# Describe to me what is happening here (Pre-algebra) I get confused on the second step. Can someone describe to me what is happening?

If I want to solve this, what should the first thing I look at in my head such that I get the correct answer in the simplest way like in the picture.

I absolutely hate fractions to an unimaginable, unparalleled extent

I don't know how to get to the second step, I should add. What process involved?

• Can you simplify $x-(x-4)$? Jul 11, 2014 at 14:07
• Serious question: why are people interpreting 'the second step' as the first equality? It's very counter-intuitive to me. Jul 11, 2014 at 14:09
• Hate is not good for your karma... Jul 11, 2014 at 14:09
• @GitGud maybe the other seemed too obvious, at least to me... Jul 11, 2014 at 14:11
• FYI, in the U.S. this wouldn't be pre-algebra. In fact, it would be one of the more advanced topics in a first year high school algebra course, and a standard exercise in a second year high school algebra course. Jul 11, 2014 at 14:31

Mostly :

$$\frac{1}{\color{red}{a}} - \frac{1}{\color{blue}{b}} = \frac{\color{blue}{b}-\color{red}{a}}{\color{red}{a}\color{blue}{b}}$$

Taking $\color{red}{a}=x-4$ and $\color{blue}{b}=x$ leads to :

$$\frac{1}{x-4}-\frac{1}{x} = \frac{x-(x-4)}{x(x-4)}$$

Now, in the fraction $\displaystyle \frac{x-(x-4)}{x(x-4)}$, the numerator simplifies to $4$. As a consequence :

$$\frac{1}{x-4}-\frac{1}{x}=\frac{4}{x(x-4)}$$

• I sort of understand you the most, but what what if the numerator was not a 1? I noticed you only dealt with the denominator hence b - a / ab and nothing to do with the numerator? @jibounet Jul 11, 2014 at 14:14
• Here, the numerator is, for both fractions, equal to $1$. If the numerator was not equal to $1$, the idea would be the same : we want both fractions to have the same denominator. If we want to simplify $\displaystyle \frac{a}{\color{red}{b}}+\frac{c}{\color{blue}{d}}$, here is what we would do : $$\frac{a}{\color{red}{b}} + \frac{c}{\color{blue}{d}} = \frac{a \color{blue}{\times d}}{\color{red}{b} \color{blue}{\times d}} + \frac{c \color{red}{\times b}}{\color{blue}{d} \color{red}{\times b}} = \frac{ad+bc}{bd}$$ Jul 11, 2014 at 14:19

Revert the step you don't understand: $\frac{x-(x-4)}{x(x-4)}=\frac x {x(x-4)}-\cdots$ and cancel appropriately...

The two fractions on the left are being combined by finding their common denominator:

$$\frac{1}{x-4} - \frac 1x = \frac {x\cdot 1}{x(x-4)} - \frac{(1)(x-4)}{x(x-4)} = \dfrac{x - (x-4)}{x(x-4)} = \frac 4{x(x-4)}$$