let's do 1, 5, 6, 10 and 13.
1)
well, the denominator is the same on each, so we simply have to look at the numerator, who is larger 3 or 5? 3 < 5, 3 is less than 5, so then........
5)

6)
we can make both denominators the same if we simply <u>multiply each fraction by the other's denominator</u>.

10)
we'll convert the mixed fractions to improper fractions first, then make their denominator the same just like we did in 6).
![\bf \stackrel{mixed}{4\frac{1}{7}}\implies \cfrac{4\cdot 7+1}{7}\implies \stackrel{improper}{\cfrac{29}{7}}~\hfill \stackrel{mixed}{3\frac{3}{18}}\implies \cfrac{3\cdot 18+3}{18}\implies \stackrel{improper}{\cfrac{57}{18}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} \cfrac{29}{7}\cdot \cfrac{18}{18}\implies \cfrac{522}{126} \\\\\\ \cfrac{57}{18}\cdot \cfrac{7}{7}\implies \cfrac{399}{126} \end{cases}\qquad \implies \cfrac{522}{126}>\cfrac{399}{126}\qquad therefore\qquad 4\frac{1}{7}>3\frac{3}{18}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B7%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%207%2B1%7D%7B7%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B29%7D%7B7%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B3%7D%7B18%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%2018%2B3%7D%7B18%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B57%7D%7B18%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20%5Ccfrac%7B29%7D%7B7%7D%5Ccdot%20%5Ccfrac%7B18%7D%7B18%7D%5Cimplies%20%5Ccfrac%7B522%7D%7B126%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B57%7D%7B18%7D%5Ccdot%20%5Ccfrac%7B7%7D%7B7%7D%5Cimplies%20%5Ccfrac%7B399%7D%7B126%7D%20%5Cend%7Bcases%7D%5Cqquad%20%5Cimplies%20%5Ccfrac%7B522%7D%7B126%7D%3E%5Ccfrac%7B399%7D%7B126%7D%5Cqquad%20therefore%5Cqquad%204%5Cfrac%7B1%7D%7B7%7D%3E3%5Cfrac%7B3%7D%7B18%7D)
13)
so both fractions are at a value from 9, so we can simply say, which is larger 2/6 or 4/12?
