• Theory and Examples
• Exercises

### Calculations with Fractions

• Reducing fractions
• Multiplication of fractions
• Division of fractions (compound fractions)

1. Reducing fractions

Rule 1:

We reduce a fraction by dividing the numerator and denominator by its greatest common divisor.

Example:

In order to reduce the fraction $\frac{15}{6}$, we are looking for the greatest common divisor of the numerator and denominator. In this example, the greatest common divisor is 3. Both the numerator and the denominator are then divided by that greatest common divisor so that the fraction $\frac{15}{6}$ simplifies to $\frac{15}{6}=\frac{\frac{15}{3}}{\frac{6}{3}}=\frac{5}{2}$.

Reducing the fraction also works with variables.

Examples:

$1. \quad \frac{ax+abx-ac}{az}=\frac{a(x+bx-c)}{az}=\frac{\frac{a(x+bx-c)}{a}}{\frac{az}{a}}=\frac{x+bx-c}{z}$

$2. \quad \frac{az}{ax+abx-ac}= \frac{az}{a(x+bx-c)}= \frac{\frac{az}{a}}{\frac{a(x+bx-c)}{a}}=\frac{z}{x+bx-c}$

$3. \quad \frac{\left( a+b \right)z+\left( a+b \right)uv}{\left( a+b \right)x+\left( a+b \right){{x}^{2}}-\left( a+b \right)c}=\frac{(a+b)(z+uv)}{(a+b)(x+x^2-c)}=\frac{z+uv}{x+{{x}^{2}}-c}$

$4. \quad \frac{{{\left( a+b \right)}^{4}}z}{\left( a+b \right)x+\left( a+b \right){{x}^{2}}-\left( a+b \right)c}=\frac{(a+b) \cdot (a+b)^3z}{(a+b)(x+x^2-c)}=\frac{{{\left( a+b \right)}^{3}}z}{x+{{x}^{2}}-c}$

Remarks:

• In the first two examples, $a$ is factored out in the numerator and in the denominator. Then, the factor $a$ is reduced. The topic factorization is covered here.
• In the third example, the term $(a+b)$ is factored out both in the numerator and the denominator. In the fourth example, the term $(a+b)$ is factored out in the numerator recognising that $(a+b)^4=(a+b) \cdot (a+b)^3$ holds. Thus, the term $(a+b)$ can be canceled as a second step both in the third and fourth example. The topic factorization is covered here.

In this subsection, we discuss how two fractions are added and described as a single fraction. Likewise, we show how to subtract two fractions and simplify it to a single fraction.

Rule 2:

Addition and subtracting fractions with identical denominators works as follows:

$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm b}{c} \quad for \quad c \neq 0$

Examples: $\frac{3}{7}+\frac{5}{7}=\frac{3+5}{7}=\frac{8}{7} \quad \quad (Addition)$ $\frac{3}{7}-\frac{5}{7}=\frac{3-5}{7}=\frac{-2}{7}=-\frac{2}{7}\quad \quad (Subtraction)$

However, the two fractions do not necessarily have of the same denominator. In this case, rule 3 helps.

Rule 3:

Addition and subtracting fractions with different denominators works as follows:

Fractions with different denominators have to be rewritten by expanding the numerator and denominator of both fractions. Once they have identical denominators, the fractions can be added or subtracted by adding or subtracting the numerators and keeping the denominator constant.

Example: $\frac{5}{6}-\frac{4}{9}=\frac{5\cdot 3}{6\cdot 3}-\frac{4\cdot 2}{9\cdot 2}=\frac{15}{18}-\frac{8}{18}=\frac{15-8}{18}=\frac{7}{18}$

Both rules also work if the dnumerator or the denominator consists of entire terms:

Example

Suppose that we add the two fractions $\frac{a}{x-y}$ and $\frac{b}{x+y}$. First, the two fractions should have an identical denominator. Therefore, we look for a possible common denominator. The easiest way is to multiply the denominators such that the common denominator becomes $(x-y)\cdot (x+y)$. Simultaneously, the numerators of each fraction must be multiplied by the denominator of the other fraction. Then, both numerators can be added, while the common denominator $(x-y)\cdot (x+y)$ remains unaltered.

