Lösung 1.2:5c

Aus Online Mathematik Brückenkurs 1

Version vom 13:30, 22. Okt. 2008

Method 1

We calculate the numerator and denominator first

310−5187−316=310−5212=103−2=110=8272−316=1614−3=1611.

Thus, the expression becomes

87−316310−51=1611110=161111161101116=161011

and because 16=2222 and 10=25, the simplified answer is

161011=25112222=855.

Method 2

If we look at the individual fractions 3/10, 1/5, 7/8 and 3/16, we see that the denominators can be factorized as

10=258=222and16=2222

and therefore 2∙2∙2∙2∙5 = 80 is the fractions' lowest common denominator.

If we multiply the top and bottom of the main fraction by 80, then it will be possible to eliminate all denominators at once,

87−316310−51=87−31680310−5180=10380−51808780−16380=103810−582587810−163165=710−338−82=855.