Solution - Absolute value equations

$$\left(r+\frac{3}{4}\right)-r=\left(r+\frac{-7}{8}\right)-r$$

Group like terms:

$$\left(r-r\right)+\frac{3}{4}=\left(r+\frac{-7}{8}\right)-r$$

Simplify the arithmetic:

$$\frac{3}{4}=\left(r+\frac{-7}{8}\right)-r$$

Group like terms:

$$\frac{3}{4}=\left(r-r\right)+\frac{-7}{8}$$

Simplify the arithmetic:

$$\frac{3}{4}=\frac{-7}{8}$$

The statement is false:

$$\frac{3}{4}=\frac{-7}{8}$$

$$\left(r+\frac{3}{4}\right)=-r+\frac{7}{8}$$

Add $$$$ to both sides:

$$\left(r+\frac{3}{4}\right)+r=\left(-r+\frac{7}{8}\right)+r$$

Group like terms:

$$\left(r+r\right)+\frac{3}{4}=\left(-r+\frac{7}{8}\right)+r$$

Simplify the arithmetic:

$$2r+\frac{3}{4}=\left(-r+\frac{7}{8}\right)+r$$

Group like terms:

$$2r+\frac{3}{4}=\left(-r+r\right)+\frac{7}{8}$$

Simplify the arithmetic:

$$2r+\frac{3}{4}=\frac{7}{8}$$

Subtract $$$$ from both sides:

$$\left(2r+\frac{3}{4}\right)-\frac{3}{4}=\left(\frac{7}{8}\right)-\frac{3}{4}$$

Combine the fractions:

$$2r+\frac{\left(3-3\right)}{4}=\left(\frac{7}{8}\right)-\frac{3}{4}$$

Combine the numerators:

$$2r+\frac{0}{4}=\left(\frac{7}{8}\right)-\frac{3}{4}$$

Reduce the zero numerator:

$$2r+0=\left(\frac{7}{8}\right)-\frac{3}{4}$$

Simplify the arithmetic:

$$2r=\left(\frac{7}{8}\right)-\frac{3}{4}$$

Find the lowest common denominator:

$$2r=\frac{7}{8}+\frac{\left(-3·2\right)}{\left(4·2\right)}$$

Multiply the denominators:

$$2r=\frac{7}{8}+\frac{\left(-3·2\right)}{8}$$

Multiply the numerators:

$$2r=\frac{7}{8}+\frac{-6}{8}$$

Combine the fractions:

$$2r=\frac{\left(7-6\right)}{8}$$

Combine the numerators:

$$2r=\frac{1}{8}$$

Divide both sides by :

$$\frac{\left(2r\right)}{2}=\frac{\left(\frac{1}{8}\right)}{2}$$

Simplify the fraction:

$$r=\frac{\left(\frac{1}{8}\right)}{2}$$

Simplify the arithmetic:

$$r=\frac{1}{\left(8·2\right)}$$

$$r=\frac{1}{16}$$