Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 1.2:3b

From Förberedande kurs i matematik 1

Revision as of 08:42, 19 September 2008 by Tek (Talk | contribs)
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
Jump to: navigation, search

If we divide the denominators in succession by 2, we see that

244016=2223=2225=2222

i.e. they all have a factor 222=8 in common,

138+158−128.

Hence we do not need to take 8 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, 2, 3 and 5,

1253825+1235823−1352835=10240+6240−15240.

The LCD is 240 and the answer is

10240+6240−15240=24010+6−15=1240.
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_1.2:3b"