Solution 1.2:5b
From Förberedande kurs i matematik 1
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{.} |
