I am to subtract two fractions:


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:


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:


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: enter image description here

  • 2
    $\begingroup$ Try multiplying the top and bottom of $\frac{1}{\frac{1}{x}+2}$ by $x$. $\endgroup$ – 79037662 Oct 18 '19 at 15:07
  • $\begingroup$ $\frac{1}{x}+2 \neq \frac{1}{x+2}$. $\endgroup$ – Eric Towers Oct 18 '19 at 15:07
  • $\begingroup$ 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 $\frac{1}{\frac{1}{y}}=y$. $\endgroup$ – 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*}

  • $\begingroup$ 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 $\endgroup$ – 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.

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

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