Solution - Absolute value equations

$$\left(y+\frac{-4}{3}\right)-y=\left(y+\frac{1}{6}\right)-y$$

Group like terms:

$$\left(y-y\right)+\frac{-4}{3}=\left(y+\frac{1}{6}\right)-y$$

Simplify the arithmetic:

$$\frac{-4}{3}=\left(y+\frac{1}{6}\right)-y$$

Group like terms:

$$\frac{-4}{3}=\left(y-y\right)+\frac{1}{6}$$

Simplify the arithmetic:

$$\frac{-4}{3}=\frac{1}{6}$$

The statement is false:

$$\frac{-4}{3}=\frac{1}{6}$$

$$\left(y+\frac{-4}{3}\right)=-y+\frac{-1}{6}$$

Add $$$$ to both sides:

$$\left(y+\frac{-4}{3}\right)+y=\left(-y+\frac{-1}{6}\right)+y$$

Group like terms:

$$\left(y+y\right)+\frac{-4}{3}=\left(-y+\frac{-1}{6}\right)+y$$

Simplify the arithmetic:

$$2y+\frac{-4}{3}=\left(-y+\frac{-1}{6}\right)+y$$

Group like terms:

$$2y+\frac{-4}{3}=\left(-y+y\right)+\frac{-1}{6}$$

Simplify the arithmetic:

$$2y+\frac{-4}{3}=\frac{-1}{6}$$

Add $$$$ to both sides:

$$\left(2y+\frac{-4}{3}\right)+\frac{4}{3}=\left(\frac{-1}{6}\right)+\frac{4}{3}$$

Combine the fractions:

$$2y+\frac{\left(-4+4\right)}{3}=\left(\frac{-1}{6}\right)+\frac{4}{3}$$

Combine the numerators:

$$2y+\frac{0}{3}=\left(\frac{-1}{6}\right)+\frac{4}{3}$$

Reduce the zero numerator:

$$2y+0=\left(\frac{-1}{6}\right)+\frac{4}{3}$$

Simplify the arithmetic:

$$2y=\left(\frac{-1}{6}\right)+\frac{4}{3}$$

Find the lowest common denominator:

$$2y=\frac{-1}{6}+\frac{\left(4·2\right)}{\left(3·2\right)}$$

Multiply the denominators:

$$2y=\frac{-1}{6}+\frac{\left(4·2\right)}{6}$$

Multiply the numerators:

$$2y=\frac{-1}{6}+\frac{8}{6}$$

Combine the fractions:

$$2y=\frac{\left(-1+8\right)}{6}$$

Combine the numerators:

$$2y=\frac{7}{6}$$

Divide both sides by :

$$\frac{\left(2y\right)}{2}=\frac{\left(\frac{7}{6}\right)}{2}$$

Simplify the fraction:

$$y=\frac{\left(\frac{7}{6}\right)}{2}$$

Simplify the arithmetic:

$$y=\frac{7}{\left(6·2\right)}$$

$$y=\frac{7}{12}$$