1. Multiply and express the result as a rational number in the standard form.
(i) \( \frac{11}{7} \) by \( \frac{-3}{8} \)
Solution
\[ \begin{align*} &= \frac{11}{7} \times \frac{-3}{8} \\ \\ &= \frac{11 \times (-3)}{7 \times 8}\\ \\ &= \frac{-33}{56} \end{align*} \]
Answer \( \color{red} \frac{-33}{56} \)
(ii) \( \frac{-7}{4} \) by \( \frac{2}{3} \)
Solution
\[ \begin{align*} &= \frac{-7}{4} \times \frac{2}{3} \\ \\ &= \frac{(-7) \times \cancelto{1}{2}}{\cancelto{2}{4} \times 3}\\ \\ &= \frac{(-7) \times 1}{2 \times 3}\\ \\ &= \frac{-7}{6} \end{align*} \]
Answer \( \color{red} \frac{-7}{6} \)
(iii) \( \frac{-3}{12} \) by \( -48 \)
Solution
\[ \begin{align*} &= \frac{-3}{12} \times \frac{-48}{1} \\ \\ &= \frac{(-3) \times (-\cancelto{4}{48})}{\cancelto{1}{12} \times 1}\\ \\ &= \frac{(-3) \times (-4)}{1}\\ \\ &= 12 \end{align*} \]
Answer \( \color{red} 12 \)
(iv) \( \frac{-14}{9} \) by \( \frac{-3}{7} \)
Solution
\[ \begin{align*} &= \frac{-14}{9} \times \frac{-3}{7} \\ \\ &= \frac{(-\cancelto{2}{14}) \times (-\cancelto{1}{3})}{\cancelto{3}{9} \times \cancelto{1}{7}}\\ \\ &= \frac{{-2} \times (-1)}{3 \times 1}\\ \\ &= \frac{2}{3} \end{align*} \]
Answer \( \color{red} \frac{2}{3} \)
(v) \( \frac{23}{5} \) by \( \frac{-25}{11} \)
Solution
\[ \begin{align*} &= \frac{23}{\cancelto{1}{5}} \times \frac{-\cancelto{5}{25}}{11} \\ \\ &= \frac{23 \times (-5)}{1 \times 11}\\ \\ &= \frac{-115}{11} \end{align*} \]
Answer \( \color{red} \frac{-115}{11} \)
(vi) \( 7 \) by \( \frac{-15}{63} \)
Solution
\[ \begin{align*} &= \frac{\cancelto{1}{7}}{1} \times \frac{-15}{\cancelto{9}{63}} \\ \\ &= \frac{-\cancelto{5}{15}}{\cancelto{3}{9}}\\ \\ &= \frac{-5}{3} \end{align*} \]
Answer \( \color{red} \frac{-5}{3} \)
2. For the following values of xand y, verify \( \color{red} x \times y = y \times x\)
(i) \( x=\frac{7}{9} , y = \frac{3}{2} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{7}{\cancelto{3}{9}} \times \frac{\cancelto{1}{3}}{2}\\ \\ & = \boxed{\frac{7}{6}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{\cancelto{1}{3}}{2} \times \frac{7}{\cancelto{3}{9}}\\ \\ & = \boxed{\frac{7}{6}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]
(ii) \( x = \frac{-2}{7}, y = \frac{5}{8} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{-\cancelto{1}{2}}{7} \times \frac{5}{\cancelto{4}{8}}\\ \\ & = \boxed{\frac{-5}{28}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{5}{\cancelto{4}{8}} \times \frac{-\cancelto{1}{2}}{7}\\ \\ & = \boxed{\frac{-5}{28}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]
(iii) \( x = \frac{4}{9}, y = \frac{-5}{11} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{4}{9} \times \frac{-5}{11}\\ \\ & = \boxed{\frac{-20}{99}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{-5}{11} \times \frac{4}{9}\\ \\ & = \boxed{\frac{-20}{99}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]
(iv) \( x = \frac{-17}{48}, y = \frac{-96}{51} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{-\cancelto{1}{17}}{\cancelto{1}{48}} \times \frac{-\cancelto{2}{96}}{\cancelto{3}{51}}\\ \\ & = \frac{-1 \times -2}{3}\\ \\ & = \boxed{\frac{ 2}{3}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{-\cancelto{2}{96}}{\cancelto{3}{51}} \times \frac{-\cancelto{1}{17}}{\cancelto{1}{48}}\\ \\ & = \frac{-2 \times -1}{3}\\ \\ & = \boxed{\frac{2}{3}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]
3. For the following values of x, yand z, find the products \( (x \times y) \times z \) and \( x \times (y \times z)\) and observe the result \( \color{red}(x \times y) \times z = x \times (y \times z) \).
