Solution 1.2:5b

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Method 1
Method 1
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One solution is to calculate the numerator and denominator in the main fraction individually:
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One solution is to calculate the numerator and denominator in the main fraction individually
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{{Displayed math||<math>\begin{align}
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\frac{1}{2}+\frac{1}{3} &= \frac{1\cdot 3}{2\cdot 3}+\frac{1\cdot 2}{3\cdot 2} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}\,,\\[10pt]
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\frac{1}{3}-\frac{1}{2} &= \frac{1\cdot 2}{3\cdot 2}-\frac{1\cdot 3}{2\cdot 3} = \frac{2}{6}-\frac{3}{6} = -\frac{1}{6}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator
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& \frac{1}{2}+\frac{1}{3}=\frac{1\centerdot 3}{2\centerdot 3}+\frac{1\centerdot 2}{3\centerdot 2}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\
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& \\
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& \frac{1}{3}-\frac{1}{2}=\frac{1\centerdot 2}{3\centerdot 2}-\frac{1\centerdot 3}{2\centerdot 3}=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6} \\
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\end{align}</math>
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{{Displayed math||<math>
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\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{6}\vphantom{\Biggl(}\,}{\,-\dfrac{1}{6}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}{\,-\dfrac{1}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}=\frac{5}{-1}=-5\,</math>.}}
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The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator:
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<math>\begin{align}
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& \frac{1}{2}+\frac{1}{3}=\frac{1\centerdot 3}{2\centerdot 3}+\frac{1\centerdot 2}{3\centerdot 2}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\
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& \\
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& \frac{1}{3}-\frac{1}{2}=\frac{1\centerdot 2}{3\centerdot 2}-\frac{1\centerdot 3}{2\centerdot 3}=\frac{2}{6}-\frac{3}{6}=-\frac{1}{6} \\
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& \frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}-\frac{1}{2}}=\frac{\frac{5}{6}}{-\frac{1}{6}}=\frac{\frac{5}{6}\centerdot 6}{-\frac{1}{6}\centerdot 6}=\frac{5}{-1}=-5 \\
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\end{align}</math>
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Method 2
Method 2
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Another way to solve the exercise is to multiplying the top and bottom of the main fraction by
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Another way to solve the exercise is to multiply the top and bottom of the main fraction by <math>3\cdot 2=6</math>, so that all denominators in the partial fractions 1/2 and 1/3 can eliminated in one step
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<math>3\centerdot 2=6</math>
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, so that all denominators in the partial fractions
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<math>{1}/{2}\;</math>
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and
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<math>{1}/{3}\;</math>
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can eliminated in one step:
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{{Displayed math||<math>
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<math>\begin{align}
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\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\left( \dfrac{1}{2}+\dfrac{1}{3} \right)\cdot 6\vphantom{\Biggl(}\,}{\,\left( \dfrac{1}{3}-\dfrac{1}{2} \right)\cdot 6\vphantom{\Biggl(}\,}=\frac{\,\dfrac{6}{2}+\dfrac{6}{3}\vphantom{\Biggl(}\,}{\,\dfrac{6}{3}-\dfrac{6}{2}\vphantom{\Biggl(}\,}=\frac{3+2}{2-3}=\frac{5}{-1}=-5\,\textrm{.}</math>}}
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& \\
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& \frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}-\frac{1}{2}}=\frac{\left( \frac{1}{2}+\frac{1}{3} \right)\centerdot 6}{\left( \frac{1}{3}-\frac{1}{2} \right)\centerdot 6}=\frac{\frac{6}{2}+\frac{6}{3}}{\frac{6}{3}-\frac{6}{2}}=\frac{3+2}{2-3}=\frac{5}{-1}=-5 \\
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\end{align}</math>
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Current revision

Method 1

One solution is to calculate the numerator and denominator in the main fraction individually

\displaystyle \begin{align}

\frac{1}{2}+\frac{1}{3} &= \frac{1\cdot 3}{2\cdot 3}+\frac{1\cdot 2}{3\cdot 2} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}\,,\\[10pt] \frac{1}{3}-\frac{1}{2} &= \frac{1\cdot 2}{3\cdot 2}-\frac{1\cdot 3}{2\cdot 3} = \frac{2}{6}-\frac{3}{6} = -\frac{1}{6}\,\textrm{.} \end{align}

The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator

\displaystyle

\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{6}\vphantom{\Biggl(}\,}{\,-\dfrac{1}{6}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}{\,-\dfrac{1}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}=\frac{5}{-1}=-5\,.


Method 2

Another way to solve the exercise is to multiply the top and bottom of the main fraction by \displaystyle 3\cdot 2=6, so that all denominators in the partial fractions 1/2 and 1/3 can eliminated in one step

\displaystyle

\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\left( \dfrac{1}{2}+\dfrac{1}{3} \right)\cdot 6\vphantom{\Biggl(}\,}{\,\left( \dfrac{1}{3}-\dfrac{1}{2} \right)\cdot 6\vphantom{\Biggl(}\,}=\frac{\,\dfrac{6}{2}+\dfrac{6}{3}\vphantom{\Biggl(}\,}{\,\dfrac{6}{3}-\dfrac{6}{2}\vphantom{\Biggl(}\,}=\frac{3+2}{2-3}=\frac{5}{-1}=-5\,\textrm{.}

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