Complex Fractions

A complex fraction is a fraction with a numerator or denominator that also contains a fraction. As an example, the following is a complex fraction:

$$\frac{2}{3-\frac{1}{2}}.$$

The easiest way to simplify complex fractions is to first express (if possible) the numerator and the denominator as fractions.

For example,

$$\frac{\frac{2}{1}}{\hspace{2mm} \frac{6-1}{2}\hspace{2mm} }.$$

Now, we can rewrite the fractions side-by-side and simplify using division:

$$\begin{aligned} \frac{2}{1} \div \frac{6-1}{2} &= \frac{2}{1} \times \frac{2}{5}\\\\ &= \frac{4}{5}. \end{aligned}$$

Simplify

$$\frac{\frac{4}{3}}{\frac{2}{3}+1} .$$

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Answer

Multiplying both the numerator and denominator by $3,$ which is the common denominator of $\frac{4}{3}$ and $\frac{2}{3},$ we have

$$\begin{aligned} \frac{\frac{4}{3}}{\frac{2}{3}+1} &=\frac{3\cdot \frac{4}{3}}{3\cdot \frac{2}{3} + 3 \cdot 1} \\ &= \frac{4}{2+3} \\ &= \frac{4}{5}.\ _\square \end{aligned}$$

We can use the same strategy from above for solving complex fractions with variables.

Simplify

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

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ANSWER

We have

$$\begin{aligned} 1+\frac{1+ \frac{1+\frac{1}{x}}{x}}{x} &= 1+ \frac{1+\frac{\hspace{2mm} \frac{x+1}{x}\hspace{2mm} }{x}}{x} \\ &= 1+\frac{1+\frac{x+1}{x^2}}{x} \\ &= 1+\frac{\hspace{2mm} \frac{x^2+x+1}{x^2}\hspace{2mm} }{x} \\ &= 1+ \frac{x^2+x+1}{x^3} \\ &= \frac{x^3+x^2+x+1}{x^3}.\ _\square \end{aligned}$$

Simplify

$$\large \frac { \frac { x+1 }{ x-1 } +\frac { x-1 }{ x+1 } }{ \frac { x+1 }{ x-1 } -\frac { x-1 }{ x+1 } }.$$

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ANSWER

We have

$$\begin{aligned} \frac{ \frac { x+1 }{ x-1 } +\frac { x-1 }{ x+1 } }{ \frac { x+1 }{ x-1 } -\frac { x-1 }{ x+1 } } &=\frac {\hspace{2mm} \frac { (x+1)^2+(x-1)^2 }{ (x-1)(x+1) } \hspace{2mm} }{ \frac { (x+1)^2-(x-1)^2 }{ (x-1)(x+1) } } \\\\ &=\frac { \frac { { x }^{ 2 }+2x+1+{ x }^{ 2 }-2x+1 }{ { x }^{ 2 }-1 } }{\hspace{2mm} \frac { { x }^{ 2 }+2x+1-({ x }^{ 2 }-2x+1) }{ { x }^{ 2 }-1} \hspace{2mm} } \\\\ &=\frac { 2{ x }^{ 2 }+2 }{ { x }^{ 2 }-1 } \times \frac { { x }^{ 2 }-1 }{ 4x }\\\\ &=\frac { { x }^{ 2 }+1 }{ 2x } \quad\text{with } x\neq 0,\ x\neq \pm 1.\ _\square \end{aligned}$$