DAV Class 7 Maths Chapter 2 Worksheet 1

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DAV Class 7 Maths Chapter 2 Worksheet 1

Operations on Rational Numbers Worksheet 1


1. Add the following:

(i) \( \displaystyle \frac{2}{7} + \frac{5}{7} \)

Solution

\[ \begin{align*} &= \frac{2 + 5}{7} \\ \\ &= \frac{7}{7} \\ \\ &= 1 \end{align*} \]

Answer \( \color{red} 1 \)

(ii) \( \displaystyle \frac{-5}{9} + \frac{7}{9} \)

Solution

\[ \begin{align*} &= \frac{-5 + 7}{9}\\ \\ &= \frac{-5+7}{9}\\ \\ &= \frac{2}{9} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{2}{9} \)

(iii) \( \displaystyle 2\frac{3}{11} + \frac{9}{-11} \)

Solution

\[ \begin{align*} &= \frac{25}{11} + \frac{(-9)}{11}\\ \\ &= \frac{25 + (-9)}{11}\\ \\ &= \frac{25 - 9}{11}\\ \\ &= \frac{16}{11} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{16}{11} \)

(iv) \( \displaystyle \frac{-5}{4} + \frac{5}{4} \)

Solution

\[ \begin{align*} &= \frac{-5 + 5}{4}\\ \\ &= \frac{0}{4}\\ \\ &= 0 \end{align*} \]

Answer \( \color{red} 0 \)

(v) \( \displaystyle -2\frac{5}{6} + \frac{13}{-6} \)

Solution

\[ \begin{align*} &= \frac{-17}{6} + \frac{-13}{6}\\ \\ &= \frac{-17 + (-13)}{6}\\ \\ &= \frac{\cancel{-30}^{-5}}{\cancel6_1}\\ \\ &= -5 \end{align*} \]

Answer \( \color{red} -5 \)

(vi) \( \displaystyle \frac{-21}{3} + \frac{18}{3} \)

Solution

\[ \begin{align*} & \color{green} \boxed{\text{Method - } 1}\\ \\ &= \frac{-21 + 18}{3}\\ \\ &= \frac{\cancel{-3}^{-1}}{\cancel3_1}\\ \\ &= -1 \\ \\ & \color{green} \boxed{\text{Method - } 2}\\ \\ &= \frac{\cancel{-21}^{-7}}{\cancel3_1} + \frac{\cancel{18}^6}{\cancel3_1}\\ \\ &= -7 + 6\\ &= -1 \end{align*} \]

Answer \( \color{red} -1 \)

2. Add the following:

(i) \( \displaystyle \frac{4}{9} + \frac{7}{4} \)

Solution

\[ \begin{align*} & \frac{4}{9} + \frac{7}{4} \\ \\ \text{LCM} & = 36 \\ \\ & = \frac{4 \color{green} \times 4}{9 \color{green} \times 4} + \frac{7 \color{green} \times 9}{4 \color{green} \times 9} \\ \\ &= \frac{16}{36} + \frac{63}{36}\\ \\ &= \frac{ 16 + 63}{36}\\ \\ &= \frac{ 79 }{36} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{ 79 }{36} \)

(ii) \( \displaystyle \frac{-7}{11} + \frac{1}{4} \)

Solution

\[ \begin{align*} &= \frac{-7}{11} + \frac{1}{4} \\ \\ \text{LCM} & = 44 \\ \\ & = \frac{-7 \color{green} \times 4}{11 \color{green} \times 4} + \frac{1 \color{green} \times 11}{4 \color{green} \times 11} \\ \\ &= \frac{ -28}{44} + \frac{11}{44} \\ \\ &= \frac{ -28 + 11}{44}\\ \\ &= \frac{ -17 }{44} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{ -17 }{44} \)

(iii) \( \displaystyle \frac{5}{8} + \frac{-3}{5} \)

Solution

\[ \begin{align*} &= \frac{5}{8} + \frac{-3}{5} \\ \\ \text{LCM} & = 40 \\ \\ &= \frac{5 \color{green} \times 5}{8 \color{green} \times 5} + \frac{-3 \color{green} \times 8}{5 \color{green} \times 8} \\ \\ &= \frac{25}{40} + \frac{-24}{40} \\ \\ &= \frac{ 25 + (-24)}{40}\\ \\ &= \frac{ 25 - 24}{40}\\ \\ &= \frac{ 1 }{40} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{ 1 }{40} \)

(iv) \( \displaystyle \frac{10}{63} + \frac{6}{7} \)

Solution

\[ \begin{align*} &= \frac{10}{63} + \frac{6}{7} \\ \\ \text{LCM} & = 63 \\ \\ &= \frac{10 \color{green} \times 1 }{63 \color{green} \times 1 } + \frac{6 \color{green} \times 9}{7 \color{green} \times 9} \\ \\ &= \frac{10}{63} + \frac{54}{63} \\ \\ &= \frac{10 + 54}{63}\\ \\ &= \frac{64}{63} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{64}{63} \)

(v) \( \displaystyle \frac{-7}{64} + \frac{3}{-16} \)

Solution

\[ \begin{align*} &= \frac{-7}{64} + \frac{-3}{16} \\ \\ \text{LCM} & = 64 \\ \\ &= \frac{-7 \color{green} \times 1}{64 \color{green} \times 1} + \frac{-3 \color{green} \times 4}{16 \color{green} \times 4} \\ \\ &= \frac{-7}{64} + \frac{-12}{16} \\ \\ &= \frac{(-7) + (-12)}{64}\\ \\ &= \frac{-7 - 12}{64}\\ \\ &= \frac{-19}{64} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-19}{64} \)

(vi) \( \displaystyle \frac{5}{12} + \frac{-9}{20} \)

