Complex Fractions with Variables

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 fractions 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 \( x \).

\[ \dfrac{2}{x} + \dfrac{3}{5} = \dfrac{2}{x} \times \color{red}{\dfrac{5}{5}} + \dfrac{3}{5} \times \color{red}{\dfrac{x}{x}} \]

Multiply the fractions:

\[ = \dfrac{2 \times \color{red}5} {x \times \color{red}5 } + \dfrac{3 \times \color{red}x} {5 \times \color{red}x} \]

Simplify (NOTE: \( x \times 5 = 5 \times x = 5x \) by commutativity of multiplication):

\[ = \dfrac{10}{5x} + \dfrac{3x}{5x} \]

Now that the two fractions have the same denominator \( 5x \), we add the numerators:

\[ = \dfrac{10 + 3x}{5x} \]

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 \( 7 \).

\[ \dfrac{1}{7} - \dfrac{3}{y} = \dfrac{1}{7} \times \color{red}{\dfrac{y}{y}} - \dfrac{3}{y} \times \color{red}{\dfrac{7}{7}} \]

Multiply the fractions:

\[ = \dfrac{1 \times \color{red} y} {7 \times \color{red}y } - \dfrac{3 \times \color{red}7} {y \times \color{red}7} \]

Simplify:

\[ = \dfrac{y}{7y} - \dfrac{21}{7y} \]

Now that they have the same denominator, we subtract the numerators:

\[ = \dfrac{y - 21}{7y} \]

Solution to Example 3
Rewrite \( 2x \) as a fraction:

\[ 2 x - \dfrac{3x}{2} = \dfrac{2 x}{1} - \dfrac{3x}{2} \]

Rewrite the fractions with the same (or common) denominator:

\[ = \dfrac{2 x}{1} \color{red} {\times \dfrac{2}{2}} - \dfrac{3x}{2} \] \[ = \dfrac{4 x}{2} - \dfrac{3x}{2} \]

Subtract numerators and keep common denominator:

\[ = \dfrac{4x - 3x}{2} \]

Simplify:

\[ = \dfrac{x}{2} \]

Solution to Example 4
To divide two fractions, we multiply the top one by the reciprocal of the bottom one. Hence:

\[ \dfrac{\dfrac{3x}{5}}{\dfrac{x}{4}} = \dfrac{3x}{5} \times \dfrac{4}{x} \]

Find common factors in the numerator and denominator:

\[ = \dfrac{3 \color{red} x}{5} \times \dfrac{4}{ \color{red}x} \]

Simplify:

\[ = \dfrac{12}{5} \]

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:

\[ \dfrac{1+\dfrac{x}{2}}{3+\dfrac{3x}{2}} = \dfrac{ (1+\dfrac{x}{2}) \times \color{red}2}{ (3+\dfrac{3x}{2}) \times \color{red}2 } \]

Distribute:

\[ = \dfrac{ 1 \times \color{red}2+\dfrac{x}{2} \times \color{red}2 }{ 3 \times \color{red}2 +\dfrac{3x}{2} \times \color{red}2 } \]

Simplify:

\[ = \dfrac{2+x}{6+3x} \]

Factor \( 3 \) out in the denominator:

\[ = \dfrac{2+x}{3(2+x)} = \dfrac{1(2+x)}{3(2+x)} \]

Simplify by dividing numerator and denominator by \( 2 + x \):

\[ = \dfrac{1}{3} \]

Solution to Example 6
There are terms with denominators \( x \) and \( 2x \) in the fraction on the right. In order to eliminate fractions, we multiply the numerator and denominator by \( 2x \). Hence:

\[ 4 + \dfrac{2+\dfrac{1}{x}}{3 - \dfrac{1}{2x}} = 4 + \dfrac{ (2+\dfrac{1}{x}) \times \color{red}{2x} }{ (3 - \dfrac{1}{2x}) \times \color{red}{2x} } \]

Distribute:

\[ = 4 + \dfrac{ 2 \times \color{red}{2x} + \dfrac{1}{x} \times \color{red}{2x} }{ 3 \times \color{red}{2x} - \dfrac{1}{2x} \times \color{red}{2x} } \]

Simplify:

\[ = 4 + \dfrac{ 4x + 2 }{ 6x - 1 } \]

Rewrite \( 4 \) as a fraction with denominator \( 6x - 1 \):

\[ = 4 \times \color{red}{\dfrac{6x-1}{6x-1}} + \dfrac{ 4x + 2 }{ 6x - 1 } \]

Expand the numerator of the fraction on the left:

\[ = \dfrac{24x-4}{6x-1} + \dfrac{ 4x + 2 }{ 6x - 1 } \]

The fractions have a common denominator and may therefore be grouped as follows:

\[ = \dfrac{24x-4+4x+2}{6x-1} = \dfrac{28x - 2}{6x-1} \]