Lösung 1.2:5c

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Method 1
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<center> [[Image:1_2_5c-1(2).gif]] </center>
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<center> [[Image:1_2_5c-2(2).gif]] </center>
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We calculate the numerator and denominator first.
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<math>\begin{align}
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& \frac{3}{10}-\frac{1}{5}=\frac{3}{10}-\frac{1\centerdot 2}{5\centerdot 2}=\frac{3-2}{10}=\frac{1}{10} \\
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& \frac{7}{8}-\frac{3}{16}=\frac{7\centerdot 2}{8\centerdot 2}-\frac{3}{16}=\frac{14-3}{16}=\frac{11}{16} \\
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\end{align}</math>
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Thus, the expression becomes
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<math>\frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{16}}=\frac{\frac{1}{10}}{\frac{11}{16}}=\frac{\frac{1}{10}\centerdot \frac{16}{11}}{\frac{11}{16}\centerdot \frac{16}{11}}=\frac{16}{10\centerdot 11}</math>
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and because
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<math>16=2\centerdot 2\centerdot 2\centerdot 2</math>
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and 10=2∙5
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<math>10=2\centerdot 5</math>
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, the simplified answer is
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<math>\frac{16}{10\centerdot 11}=\frac{2\centerdot 2\centerdot 2\centerdot 2}{2\centerdot 5\centerdot 11}=\frac{8}{55}</math>
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Method 2
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If we look at the individual fractions
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<math>{3}/{10}\;,\ \ {1}/{5,\ \ {7}/{8}\;}\;</math>
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and
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<math>{3}/{16}\;</math>
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, we see that the denominators can be factorized as
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<math>10=2\centerdot 5,\ 8=2\centerdot 2\centerdot 2</math>
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<math>10=2\centerdot 5,\quad 8=2\centerdot 2\centerdot 2</math>
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and
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<math>16=2\centerdot 2\centerdot 2\centerdot 2</math>
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and therefore 2∙2∙2∙2∙5 is the fractions' lowest common denominator.
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If we multiply the top and bottom of the main fraction by
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<math>80</math>
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, then it will be possible to eliminate all denominators at once,
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<math>\begin{align}
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& \frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{16}}=\frac{\left( \frac{3}{10}-\frac{1}{5} \right)\centerdot 80}{\left( \frac{7}{8}-\frac{3}{16} \right)\centerdot 80}=\frac{\frac{3\centerdot 80}{10}-\frac{1\centerdot 80}{5}}{\frac{7\centerdot 80}{8}-\frac{3\centerdot 80}{16}} \\
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& \\
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& =\frac{\frac{3\centerdot 8\centerdot 10}{10}-\frac{8\centerdot 2\centerdot 5}{5}}{\frac{7\centerdot 8\centerdot 10}{8}-\frac{3\centerdot 16\centerdot 5}{16}}=\frac{3\centerdot 8-8\centerdot 2}{7\centerdot 10-3\centerdot }=\frac{8}{55} \\
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& \\
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\end{align}</math>

Version vom 15:00, 11. Sep. 2008

Method 1

We calculate the numerator and denominator first.


\displaystyle \begin{align} & \frac{3}{10}-\frac{1}{5}=\frac{3}{10}-\frac{1\centerdot 2}{5\centerdot 2}=\frac{3-2}{10}=\frac{1}{10} \\ & \frac{7}{8}-\frac{3}{16}=\frac{7\centerdot 2}{8\centerdot 2}-\frac{3}{16}=\frac{14-3}{16}=\frac{11}{16} \\ \end{align}

Thus, the expression becomes


\displaystyle \frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{16}}=\frac{\frac{1}{10}}{\frac{11}{16}}=\frac{\frac{1}{10}\centerdot \frac{16}{11}}{\frac{11}{16}\centerdot \frac{16}{11}}=\frac{16}{10\centerdot 11}


and because \displaystyle 16=2\centerdot 2\centerdot 2\centerdot 2 and 10=2∙5 \displaystyle 10=2\centerdot 5 , the simplified answer is


\displaystyle \frac{16}{10\centerdot 11}=\frac{2\centerdot 2\centerdot 2\centerdot 2}{2\centerdot 5\centerdot 11}=\frac{8}{55}

Method 2

If we look at the individual fractions \displaystyle {3}/{10}\;,\ \ {1}/{5,\ \ {7}/{8}\;}\; and \displaystyle {3}/{16}\; , we see that the denominators can be factorized as


\displaystyle 10=2\centerdot 5,\ 8=2\centerdot 2\centerdot 2

\displaystyle 10=2\centerdot 5,\quad 8=2\centerdot 2\centerdot 2 and \displaystyle 16=2\centerdot 2\centerdot 2\centerdot 2


and therefore 2∙2∙2∙2∙5 is the fractions' lowest common denominator.

If we multiply the top and bottom of the main fraction by \displaystyle 80 , then it will be possible to eliminate all denominators at once,


\displaystyle \begin{align} & \frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{16}}=\frac{\left( \frac{3}{10}-\frac{1}{5} \right)\centerdot 80}{\left( \frac{7}{8}-\frac{3}{16} \right)\centerdot 80}=\frac{\frac{3\centerdot 80}{10}-\frac{1\centerdot 80}{5}}{\frac{7\centerdot 80}{8}-\frac{3\centerdot 80}{16}} \\ & \\ & =\frac{\frac{3\centerdot 8\centerdot 10}{10}-\frac{8\centerdot 2\centerdot 5}{5}}{\frac{7\centerdot 8\centerdot 10}{8}-\frac{3\centerdot 16\centerdot 5}{16}}=\frac{3\centerdot 8-8\centerdot 2}{7\centerdot 10-3\centerdot }=\frac{8}{55} \\ & \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:5c“