Solution - Absolute value equations

$$\left(-9c-7\right)+9c=\left(-9c-3\right)+9c$$

Group like terms:

$$\left(-9c+9c\right)-7=\left(-9c-3\right)+9c$$

Simplify the arithmetic:

$$-7=\left(-9c-3\right)+9c$$

Group like terms:

$$-7=\left(-9c+9c\right)-3$$

Simplify the arithmetic:

$$-7=-3$$

The statement is false:

$$-7=-3$$

$$\left(-9c-7\right)=9c+3$$

Subtract $$$$ from both sides:

$$\left(-9c-7\right)-9c=\left(9c+3\right)-9c$$

Group like terms:

$$\left(-9c-9c\right)-7=\left(9c+3\right)-9c$$

Simplify the arithmetic:

$$-18c-7=\left(9c+3\right)-9c$$

Group like terms:

$$-18c-7=\left(9c-9c\right)+3$$

Simplify the arithmetic:

$$-18c-7=3$$

Add $$$$ to both sides:

$$\left(-18c-7\right)+7=3+7$$

Simplify the arithmetic:

$$-18c=3+7$$

Simplify the arithmetic:

$$-18c=10$$

Divide both sides by :

$$\frac{\left(-18c\right)}{-18}=\frac{10}{-18}$$

Cancel out the negatives:

$$\frac{18c}{18}=\frac{10}{-18}$$

Simplify the fraction:

$$c=\frac{10}{-18}$$

Move the negative sign from the denominator to the numerator:

$$c=\frac{-10}{18}$$

Find the greatest common factor of the numerator and denominator:

$$c=\frac{\left(-5·2\right)}{\left(9·2\right)}$$

Factor out and cancel the greatest common factor:

$$c=\frac{-5}{9}$$