## Do I simplify the fractions in parenthesis?

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LoopyRider
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Joined: Thu Feb 12, 2009 12:05 am
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### Do I simplify the fractions in parenthesis?

This problem
1/2 + (fraction divided by another fraction) - (two other fractions added together) = ?

After dividing and adding the two fractions in each parenthesis.
Do I simplify the fractions in the parenthesis before continuing? Or?

DAiv
Posts: 35
Joined: Tue Dec 16, 2008 7:47 pm
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### Re: Do I simplify the fractions in parenthesis?

This problem
1/2 + (fraction divided by another fraction) - (two other fractions added together) = ?

After dividing and adding the two fractions in each parenthesis.
Do I simplify the fractions in the parenthesis before continuing? Or?
It depends on what the fractions are.

Ideally, we want all the fractions to have the same denominator (bottom number), so if part way through we had:

$\frac12\,+\,\left(\frac3{16}\right)-\left(\frac28\right)$

... we can see that we'll need to convert everything to sixteenths before proceeding:

$\frac8{16}\,+\,\left(\frac3{16}\right)-\left(\frac4{16}\right)\,=\,\frac7{16}$

If we had simplied first, we'd have had:

$\frac12\,+\,\left(\frac3{16}\right)-\left(\frac14\right)$

... which would have been a wasted step as we'd still have to convert everything to sixteenths in the next step.

Of course, had the fractions been:

$\frac12\,+\,\left(\frac{10}4\right)-\left(\frac48\right)$

... then simplifying would have given us our common denominator of 2:

$\frac12\,+\,\left(\frac52\right)-\left(\frac12\right)\,=\,\frac52$

... but getting the common denominator of 2 through simplification would have just been a coincidence.

DAiv