Solution - Absolute value equations

$$\left(-n+7\right)=-n+5$$

Add $$$$ to both sides:

$$\left(-n+7\right)+n=\left(-n+5\right)+n$$

Group like terms:

$$\left(-n+n\right)+7=\left(-n+5\right)+n$$

Simplify the arithmetic:

$$7=\left(-n+5\right)+n$$

Group like terms:

$$7=\left(-n+n\right)+5$$

Simplify the arithmetic:

$$7=5$$

The statement is false:

$$7=5$$

$$\left(-n+7\right)=n-5$$

Subtract $$$$ from both sides:

$$\left(-n+7\right)-n=\left(n-5\right)-n$$

Group like terms:

$$\left(-n-n\right)+7=\left(n-5\right)-n$$

Simplify the arithmetic:

$$-2n+7=\left(n-5\right)-n$$

Group like terms:

$$-2n+7=\left(n-n\right)-5$$

Simplify the arithmetic:

$$-2n+7=-5$$

Subtract $$$$ from both sides:

$$\left(-2n+7\right)-7=-5-7$$

Simplify the arithmetic:

$$-2n=-5-7$$

Simplify the arithmetic:

$$-2n=-12$$

Divide both sides by :

$$\frac{\left(-2n\right)}{-2}=\frac{-12}{-2}$$

Cancel out the negatives:

$$\frac{2n}{2}=\frac{-12}{-2}$$

Simplify the fraction:

$$n=\frac{-12}{-2}$$

Cancel out the negatives:

$$n=\frac{12}{2}$$

Find the greatest common factor of the numerator and denominator:

$$n=\frac{\left(6·2\right)}{\left(1·2\right)}$$

Factor out and cancel the greatest common factor:

$$n=6$$