When we can change the sign of denominator

Question

Suppose $z=\frac{-x_1}{x_2-x_3}$, find $-z$.

Which one is correct $$-z=\frac{x_1}{x_2-x_3}\ \ \ \text{or}\ \ \ -z=\frac{x_1}{-x_2+x_3}$$

Answer 1

We have $z=\frac{-x_1}{x_2-x_3}, \ -z = (-1)\cdot \frac{-x_1}{x_2-x_3}$$$-z = \frac{x_1}{x_2-x_3} =\big(\frac{-1}{1}\big)\cdot \frac{-x_1}{x_2-x_3} = \big(\frac{1}{-1}\big)\frac{-x_1}{x_2-x_3} = \frac{-x_1}{x_3-x_2}$$

Notice that in one of your proposed solutions above you would find the following: $$\frac{x_1}{x_3-x_2} = \frac{(-1)\cdot (-x_1)}{(-1)\cdot (x_2-x_3)} = \frac{-x_1}{x_2-x_3} = z$$ You may only apply the negative factor to either the numerator or the denominator, but not both because they will negate eachother.