(i) \( x=\frac{3}{5} , y = \frac{-7}{3} , z = \frac{8}{11} \)
Answer
\[ \begin{aligned} \text{LHS} & =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{\cancelto{1}{3}}{5} \times \frac{-7}{\cancelto{1}{3}}\right) \times \frac{8}{11}\\ \\ & = \frac{-7}{5} \times \frac{8}{11}\\ \\ & = \boxed{\frac{-56}{55}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{3}{5} \times \left(\frac{-7}{3} \times \frac{8}{11}\right)\\ \\ & = \frac{\cancelto{1}{3}}{5} \times \frac{-56}{\cancelto{11}{33}} \\ \\ & = \boxed{\frac{-56}{55}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]
(ii) \( x = \frac{-7}{11}, y = \frac{4}{5}, z = \frac{3}{8} \)
Answer
\[ \begin{aligned} \text{LHS} & =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{-7}{11} \times \frac{4}{5}\right) \times \frac{3}{8}\\ \\ & = \frac{-\cancelto{7}{28}}{55} \times \frac{3}{\cancelto{2}{8}}\\ \\ & = \boxed{\frac{-21}{110}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{-7}{11} \times \left(\frac{\cancelto{1}{4}}{5} \times \frac{3}{\cancelto{2}{8}}\right)\\ \\ & = \frac{-7}{11} \times \frac{3}{10} \\ \\ & = \boxed{\frac{-21}{110}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]
(iii) \( x = \frac{-4}{7}, y = \frac{-3}{8}, z = \frac{16}{5} \)
Answer
\[ \begin{aligned} \text{LHS} & =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{-4}{7} \times \frac{-3}{8}\right) \times \frac{16}{5}\\ \\ & = \left(\frac{-\cancelto{1}{4}}{7} \times \frac{-3}{\cancelto{2}{8}}\right) \times \frac{16}{5}\\ \\ & = \left(\frac{-1}{7} \times \frac{-3}{2}\right) \times \frac{16}{5}\\ \\ & = \frac{3}{\cancelto{7}{14}} \times \frac{\cancelto{8}{16}}{5} \\ \\ & = \boxed{\frac{24}{35}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{-4}{7} \times \left(\frac{-3}{8} \times \frac{16}{5}\right)\\ \\ & = \frac{-4}{7} \times \left(\frac{-3}{\cancelto{1}{8}} \times \frac{\cancelto{2}{16}}{5}\right)\\ \\ & = \frac{-4}{7} \times \left(\frac{-3}{1} \times \frac{2}{5}\right)\\ \\ & = \frac{-4}{7} \times \frac{-6}{5} \\ \\ & = \boxed{\frac{24}{35}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]
(iv) \( x = -3, y = \frac{-4}{9}, z = \frac{-7}{3} \)
Answer
\[ \begin{aligned} \text{LHS}& =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{-3}{1} \times \frac{-4}{9}\right) \times \frac{-7}{3}\\ \\ & = \left(\frac{-\cancelto{1}{3}}{1} \times \frac{-4}{\cancelto{3}{9}}\right) \times \frac{-7}{3}\\ \\ & = \frac{4}{3} \times \frac{-7}{3}\\ \\ & = \boxed{\frac{-28}{9}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{-3}{1} \times \left(\frac{-4}{9} \times \frac{-7}{3}\right)\\ \\ & = \frac{-\cancelto{1}{3}}{1} \times \frac{28}{\cancelto{9}{27}} \\ \\ & = \frac{-1 \times 28}{1 \times 9} \\ \\ & = \boxed{\frac{-28}{9}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]
4. Verify the property \( \color{red} x \times (y + z) = x \times y + x \times z \) by taking -
(i) \( x=\frac{1}{3} , y = \frac{1}{5} , z = \frac{1}{7} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} &\text{LHS}\\ & =\color{green} x \times (y + z) \\ \\ \\ & = \frac{1}{3} \times \left(\frac{1}{5} + \frac{1}{7}\right) \\ \\ \\ & = \frac{1}{3} \times \left(\frac{7 + 5}{35}\right) \\ \\ \\ & = \frac{1}{\cancelto{1}{3}} \times \frac{\cancelto{4}{12}}{35} \\ \\ \\ & = \boxed{\frac{4}{35}} \\ \\ \end{aligned} & \begin{aligned} &\text{RHS} \\ & =\color{green} x \times y + x \times z \\ \\ & = \frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{7}\\ \\ & = \frac{1}{15} + \frac{1}{21}\\ \\ &\text{LCM} = 105 \\ \\ & = \frac{7 + 5}{105} \\ \\ & = \frac{\cancelto{4}{12}}{\cancelto{35}{105}} \\ \\ & = \boxed{\frac{4}{35}} \\ \\ \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence} & \text{ Verified} \\ \color{red} x \times (y + z) &= \color{red} x \times y + x \times z \end{align*} \]
