 Marla Payton

2021-12-16

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

Expert

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
$\frac{\frac{a}{b}}{c}=\frac{1}{c}\frac{a}{b}=\frac{a}{bc}$
$\frac{a}{\frac{b}{c}}=a\frac{1}{\frac{b}{c}}=a\frac{c}{b}=\frac{ac}{b}$ Dabanka4v

Expert

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 $\frac{1}{r}$
$\frac{\frac{p}{q}}{r}=\frac{p}{q}\cdot \frac{1}{r}=\frac{p\cdot 1}{q\cdot r}=\frac{p}{qr},\left(1\right)$
Division by $\frac{t}{u}$ is equal to the multiplication by $\frac{u}{t}:$
$\frac{s}{\frac{t}{u}}=s\cdot \frac{u}{t}=\frac{s\cdot u}{t}=\frac{su}{t}.\left(2\right)$
Sum of fractions
$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.\left(3\right)$
Apply (1) to
$\frac{\frac{1}{x}}{2}=\frac{1}{x}\cdot \frac{1}{2}=\frac{1\cdot 1}{x\cdot 2}=\frac{1}{2x}$
and (2) to
$\frac{\frac{2}{3x}}{x}=\frac{2}{3x}\cdot \frac{1}{x}=\frac{2\cdot 1}{3x\cdot x}=\frac{2}{3{x}^{2}}$
So by (3) we have
$\frac{\frac{1}{x}}{2}+\frac{\frac{2}{3x}}{x}=\frac{1}{2x}+\frac{2}{3{x}^{2}}=\frac{1\cdot 3{x}^{2}+2×2x}{2x\cdot 3{x}^{2}}=\frac{3{x}^{2}+4x}{6{x}^{3}}$
$=\frac{x\left(3x+4\right)}{x\left(6{x}^{2}\right)}=\frac{3x+4}{6{x}^{2}}$
For
$\frac{x}{\frac{2}{x}}+\frac{\frac{1}{x}}{x}$
We have
$\frac{x}{\frac{2}{x}}+\frac{\frac{1}{x}}{x}=\frac{1}{x\cdot x}=\frac{{x}^{2}}{2}+\frac{1}{{x}^{2}}=\frac{{x}^{2}\cdot {x}^{2}+2\cdot 1}{2\cdot {x}^{2}}=\frac{{x}^{4}+2}{2{x}^{2}}$
We can apply the property Division by a fraction is the same as multiplication by its inverse or reciprocal to the following fraction RizerMix

Expert

Here is a start for the first one:
$\frac{\frac{1}{x}}{2}+\frac{\frac{2}{3x}}{x}=\frac{x\frac{1}{x}}{2x}+\frac{2\frac{2}{3x}}{2x}=\frac{1}{2x}+\frac{\frac{4}{3x}}{2x}=\frac{1}{2x}+\frac{4}{3x}\frac{1}{2x}=\frac{1}{2x}+\frac{4}{6{x}^{2}}=$
$=\frac{1}{2x}+\frac{2}{3{x}^{2}}$
Now try to derive $\frac{3x+4}{6{x}^{2}}$ from this.

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