$\frac{a}{x-y}+\frac{b}{x+y}=\frac{a\cdot \left( x+y \right)}{\left( x-y \right)\cdot \left( x+y \right)}+\frac{b\cdot \left( x-y \right)}{\left( x+y \right)\cdot \left( x-y \right)}=\frac{a\cdot \left( x+y \right)+b\cdot \left( x-y \right)}{\left( x+y \right)\cdot \left( x-y \right)}$

Remarks:

• In this example, both the numerator and the denominator of the first fraction are multiplied by $(x+y)$, i.e., the first fraction is multiplied by $1$ (since $\frac {x+y}{x+y}=1$,provided $x+y \neq 0$). Thus, the value of the first fraction remains unchanged. Analogously, both the numerator and the denominator of the second fraction are multiplied by $(x-y)$. Therefore, the second fraction is also multiplied by $1$ (since $\frac {x-y}{x-y}=1$ provided $x-y \neq 0$).
• Subtracting two fractions is calculated similarily, as you can see in the following example:

$\frac{a}{x-y}-\frac{b}{x+y}=\frac{a\cdot \left( x+y \right)}{\left( x-y \right)\cdot \left( x+y \right)}-\frac{b\cdot \left( x-y \right)}{\left( x+y \right)\cdot \left( x-y \right)}=\frac{a\cdot \left( x+y \right)-b\cdot \left( x-y \right)}{\left( x+y \right)\cdot \left( x-y \right)}$

3. Multiplication of fractions

Rule 4:

The multiplication of fractions works as follows:

$a\cdot \frac{c}{d}=\frac{a\cdot c}{d} \quad for \quad d \neq 0$

$\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}\quad for \quad b,d \neq 0$ Examples: $\left( -7 \right)\cdot \frac{4}{9}=\frac{\left( -7 \right)\cdot 4}{9}=\frac{-28}{9}=-\frac{28}{9}$ $\frac{-3}{4}\cdot \frac{7}{-5}=\frac{\left( -3 \right)\cdot 7}{4\cdot \left( -5 \right)}=\frac{-21}{-20}=\frac{21}{20}$

This rule also works if the numerator or the denominator consists of terms:

Examples

$a\cdot \frac{b+c}{d-e}=\frac{a\cdot \left( b+c \right)}{d-e}=\frac{ab+ac}{d-e}$ $\frac{a}{b}\cdot \frac{c+d}{e-f}=\frac{a\cdot \left( c+d \right)}{b\cdot \left( e-f \right)}=\frac{ac+ad}{be-bf}$ $\frac{a+b}{c+d}\cdot \frac{e+f}{g+h}=\frac{(a+b)\cdot \left( e+f \right)}{(c+d)\cdot \left( g+h \right)}=\frac{ae+af+be+bf}{cg+ch+dg+dh}$

4. Division of fractions (compound fractions)

Rule 5:

The division of the fraction $\frac{a}{b}$ to the fraction $\frac{c}{d}$ works as follows:

$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c} \quad for \quad b,c,d \neq 0$

Example: $\frac{\frac{-4}{3}}{\frac{5}{-7}}=\frac{-4}{3}\cdot \frac{-7}{5}=\frac{(-4)\cdot (-7)}{3 \cdot 5}=\frac{28}{15}$

Proof

Why does rule 5 hold? If the compound fraction $\frac{\frac{a}{b}}{\frac{c}{d}}$ is multiplied by $b \cdot d$ in the numerator and the denominator, i.e., the entire compound fraction is multiplied by $1=\frac {b \cdot d}{b \cdot d}$, then we get the result:

$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{\frac{a}{b}}{\frac{c}{d}}\cdot \frac{bd}{bd}=\frac{\frac{a}{b} \cdot bd}{\frac{c}{d} \cdot bd}=\frac{\frac{abd}{b}}{\frac{cbd}{d}}=\frac{ad}{cb}=\frac{a}{b}\cdot \frac{d}{c}$

There are two special cases of rule 5 if either the numerator or the denominator is not a fraction:

Two special cases of rule 5:

There is an integer c in the denominator and not a fraction:

$\frac{\left( \frac{a}{b} \right)}{c}=\frac{a}{b\cdot c}$

There is an integer a in the numerator and not a fraction:

$\frac{a}{\left( \frac{c}{d} \right)}=\frac{a\cdot d}{c}$

Reduce and simplify the following fractions:

1)

$\frac{12x}{\left( 9+3x \right)}=$

2)

$\frac{5a+20}{5}=$

3)

$\frac{28a-35b}{21}=$

4)

$\frac{{{y}^{2}}+2y-24}{{{y}^{2}}-6y+8}=$

5)

$\left( {{x}^{2}}-{{y}^{2}} \right)\cdot \frac{6x}{3xy-3{{x}^{2}}}=$

6)

$\large \frac{\frac{a-b}{{{a}^{2}}}}{\text{ }\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{3}}}}=$

7)

$\large \frac{x+\frac{1}{3}}{x-\frac{1}{3}}=$

8)

$\large \frac{\frac{4}{3}}{\frac{2}{3}}=$

9)

$\large \frac{3a}{\frac{4b}{{{c}^{2}}}}=$

10)

$\large \frac{\frac{c}{b+c}}{\frac{2c}{{{b}^{2}}+bc}}=$

An interactive calculator to reduce fractions can be found on the following website:

Examples and graphical illustration regarding the topic "addition of fractions" are presented on:

Further examples regarding the topic "multiplication of fractions" are available on:

The topic "division of fractions" is also discussed on the following website: