Aus Online Mathematik Brückenkurs 1

When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:

\displaystyle \frac{2}{3+\frac{1}{2}}\ ,\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}} and \displaystyle \frac{3}{2-\frac{2}{7}}

We can do this by multiplying the top and bottom of each fraction by \displaystyle 2 , \displaystyle 12 and \displaystyle 7 respectively, so as to get rid of the partial fractions:

\displaystyle \begin{align} & \frac{2}{3+\frac{1}{2}}=\ \frac{2\centerdot 2}{\left( 3+\frac{1}{2} \right)\centerdot 2}=\frac{4}{3\centerdot 2+\frac{1}{2}\centerdot 2}=\frac{4}{6+1}=\frac{4}{7} \\ & \\ & \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}=\frac{\frac{1}{2}\centerdot 12}{\left( \frac{1}{4}-\frac{1}{3} \right)\centerdot 12}=\frac{6}{\frac{12}{4}-\frac{12}{3}}=\frac{6}{3-4}=\frac{6}{-1}=-6 \\ & \\ & \frac{3}{2-\frac{2}{7}}=\frac{3\centerdot 7}{\left( 2-\frac{2}{7} \right)\centerdot 7}=\frac{21}{2\centerdot 7-\frac{2}{7}\centerdot 7}=\frac{21}{14-2}=\frac{21}{12} \\ \end{align}

The whole expression therefore equals

\displaystyle \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}

If we multiply the tops and bottoms of the fractions \displaystyle

\displaystyle {4}/{7}\; , \displaystyle {1}/{2}\; and \displaystyle {21}/{12}\; in the main fraction by their lowest common denominator, \displaystyle 7\centerdot 12 , we obtain integers in the numerator and denominator:

\displaystyle \begin{align} & \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}=\frac{\left( \frac{4}{7}-6 \right)\centerdot 7\centerdot 12}{\left( \frac{1}{2}-\frac{21}{12} \right)\centerdot 7\centerdot 12}=\frac{4\centerdot 12-6\centerdot 7\centerdot 12}{7\centerdot 6-21\centerdot 7} \\ & \\ & =\frac{\left( 4-6\centerdot 7 \right)\centerdot 12}{\left( 6-21 \right)\centerdot 7}=\frac{-38\centerdot 12}{-15\centerdot 7}=\frac{38\centerdot 12}{15\centerdot 7} \\ \end{align}

By factorizing \displaystyle 12 , \displaystyle 15 and \displaystyle 38 ,

\displaystyle \begin{align} & 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\ & 15=3\centerdot 5 \\ & 38=2\centerdot 19 \\ \end{align}

the answer can be simplified to

\displaystyle \frac{38\centerdot 12}{15\centerdot 7}=\frac{2\centerdot 19\centerdot 2\centerdot 2\centerdot 3}{3\centerdot 5\centerdot 7}=\frac{153}{35}