Solution - Absolute value equations

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

$$\frac{3}{4}=\frac{1}{2}$$

The statement is false:

$$\frac{3}{4}=\frac{1}{2}$$

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

Add $$$$ to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

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

Subtract $$$$ from both sides:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

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

Simplify the arithmetic:

$$2x=\left(\frac{-1}{2}\right)-\frac{3}{4}$$

Find the lowest common denominator:

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

Multiply the denominators:

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

Multiply the numerators:

$$2x=\frac{-2}{4}+\frac{-3}{4}$$

Combine the fractions:

$$2x=\frac{\left(-2-3\right)}{4}$$

Combine the numerators:

$$2x=\frac{-5}{4}$$

Divide both sides by :

$$\frac{\left(2x\right)}{2}=\frac{\left(\frac{-5}{4}\right)}{2}$$

Simplify the fraction:

$$x=\frac{\left(\frac{-5}{4}\right)}{2}$$

Simplify the arithmetic:

$$x=\frac{-5}{\left(4·2\right)}$$

$$x=\frac{-5}{8}$$