Simplifying fractions

Algebra: Adding and subtracting fractions

Simplifying fractions

Similar fractions

A fraction stays the same if #\orange{\text{numerator}}# and #\blue{\text{denominator}}#:	Examples
1. are multiplied by the same number	\[\begin{array}{rcll} \dfrac{\orange{3x+1}}{\blue{x+2}} &=& \dfrac{\orange{6x+2}}{\blue{2x+4}} \end{array}\]
2. are multiplied by the same variable	\[\begin{array}{rcll}\dfrac{\orange{x}}{\blue{y}} &=& \dfrac{\orange{x \cdot z}}{\blue{y \cdot z}} \end{array}\]
3. are divided by the same number	\[\begin{array}{rcll} \dfrac{\orange{4x+2}}{\blue{2x+2}} &=& \dfrac{\orange{2x+1}}{\blue{x+1}}\end{array}\]
4. are divided by the same variable	\[\begin{array}{rcll}\dfrac{\orange{x}}{\blue{x^2+x}} &=& \dfrac{\orange{1}}{\blue{x+1}} \end{array}\]

Simplifying fractions

The process of making the numerator and denominator smaller is called the simplification of an expression.

Example

\[\dfrac{\orange{x}}{\blue{x^2+x}} = \dfrac{\orange{1}}{\blue{x+1}} \]

Factorization

In the example on the right hand side, we see that factorization can be used to find similar terms in the numerator and denominator. These common terms can be eliminated by means of division, hence, simplifying the fraction.

Examples

#\begin{array}{rcl}\dfrac{\orange{x+2}}{\blue{x^2+5 \cdot x +6}} &=& \dfrac{\orange{x+2}}{\blue{(x+2) \cdot (x+3)}} \\ &=& \dfrac{\orange{1}}{\blue{x+3}} \end{array}#

Take another look at the reduction above:

\[\dfrac{\blue{x}}{\blue{x}^2+\blue{x}} = \dfrac{{1}}{\blue{x}+1} \]

If we enter #\blue x=\green 0# in the expression on the left, we will be dividing by zero, which is not allowed. On the other hand, the expression on the right hand side is equal to #1# if we enter #\blue x=\green 0#. The equality does not hold if #\blue x=\green 0#.

In general, the equalities when reducing only hold for numbers for which the original expression was defined. If we want to be exact, we can write down for which numbers the expression is defined but, in general, we do not do this.

Calculate #\dfrac{\blue{x}}{\blue{x}^2+\blue{x}}# for #\blue{x}=\green{0}#:

\[\dfrac{\green{0}}{\green 0^2+\green 0} = \dfrac{0}{0}\]

Calculate #\dfrac{{1}}{\blue{x}+1}# for #\blue{x}=\green{0}#:

\[\dfrac{1}{\green 0+1} = \dfrac{1}{1}\]

Simplify as much as possible (without expanding the brackets):
\[\dfrac{5\cdot a^3\cdot \left(c+1\right)^4}{a^2\cdot b^3\cdot \left(c+1\right)^3}\]

#{{5\cdot a\cdot \left(c+1\right)}\over{b^3}}#

#\begin{array}{rcl}
\dfrac{5\cdot a^3\cdot \left(c+1\right)^4}{a^2\cdot b^3\cdot \left(c+1\right)^3}
&=& \displaystyle {{5\cdot a\cdot \left(c+1\right)}\over{b^3}}
\\ && \phantom{xxx}\blue{\text{common factor divided by } a^2\cdot \left(c+1\right)^3 \text{ }}
\end{array}#

New example