0
$\begingroup$

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: enter image description here

$\endgroup$
  • 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
1
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.