Calculations with Fractions

On this page, we discuss four topics concerning fractions:

• Reducing fractions
• Adding and subtracting fractions
• Multiplication of fractions
• Division of fractions (compound fractions)

 

1. Reducing fractions

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.

 

2. Adding und subtracting fractions

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.

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

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

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)

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:

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)

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

Solution

 

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

Basic Mathematics

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

Math is fun

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

Math is fun

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

The Mathpage