# Complex Fractions with Variables

A detailed step by step approach, on how to add, subtract, multiply, divide and simplify complex fractions with variables are presented. More questions and their solutions with detailed explanations are also included.
A thorough review of fractions, equivalent fractions, and how to add, multiply and divide fractions is needed in order to understand the examples below.

## Add and Subtract Fractions with Variables

We present examples on how to simplify complex fractions including variables along with their detailed solutions. The basic ideas are very similar to simplifying numerical fractions.

Example 1
Add the fractions: $\dfrac{2}{x} + \dfrac{3}{5}$

Solution to Example 1
The denominator $x$ is not known and in general we assume that it is not equal to the denominator $5$ of the fraction $\dfrac{3}{5}$.
In order to add fractions, we first rewrite them with the same (or common) denominator.
To set both equation to a common denominator, we multiply the numerator and denominator of the first fraction by the denominator of the second fraction which is $5$ and we multiply the numerator and denominator of the second fraction by the denominator of the first fraction which is $x$.
$\quad \quad \dfrac{2}{x} + \dfrac{3}{5} = \dfrac{2}{x} \times \color{red}{\dfrac{5}{5}} + \dfrac{3}{5} \times \color{red}{\dfrac{x}{x}}$

Multiply the fractions
$\quad \quad = \dfrac{2 \times \color{red}5} {x \times \color{red}5 } + \dfrac{3 \times \color{red}x} {5 \times \color{red}x}$

Simplify (NOTE: $x \times 5 = 5 \times x = 5x$ commutativity of multiplication )
$\quad \quad = \dfrac{10}{5x} + \dfrac{3x}{5x}$

and now that they have the two fractions same denominator $5x$, we add the numerators
$\quad \quad = \dfrac{10 + 3x}{5x}$

Example 2
Subtract the fractions: $\dfrac{1}{7} - \dfrac{3}{y}$

Solution to Example 2
The denominator $y$ is unknown and in general we assume that it is not equal to the other denominator $7$.
In order to subtract fractions, we first rewrite them with the same (or common) denominator.
Multiply the numerator and denominator of the first fraction by the denominator of the second fraction which is $y$ and multiply the numerator and denominator of the second fraction by the denominator of the first fraction which is $x$.
$\quad \quad \dfrac{1}{7} - \dfrac{3}{y} = \dfrac{1}{7} \times \color{red}{\dfrac{y}{y}} - \dfrac{3}{y} \times \color{red}{\dfrac{7}{7}}$

Multiply the fractions
$\quad \quad = \dfrac{1 \times \color{red} y} {7 \times \color{red}y } - \dfrac{3 \times \color{red}7} {y \times \color{red}7}$

Simplify
$\quad \quad = \dfrac{y}{7y} - \dfrac{21}{7y}$

and now that they have the same denominator, we subtract the numerators
$\quad \quad = \dfrac{y - 21}{7y}$

Example 3
Writes as a fraction: $2 x - \dfrac{3x}{2}$

Solution to Example 3
Rewrite $2x$ as a fraction.
$\quad \quad 2 x - \dfrac{3x}{2} = \dfrac{2 x}{1} - \dfrac{3x}{2}$

Rewrite the fractions with the same (or common) denominator.
$\quad \quad = \dfrac{2 x}{1} \color{red} {\times \dfrac{2}{2}} - \dfrac{3x}{2}$

$\quad \quad = \dfrac{4 x}{2} - \dfrac{3x}{2}$

Subtract numerators and keep common denominator
$\quad \quad = \dfrac{4x - 3x}{2}$

Simplify
$\quad \quad = \dfrac{x}{2}$

Example 4
Write as a fraction: $\dfrac{\dfrac{3x}{5}}{\dfrac{x}{4}}$

Solution to Example 4
To divide two fractions, we multiply the top one by the reciprocal of the bottom one. Hence
$\quad \quad \dfrac{\dfrac{3x}{5}}{\dfrac{x}{4}} = \dfrac{3x}{5} \times \dfrac{4}{x}$
Find common factor in the numerator and denominator.

