# How to subtract a fraction where a fraction is part of the denominator already?

I am to subtract two fractions:

$$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$$

I can see the answer in my text book being just x, the only in between line I'm shown is:

$$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$$ =

$$x + 2 - 2$$ = x

I'm trying to arrive at this myself and got lost:

$$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$$

Least common denominator (lcd) is the product of the two denominators, in this case $$1(\frac{1}{x}+2)$$

So, the left side is already using the lcd so I only need to transform the right side by multiplying by $$\frac{1}{x}+2$$. I'll do this to both the numerator and denominator to keep them in proportion:

$$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$$ =

$$\frac{1}{\frac{1}{x+2}} - \frac{-2(\frac{1}{x+2})}{\frac{1}{x+2}}$$

Now they both have the same denominator so I can subtract one from the other:

$$\frac{1-2(\frac{1}{x}+2)}{\frac{1}{x+2}}$$

This is as far as I can take it.

How can I arrive at $$x + 2 - 2$$ = x?

Seeking baby, more granular steps if possible.

Screen shot of the question in full in case I've missed anything or made typos:

• Try multiplying the top and bottom of $\frac{1}{\frac{1}{x}+2}$ by $x$. – 79037662 Oct 18 '19 at 15:07
• $\frac{1}{x}+2 \neq \frac{1}{x+2}$. – Eric Towers Oct 18 '19 at 15:07
• It might help to simplify that denominator ($\frac{1}{x}+2$) first, since it itself involves fractions. You should be able to find that $\frac{1}{x}+2 = \frac{1+2x}{x}$, so $\frac{1}{\frac{1}{x}+2} = \frac{x}{1+2x}$. Now see if you can proceed. Edit: Based on the fact that the answer is $x$, it looks like your original fraction involved $\frac{1}{x+2}$, not $\frac{1}{x}+2$ as you wrote (these are different). To simplify $\frac{1}{\frac{1}{x+2}}$, recall the rule $\frac{1}{\frac{1}{y}}=y$. – Minus One-Twelfth Oct 18 '19 at 15:09

Method 1:

Whenever I see "small denominators" (nested fractions), I try to clear the small denominators.

\begin{align*} \frac{1}{\frac{1}{x}+2} - \frac{2}{1} &= \frac{1}{\frac{1}{x}+2} \cdot \underbrace{\frac{x}{x}}_{1} - \frac{2}{1} \\ &= \frac{x}{1+2x} - \frac{2}{1} \\ &= \frac{x}{1+2x} - \frac{2}{1} \cdot \underbrace{\frac{1+2x}{1+2x}}_{1} \\ &= \frac{x}{1+2x} - \frac{2+4x}{1+2x} \\ &= \frac{x - (2+4x)}{1+2x} \\ &= \frac{-2-3x}{1+2x} \text{.} \end{align*}

Method 2:

Don't try to be tricky with the common denominator. \begin{align*} \frac{1}{\frac{1}{x}+2} - \frac{2}{1} &= \frac{1}{\frac{1}{x}+2} - \frac{2}{1} \cdot \underbrace{\frac{\frac{1}{x}+2}{\frac{1}{x}+2}}_{1} \\ &= \frac{1 - 2(\frac{1}{x}+2)}{\frac{1}{x}+2} \\ &= \frac{-3 - \frac{2}{x}}{\frac{1}{x}+2} \cdot \underbrace{\frac{x}{x}}_{1} \\ &= \frac{-3x - 2}{1+2x} \text{.} \end{align*}

I notice that neither of these is "$$x$$". I suspect there is a typo' in the presented problem. Let's start from $$\frac{1}{\frac{1}{x+2}}$$ instead. \begin{align*} \frac{1}{\frac{1}{x+2}} - \frac{2}{1} &= \frac{1}{\frac{1}{x+2}} \cdot \underbrace{\frac{x+2}{x+2}}_{1} - \frac{2}{1} \\ &= \frac{x+2}{1} - \frac{2}{1} \\ &= \frac{x+2 - 2}{1} \\ &= \frac{x}{1} \\ &= x \text{.} \end{align*}

• Hi, thank you I am going to look over this closely. Meantime I added a screen shot of the question to my post in case I have typos – Doug Fir Oct 18 '19 at 15:24

The best first step is to change $$\frac{1}{\frac{1}{x}+2}$$ into $$\frac{x}{2x+1}$$.

I'm sure you can finish this off yourself now.

• Why / how did you know to do this? What made you recognise that the best thing to do here is to start with this? – Doug Fir Oct 18 '19 at 15:21
• Because it is virtually always best not to have a fraction in the denominator. – S. Dolan Oct 18 '19 at 15:23