(ii) \( x=\frac{-3}{7} , y = \frac{2}{5} , z = \frac{-4}{9} \)
Answer
\[ \begin{aligned} \text{LHS} & =\color{green} x \times (y + z) \\ \\ \\ & = \frac{-3}{7} \times \left(\frac{2}{5} + \frac{-4}{9}\right) \\ \\ & = \frac{-3}{7} \times \left(\frac{18 - 20}{45}\right) \\ \\ & = \frac{-\cancelto{1}{3}}{7} \times \frac{-2}{\cancelto{15}{45}} \\ \\ & = \frac{-1}{7} \times \frac{-2}{15} \\ \\ & = \boxed{\frac{2}{105}} \\ \\ \text{RHS} & =\color{green} x \times y + x \times z \\ \\ \\ & = \frac{-3}{7} \times \frac{2}{5} + \frac{-\cancelto{1}{3}}{7} \times \frac{-4}{\cancelto{3}{9}}\\ \\ \\ & = \frac{-6}{35} + \frac{4}{21}\\ \\ &\text{LCM} = 105 \\ \\ & = \frac{-18 + 20}{105} \\ \\ & = \boxed{\frac{2}{105}} \\ \\ \end{aligned} \] \[ \begin{align*} \text{Hence} & \text{ Verified} \\ \color{red} x \times (y + z) &= \color{red} x \times y + x \times z \end{align*} \]
5. Show that \( \color{red} \frac{-4}{3} \times \left(\frac{2}{5} + \frac{-7}{10}\right) = \left(\frac{-4}{3} \times \frac{2}{5}\right) + \left(\frac{-4}{3} \times \frac{-7}{10}\right) \)
Answer
\[ \begin{aligned} \text{LHS}& =\color{green} \frac{-4}{3} \times \left(\frac{2}{5} + \frac{-7}{10}\right) \\ \\ & = \frac{-4}{3} \times \left(\frac{4 - 7}{10}\right) \\ \\ & = \frac{-\cancelto{2}{4}}{\cancelto{1}{3}} \times \frac{-\cancelto{1}{3}}{\cancelto{5}{10}} \\ \\ & = \frac{-2 \times -1}{1 \times 5} \\ \\ & = \boxed{\frac{2}{5}} \\ \\ \text{RHS} & =\color{green} \left(\frac{-4}{3} \times \frac{2}{5}\right) + \left(\frac{-4}{3} \times \frac{-7}{10}\right) \\ \\ & =\left(\frac{-4}{3} \times \frac{2}{5}\right) + \left( \frac{-\cancelto{2}{4}}{3} \times \frac{-7}{\cancelto{5}{10}} \right) \\ \\ & = \frac{-8}{15} + \frac{14}{15} \\ \\ & = \frac{-8 + 14}{15} \\ \\ & = \frac{\cancelto{2}{6}}{\cancelto{5}{15}} \\ \\ & = \boxed{\frac{2}{5}} \\ \\ \end{aligned} \] \[ \begin{align*} \text{Hence Verified} \\ \end{align*} \] \( \color{red} \frac{-4}{3} \times \left(\frac{2}{5} + \frac{-7}{10}\right) = \left(\frac{-4}{3} \times \frac{2}{5}\right) + \left(\frac{-4}{3} \times \frac{-7}{10}\right) \)
6 . Show that \( \color{red} \frac{3}{5} \times \left(\frac{-1}{7} - \frac{5}{14}\right) = \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{5} \times \frac{5}{14}\right) \)
Answer
\[ \begin{aligned} \text{LHS} & =\color{green} \frac{3}{5} \times \left(\frac{-1}{7} - \frac{5}{14}\right) \\ \\ & = \frac{3}{5} \times \left(\frac{-2 - 5}{14}\right) \\ \\ & = \frac{3}{5} \times \frac{-\cancelto{1}{7}}{\cancelto{2}{14}} \\ \\ & = \frac{3 \times (-1)}{5 \times 2} \\ \\ & = \boxed{\frac{-3}{10}} \\ \\ \text{RHS} & =\color{green} \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{5} \times \frac{5}{14}\right) \\ \\ & = \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{\cancelto{1}{5}} \times \frac{\cancelto{1}{5}}{14}\right) \\ \\ & = \frac{-3}{35} - \frac{3}{14} \\ \\ & = \frac{-6 - 15}{70} \\ \\ & = \frac{-21}{70} \\ \\ & = \boxed{\frac{-3}{10}} \\ \\ \end{aligned} \] \[ \begin{align*} &\text{Hence Verified} \\ \end{align*} \] \( \color{red} \frac{3}{5} \times \left(\frac{-1}{7} - \frac{5}{14}\right) = \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{5} \times \frac{5}{14}\right) \)
