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.
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: $\displaystyle \frac{10}{21}$.
Hope this helps!
$\displaystyle 4\frac{1}{3}$ = $\displaystyle \frac{13}{3}$ = $\displaystyle \frac{13}{3}\frac{7}{7}$ = $\displaystyle \frac{91}{21}$
$\displaystyle 3\frac{6}{7}$ = $\displaystyle \frac{27}{7}$ = $\displaystyle \frac{27}{7}\frac{3}{3}$ = $\displaystyle \frac{81}{21}$
$\displaystyle 4\frac{1}{3} - 3\frac{6}{7}$ = $\displaystyle \frac{91}{21}-\frac{81}{21}$ = $\displaystyle \frac{10}{21}$