Solution 1.2:5b

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Method 1
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<center> [[Image:1_2_5b.gif]] </center>
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One solution is to calculate the numerator and denominator in the main fraction individually:
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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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\end{align}</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
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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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<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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<math>\begin{align}
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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>

Revision as of 14:54, 11 September 2008

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\centerdot 3}{2\centerdot 3}+\frac{1\centerdot 2}{3\centerdot 2}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\ & \\ & \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} \\ \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 \begin{align} & \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} \\ & \\ & \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} \\ & \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 \\ \end{align}


Method 2

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


\displaystyle \begin{align} & \\ & \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 \\ \end{align}

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