}{3}$$

Simplify the arithmetic:

$$2x+\frac{3}{5}=\frac{7}{3}$$

Subtract $$$$ from both sides:

$$\left(2x+\frac{3}{5}\right)-\frac{3}{5}=\left(\frac{7}{3}\right)-\frac{3}{5}$$

Combine the fractions:

$$2x+\frac{\left(3-3\right)}{5}=\left(\frac{7}{3}\right)-\frac{3}{5}$$

Combine the numerators:

$$2x+\frac{0}{5}=\left(\frac{7}{3}\right)-\frac{3}{5}$$

Reduce the zero numerator:

$$2x+0=\left(\frac{7}{3}\right)-\frac{3}{5}$$

Simplify the arithmetic:

$$2x=\left(\frac{7}{3}\right)-\frac{3}{5}$$

Find the lowest common denominator:

$$2x=\frac{\left(7·5\right)}{\left(3·5\right)}+\frac{\left(-3·3\right)}{\left(5·3\right)}$$

Multiply the denominators:

$$2x=\frac{\left(7·5\right)}{15}+\frac{\left(-3·3\right)}{15}$$

Multiply the numerators:

$$2x=\frac{35}{15}+\frac{-9}{15}$$

Combine the fractions:

$$2x=\frac{\left(35-9\right)}{15}$$

Combine the numerators:

$$2x=\frac{26}{15}$$

Divide both sides by :

$$\frac{\left(2x\right)}{2}=\frac{\left(\frac{26}{15}\right)}{2}$$

Simplify the fraction:

$$x=\frac{\left(\frac{26}{15}\right)}{2}$$

Simplify the arithmetic:

$$x=\frac{26}{\left(15·2\right)}$$

$$x=\frac{13}{15}$$