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$28\frac{16}{63}-3\frac{9}{14}$

The answer is $24\frac{11}{18}$

I did this in my head to get the answer fast:

28 - 3 = 25

$\frac{16}{63}-\frac{9}{14}$ = $\frac{-49}{126}$

So it becomes $25\frac{126}{126}-1\frac{49}{126}$ and hence the answer $24\frac{11}{18}$

Now I'd like to know how do you formally get (writing it out on paper) from $28\frac{16}{63}-3\frac{9}{14}$ to $25\frac{126}{126}-1\frac{49}{126}$ without actually the tedious multiplication of the whole numbers by the denominators ?

I tried writing (28 - 3)($\frac{16}{63}-\frac{9}{14}$) but it is incorrect because then you'd get a negative result: $-25\frac{49}{126}$

Thanks for the assistance.

dps1212
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1 Answers1

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There are no "tedious multiplications", but you can't escape reduction to a common denominator.

With a borrow, $$28\frac{16}{63}-3\frac{9}{14}=28\frac{32}{126}-3\frac{81}{126}=25\frac{32}{126}-\frac{81}{126}=24\frac{32+126}{126}-\frac{81}{126}=24\frac{77}{126}=24\frac{11}{18}.$$

More generally, when you add or subtract, the fractional parts can range in $(0,2)$ and $(-1,1)$ respectively, so that a carry or a borrow is needed in half of the cases.