Processing Math: Done
Solution 1.2:3b
From Förberedande kurs i matematik 1
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- | {{ | + | If we divide the denominators in succession by |
- | < | + | <math>2</math> |
- | {{ | + | , we see that |
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & 24=2\centerdot 2\centerdot 2\centerdot 3 \\ | ||
+ | & 40=2\centerdot 2\centerdot \centerdot 5 \\ | ||
+ | & 16=2\centerdot 2\centerdot 2\centerdot 2 \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | |||
+ | i.e. they all have a factor | ||
+ | <math>2\centerdot 2\centerdot 2=8</math> | ||
+ | in common, | ||
+ | |||
+ | |||
+ | |||
+ | <math>\frac{1}{3\centerdot 8}+\frac{1}{5\centerdot 8}-\frac{1}{2\centerdot 8}</math> | ||
+ | , | ||
+ | |||
+ | and hence we do not need to take | ||
+ | <math>8</math> | ||
+ | as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we | ||
+ | obtain the lowest common denominator by multiplying top and bottom by the other factors, | ||
+ | <math>2,\ 3</math> | ||
+ | and | ||
+ | <math>5</math> | ||
+ | : | ||
+ | |||
+ | |||
+ | <math>\frac{1\centerdot 2\centerdot 5}{3\centerdot 8\centerdot 2\centerdot 5}+\frac{1\centerdot 2\centerdot 3}{5\centerdot 8\centerdot 2\centerdot 3}-\frac{1\centerdot 3\centerdot 5}{2\centerdot 8\centerdot 3\centerdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}</math> | ||
+ | |||
+ | |||
+ | The LCD is | ||
+ | <math>240</math> | ||
+ | and the answer is | ||
+ | |||
+ | |||
+ | <math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}</math> |
Revision as of 13:10, 11 September 2008
If we divide the denominators in succession by
2
2
340=2
2
516=2
2
2
2
i.e. they all have a factor
2
2=8
8+15
8−12
8
and hence we do not need to take
3
2
53
8
2
5+1
2
35
8
2
3−1
3
52
8
3
5=10240+6240−15240
The LCD is