5
$\begingroup$

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:

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

$\endgroup$
5
  • 1
    $\begingroup$ "fractions in denominator" - for those you reciprocate the denominator. $\endgroup$ Jul 14, 2011 at 10:25
  • 2
    $\begingroup$ It might be better to use complicated instead of complex. $\endgroup$
    – Henry
    Jul 14, 2011 at 22:02
  • 5
    $\begingroup$ Or "compound fractions". $\endgroup$ Jul 14, 2011 at 22:04
  • $\begingroup$ I changed "complex fractions" to "compound fractions" in the title. $\endgroup$ Jul 15, 2011 at 7:46
  • $\begingroup$ Here's a lesson that discusses this topic in detail and offers many practice examples: lem.ma/LR $\endgroup$
    – Lemma
    Nov 13, 2017 at 21:40

4 Answers 4

8
$\begingroup$

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

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

$\endgroup$
3
  • 3
    $\begingroup$ And yet generations of students keep insisting on ambiguously stacking fractions... $\endgroup$
    – Simon
    Jul 14, 2011 at 10:52
  • 1
    $\begingroup$ I'd like to point out, that, however, $x= \frac{1}{\frac{1}{x}}$ only when $x\not=0$. $\endgroup$ Jul 14, 2011 at 14:27
  • $\begingroup$ @Bruno: Well noted. The solution has been updated to reflect that. $\endgroup$ Jul 14, 2011 at 14:38
7
$\begingroup$

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

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

For

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

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

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

$\endgroup$
5
  • 6
    $\begingroup$ 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. $\endgroup$ Jul 14, 2011 at 21:08
  • $\begingroup$ @Arturo: Thanks! I didn't know. $\endgroup$ Jul 14, 2011 at 21:10
  • $\begingroup$ I improved formatting of compound fractions, as per Arturo Magini's advice. $\endgroup$ Jul 14, 2011 at 21:32
  • 1
    $\begingroup$ @Arturo: done. (filling text) $\endgroup$ Jul 14, 2011 at 21:35
  • 1
    $\begingroup$ The simplification above of $$\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}$$ is valid if and only if $x\ne 0$. $\endgroup$ Jul 14, 2011 at 22:02
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Yet another strategy: \begin{align} \frac{\frac1x}2+\frac{\frac2{3x}}x&=\frac1{2x}+\frac2{3x^2}\\ &=\frac{3x}{6x^2}+\frac4{6x^2}=\frac{3x+4}{6x^2}\,. \end{align} What did I do? Given is the sum of two fractions, and I multiplied top-and-bottom of the first by $x$, and top-and-bottom of the second by $3x$. Second step, find the minimal common denominator, which is $6x^2$, and on each of your current fractions, multiply top-and-bottom by a suitable quantity to get the denominators equal. Now add.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .