Solution 1.2:5c

From Förberedande kurs i matematik 1

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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\cdot 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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Current revision

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}

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