Solution - Absolute value equations

$$\left(-y+5\right)=-y+5$$

Add $$$$ to both sides:

$$\left(-y+5\right)+y=\left(-y+5\right)+y$$

Group like terms:

$$\left(-y+y\right)+5=\left(-y+5\right)+y$$

Simplify the arithmetic:

$$5=\left(-y+5\right)+y$$

Group like terms:

$$5=\left(-y+y\right)+5$$

Simplify the arithmetic:

$$5=5$$

$$\left(-y+5\right)=y-5$$

Subtract $$$$ from both sides:

$$\left(-y+5\right)-y=\left(y-5\right)-y$$

Group like terms:

$$\left(-y-y\right)+5=\left(y-5\right)-y$$

Simplify the arithmetic:

$$-2y+5=\left(y-5\right)-y$$

Group like terms:

$$-2y+5=\left(y-y\right)-5$$

Simplify the arithmetic:

$$-2y+5=-5$$

Subtract $$$$ from both sides:

$$\left(-2y+5\right)-5=-5-5$$

Simplify the arithmetic:

$$-2y=-5-5$$

Simplify the arithmetic:

$$-2y=-10$$

Divide both sides by :

$$\frac{\left(-2y\right)}{-2}=\frac{-10}{-2}$$

Cancel out the negatives:

$$\frac{2y}{2}=\frac{-10}{-2}$$

Simplify the fraction:

$$y=\frac{-10}{-2}$$

Cancel out the negatives:

$$y=\frac{10}{2}$$

Find the greatest common factor of the numerator and denominator:

$$y=\frac{\left(5·2\right)}{\left(1·2\right)}$$

Factor out and cancel the greatest common factor:

$$y=5$$