Multiplication of fractions

Algebra: Adding and subtracting fractions

Multiplication of fractions

The product of two fractions

We can multiply two fractions by multiplying the numerator by the numerator and multiplying the denominator by the denominator.

\[\frac{\orange{a}}{\blue{b}} \cdot \frac{\purple{c}}{\green{d}}=\frac{\orange{a} \cdot \purple{c}}{\blue{b} \cdot \green{d}}\]

Example

\[\begin{array}{rcl} \dfrac{\orange{x}}{\blue{y}} \cdot \dfrac{\purple{5}}{\green{y^2}}&=&\dfrac{\orange{x} \cdot\purple{ 5}}{\blue{y }\cdot\green{ y^2}} \\ &=& \dfrac{{5 \cdot x}}{{y^3}}\end{array}\]

Simplification

Just as with adding and subtracting fractions, when multiplying, we simplify the answer if possible.

Example

\[\begin{array}{rcl} \dfrac{\orange{x^2}}{\blue{y^2}} \cdot \dfrac{\purple{3 \cdot y}}{\green{x}}&=&\dfrac{\orange{x^2} \cdot \purple{3 \cdot y}}{\blue{y^2} \cdot \green{x}} \\ &=& \dfrac{{3 \cdot x}}{{y}}\end{array}\]

Brackets

Just as with adding and subtracting fractions, multiplying fractions requires paying attention to the brackets. If the fractions consist of multiple terms, both the numerator and denominator must be placed in brackets.

Unless specified otherwise, brackets can remain in the final answer.

Example

\[\begin{aligned} \dfrac{\orange{x^2+1}}{\blue{y+1}} \cdot \dfrac{\purple{y+1}}{\green{x^2}}&=\dfrac{\orange{(x^2 +1)} \cdot \purple{(y+1)}}{\blue{(y+1)} \cdot \green{x^2}} \\ &= \dfrac{{x^2+1}}{{x^2}}\end{aligned}\]

Special case

\[\begin{array}{rcl}
\orange{a} \cdot \dfrac{\purple{c}}{\green{d}}
&=&
\dfrac{\orange{a} \cdot \purple{c}}{ \green{d}}
\end{array}\]

Example

#\begin{array}{rcl}
\orange{x} \cdot \dfrac{\purple{2}}{\green{xy}}
&=&
\dfrac{\orange{x} \cdot \purple{2}}{ \green{xy}}\\
&=&\dfrac{2}{y}
\end{array}#

\[\begin{array}{rcl}
\orange{a} \cdot \dfrac{\purple{c}}{\green{d}}
&=& \dfrac{\orange{a}}{\blue{1}} \cdot \dfrac{\purple{c}}{\green{d}} \\
&&\qquad\blue{\text{rule \(\tfrac{a}{1}=a\)}} \\
&=& \dfrac{\orange{a} \cdot \purple{c}}{\blue{1}\cdot \green{d}} \\
&&\qquad\blue{\text{rule \(\frac{{a}}{{b}} \cdot \frac{{c}}{{d}}=\frac{{a} \cdot {c}}{{b} \cdot {d}}\)}} \\
&=& \dfrac{\orange{a} \cdot \purple{c}}{ \green{d}} \\
&&\qquad\blue{\text{rule \(1\cdot x = x\)}}
\end{array}\]

Calculate and simplify as much as possible:
\[\dfrac{x}{z} \cdot \dfrac{y\cdot z^2}{x^2-x}\]

#{{y\cdot z}\over{x-1}}#

#\begin{array}{rcl}
\dfrac{x}{z} \cdot \dfrac{y\cdot z^2}{x^2-x} &=& \dfrac{ {x} \cdot {y\cdot z^2}}{{z} \cdot \left( x^2-x\right)}\\
&& \phantom{xxx}\blue{\text{fractions multiplied by }}\\
&& \phantom{xxx}\blue{\text{multiplying numerator by numerator and denominator by denominator}}\\
&=& \displaystyle {{y\cdot z}\over{x-1}}
\\ && \phantom{xxx}\blue{\text{simplified}}\\
\end{array}#

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