Lösung 1.2:5c

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Zeile 1: Zeile 1:
Method 1
Method 1
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We calculate the numerator and denominator first.
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We calculate the numerator and denominator first
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{3}{10}-\frac{1}{5} &= \frac{3}{10}-\frac{1\cdot 2}{5\cdot 2} = \frac{3-2}{10} = \frac{1}{10}\,,\\[10pt]
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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\cdot 2}{8\cdot 2}-\frac{3}{16} = \frac{14-3}{16} = \frac{11}{16}\,\textrm{.}
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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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\end{align}</math>
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Thus, the expression becomes
Thus, the expression becomes
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{{Displayed math||<math>\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}</math>}}
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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 <math>16=2\cdot 2\cdot 2\cdot 2</math> and <math>10=2\centerdot 5</math>, the simplified answer is
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{{Displayed math||<math>\frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,</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
Method 2
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If we look at the individual fractions
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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
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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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{{Displayed math||<math>10=2\cdot 5\,,\quad 8=2\cdot 2\cdot 2\,\quad\text{and}\quad
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16=2\cdot 2\cdot 2\cdot 2</math>}}
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If we multiply the top and bottom of the main fraction by
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and therefore 2∙2∙2∙2∙5 = 80 is the fractions' lowest common denominator.
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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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If we multiply the top and bottom of the main fraction by 80, then it will be possible to eliminate all denominators at once,
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<math>\begin{align}
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{{Displayed math||<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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\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,}
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& \\
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&= \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]
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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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&= \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{.}
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& \\
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\end{align}</math>}}
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\end{align}</math>
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Version vom 12:26, 19. Sep. 2008

Method 1

We calculate the numerator and denominator first

Vorlage:Displayed math

Thus, the expression becomes

Vorlage:Displayed math

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

Vorlage:Displayed math

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

Vorlage:Displayed math

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,

Vorlage:Displayed math

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