1.10 Fractions: order of operations

$\frac{1}{4}+\frac{5}{8}\cdot \frac{2}{\text{15}}$

1. Multiply first.
   

   $\frac{1}{4}+\frac{\stackrel{1}{\cancel{5}}}{\underset{4}{\cancel{8}}}\cdot \frac{\stackrel{1}{\cancel{2}}}{\underset{3}{\cancel{\text{15}}}}=\frac{1}{4}+\frac{1\cdot 1}{4\cdot 3}=\frac{1}{4}+\frac{1}{\text{12}}$
   

2. Now perform this addition. Find the LCD.
   

   $\left(\begin{array}{}4=2^2\\ \text{12}=2^2\cdot 3\end{array}\right\}\text{The LCD}=2^2\cdot 3=\text{12}\text{.}$
   

   $\begin{array}{ccc}\hfill \frac{1}{4}+\frac{1}{\text{12}}& =& \frac{1\cdot 3}{\text{12}}+\frac{1}{\text{12}}=\frac{3}{\text{12}}+\frac{1}{\text{12}}\hfill \\ & =& \frac{3+1}{\text{12}}=\frac{4}{\text{12}}=\frac{1}{3}\hfill \end{array}$
   

   Thus, $\frac{1}{4}+\frac{5}{8}\cdot \frac{2}{\text{15}}=\frac{1}{3}$
   

$\frac{3}{5}+\frac{9}{\text{44}}\left(\frac{5}{9}-\frac{1}{4}\right)$

1. Operate within the parentheses first, $\left(\frac{5}{9}-\frac{1}{4}\right)$ .
   

   $\left(\begin{array}{}9=3^2\\ 4=2^2\end{array}\right\}\text{The LCD}=2^2\cdot 3^2=4\cdot 9=\text{36}\text{.}$

   
   

   $\frac{5\cdot 4}{\text{36}}-\frac{1\cdot 9}{\text{36}}=\frac{\text{20}}{\text{36}}-\frac{9}{\text{36}}=\frac{\text{20}-9}{\text{36}}=\frac{\text{11}}{\text{36}}$

   
   

   Now we have

   
   

   $\frac{3}{5}+\frac{9}{\text{44}}\left(\frac{\text{11}}{\text{36}}\right)$

2. Perform the multiplication.
   

   $\frac{3}{5}+\frac{\stackrel{1}{\cancel{9}}}{\underset{4}{\cancel{\text{44}}}}\cdot \frac{\stackrel{1}{\cancel{\text{11}}}}{\underset{4}{\cancel{\text{36}}}}=\frac{3}{5}+\frac{1\cdot 1}{4\cdot 4}=\frac{3}{5}+\frac{1}{\text{16}}$

3. Now perform the addition. The LCD=80.
   

   $\frac{3}{5}+\frac{1}{\text{16}}=\frac{3\cdot \text{16}}{\text{80}}+\frac{1\cdot 5}{\text{80}}=\frac{\text{48}}{\text{80}}+\frac{5}{\text{80}}=\frac{\text{48}+5}{\text{80}}=\frac{\text{53}}{\text{80}}$

   
   

   Thus, $\frac{3}{5}+\frac{9}{\text{44}}\left(\frac{5}{9}-\frac{1}{4}\right)=\frac{\text{53}}{\text{80}}$

$8-\frac{\text{15}}{\text{426}}\left(2-1\frac{4}{\text{15}}\right)\left(3\frac{1}{5}+2\frac{1}{8}\right)$

1. Work within each set of parentheses individually.
   

   $\begin{array}{ccc}2-1\frac{4}{\text{15}}& =& 2\frac{1\cdot \text{15}+4}{\text{15}}=2-\frac{\text{19}}{\text{15}}\hfill \\ & =& \frac{\text{30}}{\text{15}}-\frac{\text{19}}{\text{15}}=\frac{\text{30}-\text{19}}{\text{15}}=\frac{\text{11}}{\text{15}}\hfill \\ 3\frac{1}{5}+2\frac{1}{8}& =& \frac{3\cdot 5+1}{5}+\frac{2\cdot 8+1}{8}\hfill \\ & =& \frac{\text{16}}{5}+\frac{\text{17}}{8}\text{LCD}=\text{40}\hfill \\ & =& \frac{\text{16}\cdot 8}{\text{40}}+\frac{\text{17}\cdot 5}{\text{40}}\hfill \\ & =& \frac{\text{128}}{\text{40}}+\frac{\text{85}}{\text{40}}\hfill \\ & =& \frac{\text{128}+\text{85}}{\text{40}}\hfill \\ & =& \frac{\text{213}}{\text{40}}\hfill \end{array}$

   
   

   Now we have

   $8-\frac{\text{15}}{\text{426}}\left(\frac{\text{11}}{\text{15}}\right)\left(\frac{\text{213}}{\text{40}}\right)$

2. Now multiply.
   

   $8-\frac{\stackrel{1}{\cancel{\text{15}}}}{\underset{2}{\cancel{\text{426}}}}\cdot \frac{\text{11}}{\underset{1}{\cancel{\text{15}}}}\cdot \frac{\stackrel{1}{\cancel{\text{213}}}}{\text{40}}=8-\frac{1\cdot \text{11}\cdot 1}{2\cdot 1\cdot \text{40}}=8-\frac{\text{11}}{\text{80}}$

3. Now subtract.
   

   $8-\frac{\text{11}}{\text{80}}=\frac{\text{80}\cdot 8}{\text{80}}-\frac{\text{11}}{\text{80}}=\frac{\text{640}}{\text{80}}-\frac{\text{11}}{\text{80}}=\frac{\text{640}-\text{11}}{\text{80}}=\frac{\text{629}}{\text{80}}\text{or}7\frac{\text{69}}{\text{80}}$

   
   

   Thus, $8-\frac{15}{426}\left(2-1\frac{4}{15}\right)\left(3\frac{1}{5}+2\frac{1}{8}\right)=7\frac{69}{80}$

${\left(\frac{3}{4}\right)}^2\cdot \frac{8}{9}-\frac{5}{\text{12}}$

1. Square $\frac{3}{4}$ .
   

   ${\left(\frac{3}{4}\right)}^2=\frac{3}{4}\cdot \frac{3}{4}=\frac{3\cdot 3}{4\cdot 4}=\frac{9}{\text{16}}$

   Now we have

   $\frac{9}{\text{16}}\cdot \frac{8}{9}-\frac{5}{\text{12}}$

2. Perform the multiplication.
   

   $\frac{\stackrel{1}{\cancel{9}}}{\underset{2}{\cancel{\text{16}}}}\cdot \frac{\stackrel{1}{\cancel{8}}}{\underset{1}{\cancel{9}}}-\frac{5}{\text{12}}=\frac{1\cdot 1}{2\cdot 1}-\frac{5}{\text{12}}=\frac{1}{2}-\frac{5}{\text{12}}$

3. Now perform the subtraction.
   

   $\frac{1}{2}-\frac{5}{\text{12}}=\frac{6}{\text{12}}-\frac{5}{\text{12}}=\frac{6-5}{\text{12}}=\frac{1}{\text{12}}$

   Thus, ${\left(\frac{4}{3}\right)}^2\cdot \frac{8}{9}-\frac{5}{\text{12}}=\frac{1}{\text{12}}$

$2\frac{7}{8}+\sqrt{\frac{\text{25}}{\text{36}}}÷\left(2\frac{1}{2}-1\frac{1}{3}\right)$

1. Begin by operating inside the parentheses.
   

   $\begin{array}{ccc}2\frac{1}{2}-1\frac{1}{3}& =& \frac{2\cdot 2+1}{2}-\frac{1\cdot 3+1}{3}=\frac{5}{2}-\frac{4}{3}\hfill \\ & =& \frac{\text{15}}{6}-\frac{8}{6}=\frac{\text{15}-8}{6}=\frac{7}{6}\hfill \end{array}$

2. Now simplify the square root.
   

   $\sqrt{\frac{\text{25}}{\text{36}}}=\frac{5}{6}\left(\text{since}{\left(\frac{5}{6}\right)}^2=\frac{\text{25}}{\text{36}}\right)$

   Now we have

   $2\frac{7}{8}+\frac{5}{6}÷\frac{7}{6}$

3. Perform the division.
   

   $2\frac{7}{8}+\frac{5}{\underset{1}{\cancel{6}}}\cdot \frac{\stackrel{1}{\cancel{6}}}{7}=2\frac{7}{8}+\frac{5\cdot 1}{1\cdot 7}=2\frac{7}{8}+\frac{5}{7}$

4. Now perform the addition.
   

   $\begin{array}{ccc}2\frac{7}{8}+\frac{5}{7}& =& \frac{2\cdot 8+7}{8}+\frac{5}{7}=\frac{\text{23}}{8}+\frac{5}{7}\text{LCD}=\text{56}\text{.}\hfill \\ & =& \frac{\text{23}\cdot 7}{\text{56}}+\frac{5\cdot 8}{\text{56}}=\frac{\text{161}}{\text{56}}+\frac{\text{40}}{\text{56}}\hfill \\ & =& \frac{\text{161}+\text{40}}{\text{56}}=\frac{\text{201}}{\text{56}}\text{ or }3\frac{\text{33}}{\text{56}}\hfill \end{array}$

   Thus, $2\frac{7}{8}+\sqrt{\frac{\text{25}}{\text{36}}}÷\left(2\frac{1}{2}-1\frac{1}{3}\right)=3\frac{\text{33}}{\text{56}}$