Lösung 1.2:5c
Aus Online Mathematik Brückenkurs 1
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Version vom 13:30, 22. Okt. 2008
Method 1
We calculate the numerator and denominator first
| \displaystyle \begin{align}
\frac{3}{10}-\frac{1}{5} &= \frac{3}{10}-\frac{1\cdot 2}{5\cdot 2} = \frac{3-2}{10} = \frac{1}{10}\,,\\[10pt] \frac{7}{8}-\frac{3}{16} &= \frac{7\cdot 2}{8\cdot 2}-\frac{3}{16} = \frac{14-3}{16} = \frac{11}{16}\,\textrm{.} \end{align} |
Thus, the expression becomes
| \displaystyle \frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\vphantom{\Biggl(}\,}{\,\dfrac{11}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\cdot \dfrac{16}{11}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{\,/}11}{\rlap{\,/}16}\cdot \dfrac{\rlap{\,/}16}{\rlap{\,/}11}\vphantom{\Biggl(}\,} = \dfrac{16}{10\cdot 11} |
and because \displaystyle 16=2\cdot 2\cdot 2\cdot 2 and \displaystyle 10=2\cdot 5, the simplified answer is
| \displaystyle \frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,. |
Method 2
If we look at the individual fractions 3/10, 1/5, 7/8 and 3/16, we see that the denominators can be factorized as
| \displaystyle 10=2\cdot 5\,,\quad 8=2\cdot 2\cdot 2\,\quad\text{and}\quad
16=2\cdot 2\cdot 2\cdot 2 |
and therefore 2∙2∙2∙2∙5 = 80 is the fractions' lowest common denominator.
If we multiply the top and bottom of the main fraction by 80, then it will be possible to eliminate all denominators at once,
| \displaystyle \begin{align}
\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} &= \frac{\,\left( \dfrac{3}{10}-\dfrac{1}{5} \right)\cdot 80\vphantom{\Biggl(}\,}{\,\left( \dfrac{7}{8}-\dfrac{3}{16} \right)\cdot 80\vphantom{\Biggl(}\,} = \frac{\dfrac{3\cdot 80}{10}-\dfrac{1\cdot 80}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7\cdot 80}{8}-\dfrac{3\cdot 80}{16}\vphantom{\Biggl(}\,}\\[10pt] &= \frac{\,\dfrac{3\cdot 8\cdot{}\rlap{\,/}10}{\rlap{\,/}10}-\dfrac{8\cdot 2\cdot{}\rlap{/}5}{\rlap{/}5}\vphantom{\Biggl(}\,}{\,\dfrac{7\cdot{}\rlap{/}8\cdot 10}{\rlap{/}8}-\dfrac{3\cdot{}\rlap{\,/}16\cdot 5}{\rlap{\,/}16}\vphantom{\Biggl(}\,} = \dfrac{3\cdot 8-8\cdot 2}{7\cdot 10-3\cdot } = \frac{8}{55}\,\textrm{.} \end{align} |
