Solution 1.2:2d
From Förberedande kurs i matematik 1
(Difference between revisions)
m |
m |
||
| Line 8: | Line 8: | ||
the expression can be written as | the expression can be written as | ||
| - | {{Displayed math||<math>\frac{ | + | {{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 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 | ||
| Line 14: | Line 14: | ||
{{Displayed math||<math>\begin{align} | {{Displayed math||<math>\begin{align} | ||
| - | \frac{2}{ | + | \frac{2}{3\cdot 3\cdot 5}\cdot \frac{5}{5}+\frac{1}{3\cdot 5\cdot 5}\cdot |
| - | \frac{3}{3} &=\frac{2}{ | + | \frac{3}{3} &=\frac{2}{3\cdot 3\cdot 5\cdot 5} |
+\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt] | +\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt] | ||
&= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\ | &= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\ | ||
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.
