Solution - Absolute value equations

$$\left(-z+9\right)=-z-1$$

Add $$$$ to both sides:

$$\left(-z+9\right)+z=\left(-z-1\right)+z$$

Group like terms:

$$\left(-z+z\right)+9=\left(-z-1\right)+z$$

Simplify the arithmetic:

$$9=\left(-z-1\right)+z$$

Group like terms:

$$9=\left(-z+z\right)-1$$

Simplify the arithmetic:

$$9=-1$$

The statement is false:

$$9=-1$$

$$\left(-z+9\right)=z+1$$

Subtract $$$$ from both sides:

$$\left(-z+9\right)-z=\left(z+1\right)-z$$

Group like terms:

$$\left(-z-z\right)+9=\left(z+1\right)-z$$

Simplify the arithmetic:

$$-2z+9=\left(z+1\right)-z$$

Group like terms:

$$-2z+9=\left(z-z\right)+1$$

Simplify the arithmetic:

$$-2z+9=1$$

Subtract $$$$ from both sides:

$$\left(-2z+9\right)-9=1-9$$

Simplify the arithmetic:

$$-2z=1-9$$

Simplify the arithmetic:

$$-2z=-8$$

Divide both sides by :

$$\frac{\left(-2z\right)}{-2}=\frac{-8}{-2}$$

Cancel out the negatives:

$$\frac{2z}{2}=\frac{-8}{-2}$$

Simplify the fraction:

$$z=\frac{-8}{-2}$$

Cancel out the negatives:

$$z=\frac{8}{2}$$

Find the greatest common factor of the numerator and denominator:

$$z=\frac{\left(4·2\right)}{\left(1·2\right)}$$

Factor out and cancel the greatest common factor:

$$z=4$$