Solution 1.2:6
From Förberedande kurs i matematik 1
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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 |
| - | < | + | |
| - | < | + | {{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>.}} |
| - | {{ | + | |
| + | 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 | ||
| + | |||
| + | {{Displayed math||<math>\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}</math>}} | ||
| + | |||
| + | The whole expression therefore equals | ||
| + | |||
| + | {{Displayed math||<math>\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}}\,</math>.}} | ||
| + | |||
| + | 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 | ||
| + | |||
| + | {{Displayed math||<math>\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}</math>}} | ||
| + | |||
| + | By factorizing 12, 15 and 38, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | 12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\ | ||
| + | 15 &= 3\cdot 5\,,\\ | ||
| + | 38 &= 2\cdot 19\,,\\ | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | the answer can be simplified to | ||
| + | |||
| + | {{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{152}{35}\,</math>.}} | ||
Current revision
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{152}{35}\,. |
