Solution 1.2:6

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When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:
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When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts
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{{Displayed math||<math>\frac{2}{\,3+\dfrac{1}{2}\vphantom{\Biggl(}\,}\ ,\quad \frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\,\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}\,}\quad\text{and}\quad\frac{3}{\,2-\dfrac{2}{7}\vphantom{\Biggl(}\,}\,</math>.}}
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<math>\frac{2}{3+\frac{1}{2}}\ ,\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}</math>
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We can do this by multiplying the top and bottom of each fraction by 2, 12 and 7 respectively, so as to get rid of the partial fractions
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and
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<math>\frac{3}{2-\frac{2}{7}}</math>
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{{Displayed math||<math>\begin{align}
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We can do this by multiplying the top and bottom of each fraction by
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\frac{2}{3+\dfrac{1}{2}\vphantom{\Biggl(}} &= \frac{2\cdot 2}{\left( 3+\dfrac{1}{2} \right)\cdot 2\vphantom{\Biggl(}} = \frac{4}{3\cdot 2+\dfrac{1}{2}\cdot 2\vphantom{\Biggl(}} = \frac{4}{6+1} = \frac{4}{7}\,,\\[5pt]
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<math>2</math>
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\frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}} &= \frac{\dfrac{1}{2}\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{4}-\dfrac{1}{3} \right)\cdot 12\vphantom{\Biggl(}} = \frac{6}{\dfrac{12}{4}-\dfrac{12}{3}\vphantom{\Biggl(}} = \frac{6}{3-4} = \frac{6}{-1} = -6\,,\\[10pt]
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,
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\frac{3}{2-\dfrac{2}{7}\vphantom{\Biggl(}} &= \frac{3\cdot 7}{\left( 2-\dfrac{2}{7} \right)\cdot 7\vphantom{\Biggl(}} = \frac{21}{2\cdot 7-\dfrac{2}{7}\cdot 7\vphantom{\Biggl(}} = \frac{21}{14-2} = \frac{21}{12}\,\textrm{.}
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<math>12</math>
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\end{align}</math>}}
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and
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<math>7</math>
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respectively, so as to get rid of the partial fractions:
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<math>\begin{align}
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& \frac{2}{3+\frac{1}{2}}=\ \frac{2\centerdot 2}{\left( 3+\frac{1}{2} \right)\centerdot 2}=\frac{4}{3\centerdot 2+\frac{1}{2}\centerdot 2}=\frac{4}{6+1}=\frac{4}{7} \\
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& \\
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& \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}=\frac{\frac{1}{2}\centerdot 12}{\left( \frac{1}{4}-\frac{1}{3} \right)\centerdot 12}=\frac{6}{\frac{12}{4}-\frac{12}{3}}=\frac{6}{3-4}=\frac{6}{-1}=-6 \\
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& \\
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& \frac{3}{2-\frac{2}{7}}=\frac{3\centerdot 7}{\left( 2-\frac{2}{7} \right)\centerdot 7}=\frac{21}{2\centerdot 7-\frac{2}{7}\centerdot 7}=\frac{21}{14-2}=\frac{21}{12} \\
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\end{align}</math>
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The whole expression therefore equals
The whole expression therefore equals
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{{Displayed math||<math>\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}}\,</math>.}}
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<math>\frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}</math>
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If we multiply the tops and bottoms of the fractions 4/7, 1/2 and 21/12 in the main fraction by their lowest common denominator, <math>7\cdot 12</math>, we obtain integers in the numerator and denominator
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{{Displayed math||<math>\begin{align}
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\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}} &= \frac{\left( \dfrac{4}{7}-6 \right)\cdot 7\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{2}-\dfrac{21}{12} \right)\cdot 7\cdot 12\vphantom{\Biggl(}} = \frac{4\cdot 12-6\cdot 7\cdot 12}{7\cdot 6-21\cdot 7}\\[10pt]
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& =\frac{( 4-6\cdot 7)\cdot 12}{( 6-21)\cdot 7} = \frac{-38\cdot 12}{-15\cdot 7} = \frac{38\cdot 12}{15\cdot 7}\,\textrm{.}
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\end{align}</math>}}
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By factorizing 12, 15 and 38,
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If we multiply the tops and bottoms of the fractions
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{{Displayed math||<math>\begin{align}
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<math></math>
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12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\
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15 &= 3\cdot 5\,,\\
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<math>{4}/{7}\;</math>
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38 &= 2\cdot 19\,,\\
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,
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\end{align}</math>}}
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<math>{1}/{2}\;</math>
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and
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<math>{21}/{12}\;</math>
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in the main fraction by
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their lowest common denominator,
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<math>7\centerdot 12</math>
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, we obtain integers in the numerator and denominator:
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<math>\begin{align}
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& \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}=\frac{\left( \frac{4}{7}-6 \right)\centerdot 7\centerdot 12}{\left( \frac{1}{2}-\frac{21}{12} \right)\centerdot 7\centerdot 12}=\frac{4\centerdot 12-6\centerdot 7\centerdot 12}{7\centerdot 6-21\centerdot 7} \\
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& \\
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& =\frac{\left( 4-6\centerdot 7 \right)\centerdot 12}{\left( 6-21 \right)\centerdot 7}=\frac{-38\centerdot 12}{-15\centerdot 7}=\frac{38\centerdot 12}{15\centerdot 7} \\
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\end{align}</math>
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By factorizing
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<math>12</math>
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,
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<math>15</math>
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and
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<math>38</math>
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,
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<math>\begin{align}
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& 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\
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& 15=3\centerdot 5 \\
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& 38=2\centerdot 19 \\
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\end{align}</math>
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the answer can be simplified to
the answer can be simplified to
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{{Displayed math||<math>\frac{38\cdot 12}{15\cdot 7}=\frac{2\cdot 19\cdot 2\cdot 2\cdot{}\rlap{/}3}{\rlap{/}3\cdot 5\cdot 7}=\frac{153}{35}\,</math>.}}
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<math>\frac{38\centerdot 12}{15\centerdot 7}=\frac{2\centerdot 19\centerdot 2\centerdot 2\centerdot 3}{3\centerdot 5\centerdot 7}=\frac{153}{35}</math>
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Revision as of 14:22, 19 September 2008

