Recently I was given a problem where I had a quotient with variables and a regular quotient and had to simplify it, but had no clue how to do it. For example, $$ \left|\frac{x-1}{x+3}-\frac{1}{5}\right|=\frac{4}{5}\left|\frac{x-2}{x+3}\right|$$
Can you please explain how this works? I am having trouble understanding this, any help will be appreciated. Original Image
As Nick indicates in the comment, you do this with variables the exact same way you do it without variables. Remember that variables are just place-holders for numbers (or whatever else they are varying over). They behave just like those numbers.
There is one rule for mixing multiplication/division with addition/subtraction. That is the distributive law:
$$A(B + C) = AB + AC$$
To apply this to adding fractions, you need to rewrite them into a form where it applies. If you have $$\frac ab + \frac cd$$ you don't have a common multiplier. But, you can rewrite the fractions in a form where they do. Note that $\frac AA = 1$ and multiplying by $1$ does not change values. So you can multiply the first term by $\frac dd$ and the second term by $\frac bb$: $$\begin{align}\frac ab + \frac cd &= \frac dd\frac ab + \frac bb\frac cd \\&= \frac{ad}{bd} + \frac{bc}{bd} \\&= \frac 1{bd}ad +\frac 1{bd}bc \\&= \frac{1}{bd}(ad + bc) \\&= \frac{ad+bc}{bd}\end{align}$$
In your example, $a = x-1, b = x+3, c = -1, d = 5$ $$\begin{align}\frac{x-1}{x+3}-\frac{1}{5} &= \frac 55 \frac{x-1}{x+3} + \frac{x+3}{x+3}\frac{-1}5\\ &=\frac{5x-5}{5x+15} + \frac{-x-3}{5x+15}\\ &=\frac 1{5x+15}\left((5x-5) + (-x - 3)\right)\\ &=\frac {4x-8}{5x+15}\end{align}$$
That combines the fractions, but then your calculation goes a step further, factoring the combined fraction: $$\frac {4x-8}{5x+15} = \frac{4(x-2)}{5(x+3)} = \frac 45\frac{x-2}{x+3}$$
And finally also handles the absolute values: $$\left|\frac{x-1}{x+3}-\frac15\right|=\left|\frac45\frac{x-2}{x+3}\right| = \frac45\left|\frac{x-2}{x+3}\right|$$
Of course, one normally does not write out the steps as I've done here - particularly in pulling out the denominators as a multiplication by the inverse. I choose to show it this way to tie it directly to the distributive law, so that you can see how this is justified.