simplifying complex frac's: [(5x)/(27)]/[(-10xy)/(21)], etc

[tseday871221]

Problem 1: [ (5x) / (27) ] / [ (-10xy) / (21) ]

Problem 2: [ (3/5) + (2/7) ] / [ (1/5) + (5/6) ]

I would like to know an easy method for simplifying these sorts of expressions. Any information would help a lot. Thank you!

 

[arthur ohlsten]

[5x/27] / [-10xy/21]
to divide fractions invert denominator and multiply
[5x/27][21/-10xy]
{[5x][21]} /{[27][-10xy]}
105x/[-270xy] multiply by -/-
-105x/[270xy] cancel likes
-105/[270y]
-21/[54y]
-7/[18y] answer
=================================================
is it?
[3/5 + 2/7] / [1/5+5/6]

[3/5+2/7]=?
place over lowest common denominator 35
[21+10]/35
31/35

1/5+5/6=?
[6+25]/30
31/30

[3/5+2/7] / [1/5 + 5/6]
[31/35] / [ 31/30]
invert and multiply
[31/35][30/31]
{[31][30]} / { 35[31]}
cancel likes
30/35
reduce
6/7 answer

Arthur

 

[soroban]

Re: HELP SOLVING COMPLEX FRACTIONS

Hello, tseday871221!

\(\displaystyle 2)\;\;\L\frac{\frac{3}{5}\,+\,\frac{2}{7}}{\frac{1}{5}\,+\,\frac{5}{6}}\)

Given a "complex" fraction (a fraction with more than two "levels",
. . multiply top and bottom by the LCD of all the denominators.

The denominators are: \(\displaystyle \,5,\,7,\,5,\,6.\;\;\)Their LCD is: \(\displaystyle 210\)


\(\displaystyle \L\frac{210\,\left(\frac{3}{5}\,+\,\frac{2}{7}\right)}{210\,\left(\frac{1}{5}\,+\,\frac{5}{6}\right)} \;=\;
\frac{\sout{210}^{^{42}}\left(\frac{3}{\not{5}}\right)\,+\,\sout{210}^{^{30}}\left(\frac{2}{\not{7}}\right)}{\sout{210}^{^{42}}\left(\frac{1}{\not{5}}\right)\,+\,\sout{210}^{^{35}}\left(\frac{5}{\not{6}}\right)} \;=\;
\frac{126\,+\,60}{42\,+\,175}\;=\;\frac{186}{217}\;=\;\fbox{\frac{6}{7}}\)

 

[Denis]

Or you can go in 2 steps (which often eliminates a large LCD):

step 1 (LCD = 35): 3/5 + 2/7 = 21/35 + 10/35 = 31/35

step 2 (LCD = 30): 1/5 + 5/6 = 6/30 + 25/30 = 31/30

Now you have:
31/35 / (31/30)
= 31/35 * (30/31)
= 30/35
= 6/7