1. Find the value of
(i) \( \frac{6}{7} - \frac{-5}{7} \)
Solution
\[ \begin{align*} &= \frac{6 - (-5)}{7}\\ \\ &= \frac{6 + 5}{7}\\ \\ &= \frac{11}{7} \end{align*} \]
Answer \( \color{red} \frac{11}{7} \)
(ii) \( \frac{5}{24} - \frac{7}{36} \)
Solution
\[ \begin{array}{c|c} 2 & 24, 36 \\ \hline 2 & 12, 18 \\ \hline 2 & 6, 9 \\ \hline 3 & 3, 9 \\ \hline 3 & 1, 3\\ \hline & 1, 1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 72 \\ \\ &= \frac{(5 \times 3) - (7 \times 2)}{72} \\ \\ &= \frac{15 - 14}{72} \\ \\ &= \frac{1}{72} \end{align*} \]
Answer \( \color{red} \frac{1}{72} \)
(iii) \( \frac{9}{10} - \frac{7}{-15} \)
Solution
\[ \begin{align*} &\frac{9}{10} - \frac{-7}{15} \end{align*} \] \[ \begin{array}{c|c} 2 & 10, 15 \\ \hline 3 & 5, 15 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 30 \\ \\ &= \frac{(9 \times 3) + (-7 \times 2)}{30} \\ \\ &= \frac{27 - (-14)}{30} \\ \\ &= \frac{27 + 14}{30} \\ \\ &= \frac{41}{30} \end{align*} \]
Answer \( \color{red} \frac{41}{30} \)
(iv) \( \frac{-3}{8} - \frac{-6}{20} \)
Solution
\[ \begin{array}{c|c} 2 & 8, 20 \\ \hline 2 & 4, 10 \\ \hline 2 & 2, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 40 \\ \\ &= \frac{(-3 \times 5) - (-6 \times 2)}{40} \\ \\ &= \frac{-15 - (-12)}{40} \\ \\ &= \frac{-15 + 12}{40} \\ \\ &= \frac{-3}{40} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{-3}{40} \)
2. Subtract
(i) \( \frac{5}{9} \) from \( \frac{-7}{9} \)
Solution
\[ \begin{align*} & = \frac{-7}{9} - \frac{5}{9} \\ \\ &= \frac{-7 - 5}{9}\\ \\ &= \frac{-\cancelto{4}{12}}{\cancelto{3}9}\\ \\ &= \frac{-4}{3}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{-4}{3} \)
(ii) \( \frac{-5}{7} \) from \( 0 \)
Solution
\[ \begin{align*} & = 0 - \frac{-5}{7} \\ \\ & = 0 + \frac{5}{7} \\ \\ & = \frac{5}{7} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{5}{7} \)
(iii) \( \frac{5}{11} \) from \( \frac{-8}{23} \)
Solution
\[ \begin{align*} &= \frac{-8}{23} - \frac{5}{11} \\ \\ \text{LCM } &= 253 \\ \\ &= \frac{(-8 \times 11) - (5 \times 23)}{253} \\ \\ &= \frac{-88 - 115}{253} \\ \\ &= \frac{-203}{253} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{-203}{253} \)
(iv) \( \frac{-2}{9} \) from \( \frac{7}{6} \)
Solution
\[ \begin{align*} &= \frac{7}{6} - \frac{-2}{9} \\ \\ \text{LCM } &= 18 \\ \\ &= \frac{(7 \times 3) - (-2 \times 2)}{18} \\ \\ &= \frac{21 - (-4)}{18} \\ \\ &= \frac{21 +4}{18} \\ \\ &= \frac{25}{18} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{25}{18} \)
(3) The sum of two rational numbers is - 5. If one of the number is \( \frac{2}{3} \), find the other.
Solution
\[ \begin{align*} \text{Sum of two numbers} &= -5 \\ \\ \text{One of the number} &= \frac{2}{3} \\ \\ \text{Other number} &= -5 - \frac{2}{3} \\ \\ &= \frac{-5}{1} - \frac{2}{3} \\ \\ &= \frac{-15 -2}{3} \\ \\ &= \frac{-17}{3} \\ \end{align*} \]
Answer The other number is \( \color{red} \frac{-17}{3} \)
(4) What number should be added to \( \frac{-3}{7} \) so as to get 1?
Solution
\( \text{Let the required number be} = x \) \[ \begin{align*} \frac{-3}{7} + x &= 1 \\ \\ x &= 1 + \frac{3}{7} \\ \\ x &= \frac{1}{1} + \frac{3}{7} \\ \\ x &= \frac{7 + 3}{7} \\ \\ x &= \frac{10}{7} \\ \\ \end{align*} \]
Answer Required number is \( \color{red} \frac{10}{7} \)
(5) What number should be subtracted from - 1 so as to get \( \frac{5}{3} \) ?
