Solution 1.2:2d
From Förberedande kurs i matematik 1
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| - | {{ | + | If we divide up the denominators into their smallest possible integer factors, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | 45&=5\cdot 9=5\cdot 3\cdot 3\,, \\ | ||
| + | 75&=3\cdot 25=3\cdot 5\cdot 5\,, \\ | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | the expression can be written as | ||
| + | |||
| + | {{Displayed math||<math>\frac{2}{3\cdot 3\cdot 5}+\frac{1}{3\cdot 5\cdot 5}</math>}} | ||
| + | |||
| + | and then we see that the denominators have <math>3\cdot 5</math> as a common factor. Therefore, if we multiply the top and bottom of the first fraction by 5 | ||
| + | and the second by 3, the result is the lowest possible denominator | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{2}{3\cdot 3\cdot 5}\cdot \frac{5}{5}+\frac{1}{3\cdot 5\cdot 5}\cdot | ||
| + | \frac{3}{3} &=\frac{2}{3\cdot 3\cdot 5\cdot 5} | ||
| + | +\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt] | ||
| + | &= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\ | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The lowest common denominator is 225. | ||
Current revision
If we divide up the denominators into their smallest possible integer factors,
\displaystyle \begin{align}
45&=5\cdot 9=5\cdot 3\cdot 3\,, \\ 75&=3\cdot 25=3\cdot 5\cdot 5\,, \\ \end{align} |
the expression can be written as
| \displaystyle \frac{2}{3\cdot 3\cdot 5}+\frac{1}{3\cdot 5\cdot 5} |
and then we see that the denominators have \displaystyle 3\cdot 5 as a common factor. Therefore, if we multiply the top and bottom of the first fraction by 5 and the second by 3, the result is the lowest possible denominator
\displaystyle \begin{align}
\frac{2}{3\cdot 3\cdot 5}\cdot \frac{5}{5}+\frac{1}{3\cdot 5\cdot 5}\cdot
\frac{3}{3} &=\frac{2}{3\cdot 3\cdot 5\cdot 5}
+\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt]
&= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\
\end{align} |
The lowest common denominator is 225.
