Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to add two compound fractions with fractions in numerator like this one:

$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$

or fractions with fractions in denominator like this one:

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

share|cite|improve this question
"fractions in denominator" - for those you reciprocate the denominator. – J. M. Jul 14 '11 at 10:25
It might be better to use complicated instead of complex. – Henry Jul 14 '11 at 22:02
Or "compound fractions". – Américo Tavares Jul 14 '11 at 22:04
I changed "complex fractions" to "compound fractions" in the title. – Américo Tavares Jul 15 '11 at 7:46
up vote 6 down vote accepted

The multiplicative inverse of a fraction a/b is b/a. (Wikipedia)

Let us start with the properties:

  • Division by a number or fraction is the same as multiplication by its inverse or reciprocal.

    Division by $r$ is equal to the multiplication by $\dfrac{1}{r}$: $$\dfrac{\ \dfrac{p}{q}\ }{r}=\dfrac{p}{q}\cdot \dfrac{1}{r}=\dfrac{p\cdot 1}{q\cdot r}=\dfrac{p}{q r}, \quad (1)$$

    Division by $\dfrac{t}{u}$ is equal to the multiplication by $\dfrac{u}{t}$:

$$\dfrac{\ s}{\ \dfrac{t}{u}\ }=s\cdot\dfrac{u}{t}=\dfrac{s\cdot u}{t}=\dfrac{su}{t}.\quad (2)$$

  • Sum of fractions

$$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}.\quad (3)$$

Apply $(1)$ to

$$\dfrac{\ \dfrac{1}{x}\ }{2}=\dfrac{1}{x}\cdot\dfrac{1}{2}=\dfrac{1\cdot 1}{x\cdot 2}=\dfrac{1}{2x}$$

and $(2)$ to

$$\frac{\ \dfrac{2}{3x}\ }{x}=\dfrac{2}{3x}\cdot\dfrac{1}{x}=\dfrac{2\cdot 1}{3x\cdot x}=\dfrac{2}{3x^2}.$$

So by $(3)$ we have

$$\dfrac{\ \dfrac{1}{x}\ }{2} + \dfrac{\ \dfrac{2}{3x}\ }{x}=\dfrac{1}{2x}+\dfrac{2}{3x^2}=\dfrac{1\cdot 3x^2+2\times 2x}{2x\cdot 3x^2 }=\dfrac{3x^2+4x}{6x^3}=\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}.$$


$$\dfrac{x}{\; \dfrac{2}{x}\ } + \dfrac{\; \dfrac{1}{x}\; }{x}$$

we have

$$\dfrac{\; x\; }{\dfrac{2}{x}} + \dfrac{\; \dfrac{1}{x}\; }{x}=\dfrac{x\cdot x}{2} + \dfrac{1}{x\cdot x}=\dfrac{x^2}{2}+\cfrac{1}{x^2}=\dfrac{x^2\cdot x^2+2\cdot 1}{2\cdot x^2}=\dfrac{x^4+2}{2x^2}.$$

We can apply the property Division by a fraction is the same as multiplication by its inverse or reciprocal to the following fraction

$$\dfrac{\;\dfrac{a}{b}\;}{\dfrac{c}{d}}=\dfrac{a}{b}\cdot \dfrac{d}{c}=\dfrac{a\cdot d}{b\cdot c}=\dfrac{ad}{bc}\qquad (4).$$

share|cite|improve this answer
You can clarify compound fractions in LaTeX by adding dummy space in either denominator or numerator; e.g., \frac{\ \frac{a}{b}\ }{\frac{c}{d}} vs. \frac{\frac{a}{b}}{\frac{c}{d}} to get $$\frac{\ \frac{a}{b}\ }{\frac{c}{d}}\quad\mathrm{vs.}\quad \frac{\frac{a}{b}}{\frac{c}{d}}.$$This might be useful in some of your expressions. – Arturo Magidin Jul 14 '11 at 21:08
@Arturo: Thanks! I didn't know. – Américo Tavares Jul 14 '11 at 21:10
I improved formatting of compound fractions, as per Arturo Magini's advice. – Américo Tavares Jul 14 '11 at 21:32
@Arturo: done. (filling text) – Américo Tavares Jul 14 '11 at 21:35
The simplification above of $$\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}$$ is valid if and only if $x\ne 0$. – Américo Tavares Jul 14 '11 at 22:02

One easy way to figure this out is that dividing by $x$ is the same as multiplying by $1/x$ (but all bets are off when $x=0$, as division by $0$ is undefined). So

$$ \begin{align*} \frac{ \frac{a}{b} }{c} &= \frac{1}{c} \frac{a}{b} = \frac{a}{bc} \\ \\ \frac{a}{\frac{b}{c}} &= a \frac{1}{\frac{b}{c}} = a \frac{c}{b} = \frac{ac}{b} \end{align*} $$

share|cite|improve this answer
And yet generations of students keep insisting on ambiguously stacking fractions... – Simon Jul 14 '11 at 10:52
I'd like to point out, that, however, $x= \frac{1}{\frac{1}{x}}$ only when $x\not=0$. – lentic catachresis Jul 14 '11 at 14:27
@Bruno: Well noted. The solution has been updated to reflect that. – Michael Chen Jul 14 '11 at 14:38

Here is a start for the first one:

$$\frac{\frac{1}{x}}{2} + \frac{\frac{2}{3x}}{x} = \frac{x\frac{1}{x}}{2x} + \frac{2\frac{2}{3x}}{2x} = \frac{1}{2x} + \frac{\frac{4}{3x}}{2x} = \frac{1}{2x} + \frac{4}{3x}\frac{1}{2x} = \frac{1}{2x} + \frac{4}{6x^2} = \frac{1}{2x} + \frac{2}{3x^2}$$

Now try to derive $\displaystyle\frac{3x + 4}{6x^2}$ from this.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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