Solution
\( \text{Let the required number be} = x \) \[ \begin{align*} -1 -x &= \frac{5}{3} \\ \\ -x &= \frac{5}{3} + 1 \\ \\ -x &= \frac{5}{3} + \frac{1}{1} \\ \\ -x &= \frac{5 + 3}{3} \\ \\ -x &= \frac{8}{3} \\ \\ x &= \frac{-8}{3} \\ \\ \end{align*} \]
Answer Required number is \( \color{red} \frac{-8}{3} \)
6. Simplify
(i) \( \frac{-4}{5} - \frac{3}{15} + \frac{7}{20} \)
Solution
\[ \begin{array}{c|c} 5 & 5, 15, 20 \\ \hline 3 & 1, 3, 4 \\ \hline 2 & 1,1,4 \\ \hline 2 & 1,1,2 \\ \hline & 1, 1,1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 60\\ \\ & = \frac{(-4 \times 12) - (3 \times 4) + (7 \times 3)}{60} \\ \\ & = \frac{-48 - 12 + 21}{60} \\ \\ & = \frac{-60 + 21}{60} \\ \\ & = \frac{-39}{60} \\ \\ & = \frac{-13}{20} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{-13}{20} \)
(ii) \( \frac{-5}{13} - \frac{-3}{26} - \frac{9}{-52} \)
Solution
\[ \begin{align*} &= \frac{-5}{13} - \frac{-3}{26} - \frac{-9}{52} \end{align*} \] \[ \begin{array}{c|c} 13 & 13, 26, 52 \\ \hline 2 & 1, 2, 4 \\ \hline 2 & 1,1,2 \\ \hline & 1, 1,1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 52\\ \\ & = \frac{(-5 \times 4) - (-3 \times 2) - (-9 \times 1)}{52} \\ \\ & = \frac{-20 - (-6) - (-9)}{52} \\ \\ & = \frac{-20 + 6 + 9}{52} \\ \\ & = \frac{-20 + 15}{52} \\ \\ & = \frac{-5}{52} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{-5}{52} \)
(iii) \( \frac{7}{24} + \frac{5}{12} - \frac{11}{18} \)
Solution
\[ \begin{array}{c|c} 2 & 24, 12, 18 \\ \hline 2 & 12, 6, 9 \\ \hline 2 & 6, 3, 9 \\ \hline 3 & 3, 3, 9 \\ \hline 3 & 1, 1, 3 \\ \hline & 1, 1, 1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 72\\ \\ & = \frac{(7 \times 3) + (5 \times 6) - (11 \times 4)}{72} \\ \\ & = \frac{21 + 30 - 44}{72} \\ \\ & = \frac{51 - 44}{72} \\ \\ & = \frac{7}{72} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{7}{72} \)
(iv) \( \frac{-11}{30} - \frac{8}{15} + \frac{7}{6} + \frac{-2}{5} \)
Solution
\[ \begin{array}{c|c} 2 & 30, 15, 6, 5 \\ \hline 3 & 15, 15, 3, 5 \\ \hline 5 & 5, 5, 1, 5 \\ \hline & 1, 1, 1, 1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 30\\ \\ & = \frac{(-11 \times 1) - (8 \times 2) + (7 \times 5) + (-2 \times 6)}{30} \\ \\ & = \frac{-11 - 16 + 35 - 12}{30} \\ \\ & = \frac{-27 + 35 - 12}{30} \\ \\ & = \frac{8 - 12}{30} \\ \\ & = \frac{-4}{30} \\ \\ & = \frac{-2}{15} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{-2}{15} \)
7. Find the values of \( \color{red} x - y \) and \( \color{red} y - x \) for \( \color{red} x =\frac{2}{3} , y = \frac{5}{9} \). Are they equal ?
Solution
\[ \begin{array}{c|c} \begin{aligned} & \color{green} x -y \\ \\ &= \frac{2}{3} - \frac{5}{9} \\ \\ &= \frac{6 - 5}{9} \\ \\ &= \frac{1}{9} \\ \\ \end{aligned} & \begin{aligned} & \color{green} y - x \\ \\ &= \frac{5}{9} - \frac{2}{3} \\ \\ &= \frac{5 - 6}{9} \\ \\ &= \frac{-1}{9} \\ \\ \end{aligned}\\ \end{array}\\ \]
Answer No. They are not equal.
8. For \( \color{red} x=\frac{1}{10} , y= \frac{-3}{5} , z= \frac{7}{20} \), find the values of the expressions \( \color{red} (x - y) - z \) and \( \color{red} x - (y - z) \). Are they equal ?
Solution
\[ \begin{align*} & \color{green} (x -y) - z \\ \\ &= \left(\frac{1}{10} - \frac{-3}{5} \right) - \frac{7}{20} \\ \\ &= \left(\frac{1}{10} + \frac{3}{5} \right) - \frac{7}{20} \\ \\ &= \left(\frac{1 + 6}{10} \right) - \frac{7}{20} \\ \\ &= \frac{7}{10} - \frac{7}{20} \\ \\ &= \frac{14 - 7}{20} \\ \\ &= \frac{7}{20} \\ \\ & \color{green} x - (y - z) \\ \\ &= \frac{1}{10} - \left(\frac{-3}{5} - \frac{7}{20} \right)\\ \\ &= \frac{1}{10} - \left(\frac{-12 - 7}{20} \right)\\ \\ &= \frac{1}{10} - \left(\frac{-19}{20} \right)\\ \\ &= \frac{1}{10} + \frac{19}{20} \\ \\ &= \frac{2 + 19}{20} \\ \\ &= \frac{21}{20} \\ \\ \color{green} (x -y) - z & \neq \color{green} x - (y - z) \\ \end{align*}\\ \]
Answer No. They are not equal.