7. Simplify and express the result in standard form.
(i) \( -4 \times \left(\frac{7}{3} - \frac{9}{10}\right) \)
Solution
\[ \begin{aligned} & = \frac{-4}{1} \times \left(\frac{7}{3} - \frac{9}{10}\right) \\ \\ & = \frac{-4}{1} \times \left(\frac{70-27}{30}\right) \\ \\ & = \frac{-\cancelto{2}{4}}{1} \times \frac{43}{\cancelto{15}{30}} \\ \\ & = \boxed{\frac{-86}{15}} \\ \\ \end{aligned} \]
Answer \( \color{red}{\frac{-86}{15}} \)
(ii) \( \frac{7}{3} \times \left(\frac{9}{8} + 3\right) \)
Solution
\[ \begin{aligned} & = \frac{7}{3} \times \left(\frac{9}{8} + \frac{3}{1}\right) \\ \\ & = \frac{7}{3} \times \left(\frac{9+24}{8}\right) \\ \\ & = \frac{7}{\cancelto{1}{3}} \times \frac{\cancelto{11}{33}}{8} \\ \\ & = \boxed{\frac{77}{8}} \\ \\ \end{aligned} \]
Answer \( \color{red}{\frac{77}{8}} \)
(iii) \( \left(\frac{-4}{3} + \frac{5}{7}\right) \times \frac{7}{9} \)
Solution
\[ \begin{aligned} & = \left(\frac{-4}{3} + \frac{5}{7}\right) \times \frac{7}{9} \\ \\ & = \left(\frac{-28 + 15}{21}\right) \times \frac{7}{9} \\ \\ & = \frac{-13}{\cancelto{3}{21}} \times \frac{\cancelto{1}{7}}{9} \\ \\ & = \boxed{\frac{-13}{27}} \\ \\ \end{aligned} \]
Answer \( \color{red}{\frac{-13}{27}} \)
(iv) \( \left(\frac{5}{4} - \frac{6}{20}\right) \times \frac{8}{11} \)
Solution
\[ \begin{aligned} & = \left(\frac{5}{4} - \frac{6}{20}\right) \times \frac{8}{11} \\ \\ & = \left(\frac{25 - 6}{20}\right) \times \frac{8}{11} \\ \\ & = \frac{19}{\cancelto{5}{20}} \times \frac{\cancelto{2}{8}}{11} \\ \\ & = \boxed{\frac{38}{55}} \\ \\ \end{aligned} \]
Answer \( \color{red}{\frac{38}{55}} \)
8. Fill in the blanks
(i) \( \frac{-4}{7} \times \) \( 1 \) \( = \frac{-4}{7} \)
(ii) \( \frac{3}{8} \times \) \( -1 \) \( = \frac{-3}{8} \)
(iii) \(\left(\frac{-1}{3} \times \frac{4}{5}\right) \times \frac{6}{7} = \) \( \frac{-1}{3} \) \( \times \left(\frac{4}{5} \times \frac{6}{7} \right)\)
(iv) \( \frac{5}{3} \times \left(\frac{-7}{8} \times \frac{11}{3}\right) = (\frac{5}{3} \times \) \( \frac{-7}{8} \) \( )\times \frac{11}{3} \)
(v) \( \frac{3}{7} \times \frac{-6}{11} = \frac{-6}{11} \times \) \( \frac{3}{7} \)
(vi) \( \frac{4}{3} \times \) \( 0 \) \( = 0 \)
(vii) For any rational number \( x \), \( x \times 5 = x + x ... = \) \( 5 \) times.
(viii) \( \frac{2}{3} \times \left(\frac{7}{5} - \frac{2}{9}\right) = \frac{2}{3} \times \frac{7}{5} - \) \( \frac{2}{3} \times \frac{2}{9} \)
(ix) \( \frac{-5}{7} \times \frac{1}{3} + \frac{-5}{7} \times \frac{1}{6} = \frac{-5}{7} \times (\frac{1}{3} + \) \( \frac{1}{6} \) )
(x) \( \frac{4}{7} \times \frac{2}{3} - \frac{4}{5} \times \frac{5}{6} = \frac{4}{7} \times \) \( \left(\frac{2}{3} - \frac{5}{6}\right) \)