Practice Solutions

1. $–4\frac{2}{3}–\left(–\frac{4}{5}\right)$

$$\begin{gathered} –4\frac{2}{3}+\frac{4}{5}=–\frac{14}{3}+\frac{4}{5} \mathrm{LCD}\left(3, 5\right)=15 \\ –\frac{14}{3}\left(\frac{5}{5}\right)+\frac{4}{5}\left(\frac{3}{3}\right) \\ –\frac{70}{15}+\frac{12}{15}=\overline{)–\frac{58}{15}} \end{gathered}$$

2. $3\frac{3}{4}·2\frac{2}{13}$

$\frac{15}{{}_1\cancel{4}}·\frac{{\cancel{28}}^7}{13}=\overline{)\frac{105}{13}}$

3. $\frac{7}{12}–\frac{1}{3}$

$$\begin{gathered} \mathrm{LCD}\left(12, 3\right)=12 \\ \frac{7}{12}–\frac{1}{3}\left(\frac{4}{4}\right) \\ \frac{7}{12}–\frac{4}{12}=\frac{3}{12}=\overline{)\frac{1}{4}} \end{gathered}$$

4. $\frac{9}{10}÷1\frac{1}{2}$

$\frac{9}{10}÷\frac{3}{2}=\frac{{}^3\cancel{9}}{{}_5\cancel{10}}·\frac{{\cancel{2}}^1}{{\cancel{3}}_1}=\overline{)\frac{3}{5}}$

5. $2\frac{3}{4}–\frac{5}{6}$

$$\begin{gathered} \frac{11}{4}–\frac{5}{6} \mathrm{LCD}\left(4, 6\right)=12 \\ \frac{11}{4}\left(\frac{3}{3}\right)–\frac{5}{6}\left(\frac{2}{2}\right)=\frac{33}{12}–\frac{10}{12}=\overline{)\frac{23}{12}} \end{gathered}$$

6. $–3\frac{2}{5}÷1\frac{2}{7}$

$–\frac{17}{5}÷\frac{9}{7}=–\frac{17}{5}·\frac{7}{9}=\overline{)–\frac{119}{45}}$

7. $\frac{3}{4}· 2\frac{2}{9}$

$\frac{3}{4}·\frac{20}{9}=\frac{{}^1\cancel{3}}{{}_1 \cancel{4} }·\frac{{\cancel{20}}^5}{{\cancel{9}}_3}=\overline{)\frac{5}{3}}$

8. $–\frac{2}{5}–5\frac{1}{3}$

$$\begin{gathered} –\frac{2}{5}–\frac{16}{3} \mathrm{LCD}\left(3, 5\right)=15 \\ –\frac{2}{5}\left(\frac{3}{3}\right)–\frac{16}{3}\left(\frac{5}{5}\right)=–\frac{6}{15}–\frac{80}{15}=\overline{)–\frac{86}{15}} \end{gathered}$$

9. $\frac{5}{9}–\frac{5}{6}$

$$\begin{gathered} \mathrm{LCD}\left(9, 6\right)=18 \\ \frac{5}{9}\left(\frac{2}{2}\right)–\frac{5}{6}\left(\frac{3}{3}\right)=\frac{10}{18}–\frac{15}{18}=\overline{)–\frac{5}{18}} \end{gathered}$$

10. $\frac{8}{21}÷\frac{9}{14}$

$\frac{8}{{}_3\cancel{21}}·\frac{{\cancel{14}}^2}{9}=\overline{)\frac{16}{27}}$

11. $2\frac{2}{5}·\frac{9}{16}$

$\frac{{}^3\cancel{12}}{5}·\frac{9}{{\cancel{16}}_4}=\overline{)\frac{27}{20}}$

12. $4\frac{2}{3}–1\frac{1}{8}$

$$\begin{gathered} \frac{14}{3}–\frac{9}{8} \mathrm{LCD}\left(3, 8\right)=24 \\ \frac{14}{3}\left(\frac{8}{8}\right)–\frac{9}{8}\left(\frac{3}{3}\right) \\ \frac{112}{24}–\frac{27}{24}=\overline{)\frac{85}{24}} \end{gathered}$$