When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts

\displaystyle \frac{2}{\,3+\dfrac{1}{2}\vphantom{\Biggl(}\,}\ ,\quad \frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\,\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}\,}\quad\text{and}\quad\frac{3}{\,2-\dfrac{2}{7}\vphantom{\Biggl(}\,}\,.

We can do this by multiplying the top and bottom of each fraction by 2, 12 and 7 respectively, so as to get rid of the partial fractions

\displaystyle \begin{align}

\frac{2}{3+\dfrac{1}{2}\vphantom{\Biggl(}} &= \frac{2\cdot 2}{\left( 3+\dfrac{1}{2} \right)\cdot 2\vphantom{\Biggl(}} = \frac{4}{3\cdot 2+\dfrac{1}{2}\cdot 2\vphantom{\Biggl(}} = \frac{4}{6+1} = \frac{4}{7}\,,\\[5pt] \frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}} &= \frac{\dfrac{1}{2}\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{4}-\dfrac{1}{3} \right)\cdot 12\vphantom{\Biggl(}} = \frac{6}{\dfrac{12}{4}-\dfrac{12}{3}\vphantom{\Biggl(}} = \frac{6}{3-4} = \frac{6}{-1} = -6\,,\\[10pt] \frac{3}{2-\dfrac{2}{7}\vphantom{\Biggl(}} &= \frac{3\cdot 7}{\left( 2-\dfrac{2}{7} \right)\cdot 7\vphantom{\Biggl(}} = \frac{21}{2\cdot 7-\dfrac{2}{7}\cdot 7\vphantom{\Biggl(}} = \frac{21}{14-2} = \frac{21}{12}\,\textrm{.} \end{align}

The whole expression therefore equals

\displaystyle \frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}}\,.

If we multiply the tops and bottoms of the fractions 4/7, 1/2 and 21/12 in the main fraction by their lowest common denominator, \displaystyle 7\cdot 12, we obtain integers in the numerator and denominator

\displaystyle \begin{align}

\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}} &= \frac{\left( \dfrac{4}{7}-6 \right)\cdot 7\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{2}-\dfrac{21}{12} \right)\cdot 7\cdot 12\vphantom{\Biggl(}} = \frac{4\cdot 12-6\cdot 7\cdot 12}{7\cdot 6-21\cdot 7}\\[10pt] & =\frac{( 4-6\cdot 7)\cdot 12}{( 6-21)\cdot 7} = \frac{-38\cdot 12}{-15\cdot 7} = \frac{38\cdot 12}{15\cdot 7}\,\textrm{.} \end{align}

By factorizing 12, 15 and 38,

\displaystyle \begin{align}

12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\ 15 &= 3\cdot 5\,,\\ 38 &= 2\cdot 19\,,\\ \end{align}

the answer can be simplified to

\displaystyle \frac{38\cdot 12}{15\cdot 7}=\frac{2\cdot 19\cdot 2\cdot 2\cdot{}\rlap{/}3}{\rlap{/}3\cdot 5\cdot 7}=\frac{153}{35}\,.
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