Lesson 1
Practice Problem Set 4 - Solutions

15. $\frac{\left(-14\right)\left(12\right)}{\left(-4\right)\left(-7\right)}=\frac{\stackrel{2}{\cancel{\left(14\right)}}\stackrel{3}{\cancel{\left(12\right)}}}{\underset{1}{\cancel{\left(4\right)}}\underset{1}{\cancel{\left(7\right)}}}=\frac{2·3}{1·1}=-6$

16. $\frac{5}{12}-\frac{1}{8}=\frac{5}{3·4}-\frac{1}{8}=\frac{5·2}{3·4·2}-\frac{1·3}{8·3}=\frac{10}{24}-\frac{3}{24}=\frac{7}{24}$

17. $\frac{5}{9}+\frac{2}{3}=\frac{5}{9}+\frac{2·3}{3·3}=\frac{5}{9}+\frac{6}{9}=\frac{11}{9}$

18. $\frac{7}{16}÷\frac{1}{20}=\frac{7}{\underset{4}{\cancel{16}}}·\frac{\stackrel{5}{\cancel{20}}}{1}=\frac{7·5}{4·1}=\frac{35}{4}$

19. $\frac{3}{8}÷9=\frac{3}{8}÷\frac{9}{1}=\frac{\stackrel{1}{\cancel{3}}}{8}·\frac{1}{\underset{3}{\cancel{9}}}=\frac{1·1}{8·3}=\frac{1}{24}$

20. $\frac{3}{8}·\frac{7}{12}=\frac{\stackrel{1}{\cancel{3}}}{8}·\frac{7}{\underset{4}{\cancel{12}}}=\frac{1·7}{8·4}=\frac{7}{32}$