got this question and some others.

i got 4and 1 third - 3 and 6 sevenths i need to know the answer and how to do it if you can help please post thanks.

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- Nov 10th 2008, 07:22 AMandyboy179Subtracting Mixed Fractions
got this question and some others.

i got 4and 1 third - 3 and 6 sevenths i need to know the answer and how to do it if you can help please post thanks. - Nov 10th 2008, 07:59 AMHallsofIvy
Okay, I'll type as fast as I can!

1) Change to "improper fractions" 4 is $\displaystyle \frac{12}{3}$ because 12/3= 4. "4 and 1/3" is $\displaystyle \frac{12}{3}+\frac{1}{3}= \frac{13}{3}$. Similarly 3 is $\displaystyle \frac{21}{7}$ because 21/7= 3. "3 and 6 sevenths" is $\displaystyle \frac{21}{7}+ \frac{6}{7}= \frac{27}{6}$.

So "4 and 1/3 minus 3 and 6/7" is $\displaystyle \frac{13}{3}- \frac{27}{7}$. To add or subtract two fractions you need a "common denominator". Since 3 and 7 have no factors in common, we can just multiply the numerator and denominator of the first fraction by 7, $\displaystyle \frac{7(13)}{7(3)}= \frac{91}{21}$, and multiply the numerator and denominator of the second fraction by 3, $\displaystyle \frac{3(27)}{3(7)}= \frac{81}{21}$.

Now the problem is $\displaystyle \frac{91}{21}-\frac{81}{21}$ and we can simply subtract the numerators: $\displaystyle \frac{10}{21}$. - Nov 10th 2008, 08:24 AMandyboy179
thanks for the answer, but i don't really get what you have said in the post. is there any way you can explain the answer in an easierway, if so post again if not don't post. thnaks again for the help.

- Nov 10th 2008, 07:23 PMfrankdent1
Hope this helps!

$\displaystyle 4\frac{1}{3}$ = $\displaystyle \frac{13}{3}$ = $\displaystyle \frac{13}{3}\frac{7}{7}$ = $\displaystyle \frac{91}{21}$

$\displaystyle 3\frac{6}{7}$ = $\displaystyle \frac{27}{7}$ = $\displaystyle \frac{27}{7}\frac{3}{3}$ = $\displaystyle \frac{81}{21}$

$\displaystyle 4\frac{1}{3} - 3\frac{6}{7}$ = $\displaystyle \frac{91}{21}-\frac{81}{21}$ = $\displaystyle \frac{10}{21}$

- Nov 11th 2008, 06:17 AMandyboy179
thanks alot for your help

- Dec 21st 2008, 12:19 AMrick_rine@yahoo.comEstimate first!