$\quad \quad = \dfrac{3 \color{red} x}{5} \times \dfrac{4}{ \color{red}x}$

Simplify
$\quad \quad = \dfrac{12}{5}$

Example 5
Write as a fraction: $\dfrac{1+\dfrac{x}{2}}{3+\dfrac{3x}{2}}$

Solution to Example 5
There are terms with the denominator $2$ in both numerator and denominator. Hence we multiply the numerator and denominator by $2$ to eliminate fractions.

$\quad \quad \dfrac{1+\dfrac{x}{2}}{3+\dfrac{3x}{2}} = \dfrac{ (1+\dfrac{x}{2}) \times \color{red}2}{ (3+\dfrac{3x}{2}) \times \color{red}2 }$

Distribute.
$\quad \quad = \dfrac{ 1 \times \color{red}2+\dfrac{x}{2} \times \color{red}2 }{ 3 \times \color{red}2 +\dfrac{3x}{2} \times \color{red}2 }$

Simplify
$\quad \quad = \dfrac{2+x}{6+3x}$

Factor $3$ out in the denominator
$\quad \quad = \dfrac{2+x}{3(2+x)} = \dfrac{1(2+x)}{3(2+x)}$

Simplify by dividing numerator and denominator by $2 + x$.
$\quad \quad = \dfrac{1}{3}$

Example 6
Write as a fraction: $4 + \dfrac{2+\dfrac{1}{x}}{3 - \dfrac{1}{2x}}$

Solution to Example 6
There are two terms with denominator $x$ and $2x$ in the fraction on the right. In order to eliminate fractions, we multiply the numerator and denominator by $2x$. Hence
$\quad \quad 4 + \dfrac{2+\dfrac{1}{x}}{3 - \dfrac{1}{2x}} = 4 + \dfrac{ (2+\dfrac{1}{x}) \times \color{red}{2x} }{ (3 - \dfrac{1}{2x}) \times \color{red}{2x} }$

Distribute.
$\quad \quad = 4 + \dfrac{ 2 \times \color{red}{2x} + \dfrac{1}{x} \times \color{red}{2x} }{ 3 \times \color{red}{2x} - \dfrac{1}{2x} \times \color{red}{2x} }$

Simplify
$\quad \quad = 4 + \dfrac{ 4x + 2 }{ 6x - 1 }$

Rwrite $4$ as a fraction with denominator $6x - 1$
$\quad \quad = 4 \times \color{red}{\dfrac{6x-1}{6x-1}} + \dfrac{ 4x + 2 }{ 6x - 1 }$

Expand the numerator of the fraction on the left
$\quad \quad = \dfrac{24x-4}{6x-1} + \dfrac{ 4x + 2 }{ 6x - 1 }$

The fractions have a common denominator and may therefore be grouped as follows
$\quad \quad = \dfrac{24x-4+4x+2}{6x-1} = \dfrac{28x - 2}{6x-1}$

## Questions

Rewrite the following expressions as a single fraction.

1. ) $\dfrac{x}{3} - 6$

2. ) $\dfrac{1}{3x} - \dfrac{x}{3}$

3. ) $2+\dfrac{5x-\dfrac{5}{2}}{6x-3}$

## Solutions to the Above Questions

1. ) $\quad \quad \dfrac{x}{3} - 6 = \dfrac{x}{3} - 6 \color{red}{\times\dfrac{3}{3}} = \dfrac{x-18}{3}$

2. ) $\quad \quad \dfrac{1}{3x} - \dfrac{x}{3} = \dfrac{1}{3x} - \dfrac{x}{3} \color{red}{\times \dfrac{x}{x}} = \dfrac{1}{3x} - \dfrac{x^2}{3x } = \dfrac{1 - x^2}{3x}$

3. ) $\quad \quad 2+\dfrac{5x-\dfrac{5}{2}}{6x-3} = 2 + \dfrac{ (5x-\dfrac{5}{2}) \color{red}{\times 2}}{(6x-3) \color{red}{\times 2}} = 2 + \dfrac{10x-5}{12x-6}$

$\quad \quad = 2 \times \dfrac{12x-6}{12x-6} + \dfrac{10x-5}{12x-6} = \dfrac{24x-12+10x-5}{12x-6}$

$\quad \quad = \dfrac{34x - 17}{12x-6} = \dfrac{17(2x-1)}{6(2x-1)} = \dfrac{17}{6}$