Solution

\[ \begin{align*} &= \frac{5}{12} + \frac{-9}{20} \\ \\ \text{LCM} & = 60 \\ \\ &= \frac{5 \color{green} \times 5}{12 \color{green} \times 5} + \frac{-9 \color{green} \times 3}{20 \color{green} \times 3} \\ \\ &= \frac{25}{60} + \frac{-27}{60} \\ \\ &= \frac{25 + (-27)}{60}\\ \\ &= \frac{25 - 27}{60}\\ \\ &= \frac{\cancel{-2}^{-1}}{\cancel{60}_{30}}\\ \\ &= \frac{-1}{30} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-1}{30} \)

3. Verify \(\color{red} x + y = y + x \) for following values of \( x \text{ and } y \).

(i) \( \displaystyle x=\frac{5}{7} , y = \frac{-3}{2} \)

Answer

\[ \begin{align*} \text{LHS} & = x + y \\ \\ & = \frac{5}{7} + \frac{-3}{2}\\ \\ \text{LCM} & = 14 \\ \\ & = \frac{5 \color{green} \times 2 }{7 \color{green} \times 2 } + \frac{-3\color{green} \times 7}{2\color{green} \times 7}\\ \\ & = \frac{10}{14} + \frac{-21}{14}\\ \\ & = \frac{10 + (-21)}{14} \\ \\ & = \frac{10 -21}{14} \\ \\ & = \color{green} \boxed{\frac{-11}{14}} \\ \\ \text{RHS} & = y + x \\ \\ & = \frac{-3}{2} + \frac{5}{7} \\ \\ \text{LCM} & = 14 \\ \\ & = \frac{-3 \color{green} \times 7}{2 \color{green} \times 7} + \frac{5 \color{green} \times 2}{7 \color{green} \times 2} \\ \\ & = \frac{-21 + 10 }{14} \\ \\ & = \color{green} \boxed{\frac{-11}{14}} \\ \\ &\text{Hence verified} \\ & \color{red} x + y = y + x \end{align*} \]

(ii) \( \displaystyle x=5, y=\frac{3}{2} \)

Answer

\[ \begin{align*} \text{LHS} &= x + y \\ \\ & = \frac{5}{1} + \frac{3}{2} \\ \\ \text{LCM} & = 2 \\ \\ & = \frac{5 \color{green} \times 2}{1 \color{green} \times 2} + \frac{3 \color{green} \times 1}{2 \color{green} \times 1}\\ \\ & = \frac{10 + 3}{2} \\ \\ & = \color{green} \boxed{\frac{13}{2}} \\ \\ \text{RHS} &= y + x \\ \\ & = \frac{3}{2} + \frac{5}{1} \\ \\ \text{LCM} & = 2 \\ \\ & = \frac{(3 \times 1) + (5 \times 2) }{2} \\ \\ & = \frac{3 + 10}{2} \\ \\ & = \color{green} \boxed{\frac{13}{2}} \\ \\ &\text{Hence verified} \\ & \color{red} x + y = y + x \end{align*} \]

(iii) \( \displaystyle x=\frac{-5}{14}, y=\frac{-1}{21} \)

Answer

\[ \begin{align*} \text{LHS} &= x + y \\ \\ & = \frac{-5}{14} + \frac{-1}{21} \\ \\ \text{LCM} & = 42 \\ \\ & = \frac{-5 \color{green} \times 3}{14 \color{green} \times 3} + \frac{-1 \color{green} \times 2}{21 \color{green} \times 2}\\ \\ & = \frac{-15}{42} + \frac{-2}{42}\\ \\ & = \frac{-15 +(-2)}{42} \\ \\ & = \frac{-15 - 2}{42} \\ \\ & = \color{green} \boxed{\frac{-17}{42}} \\ \\ \text{RHS} &= y + x \\ \\ & = \frac{-1}{21} + \frac{-5}{14} \\ \\ \text{LCM} & = 42 \\ \\ & = \frac{-1 \color{green} \times 2}{21 \color{green} \times 2} + \frac{-5 \color{green} \times 3}{14 \color{green} \times 3} \\ \\ & =\frac{-2}{42} + \frac{-15}{42} \\ \\ & = \frac{-2 + (-15)}{42} \\ \\ & = \frac{-2 - 15}{42} \\ \\ & = \color{green} \boxed{\frac{-17}{42}} \\ \\ &\text{Hence verified} \\ & \color{red} x + y = y + x \end{align*} \]

(iv) \( \displaystyle x= -8, y=\frac{9}{2} \)

Answer

\[ \begin{align*} \text{LHS} &= x + y \\ \\ & = \frac{-8}{1} + \frac{9}{2} \\ \\ \text{LCM} & = 2 \\ \\ & = \frac{-8 \color{green} \times 2}{1 \color{green} \times 2} + \frac{9 \color{green} \times 1}{2 \color{green} \times 1}\\ \\ & = \frac{-16}{2} + \frac{9}{2} \\ \\ & = \frac{-16 + 9}{2} \\ \\ & = \color{green} \boxed{\frac{-7}{2}} \\ \\ \text{RHS} &= y + x \\ \\ & = \frac{9}{2} + \frac{-8}{1} \\ \\ \text{LCM} & = 2 \\ \\ & = \frac{9 \color{green} \times 1 }{2 \color{green} \times 1 } + \frac{-8 \color{green} \times 2}{1 \color{green} \times 2} \\ \\ & = \frac{9}{2} + \frac{-16}{2} \\ \\ & = \frac{9 + (-16)}{2} \\ \\ & = \frac{9 - 16}{2} \\ \\ & = \color{green} \boxed{\frac{-7}{2}} \\ \\ &\text{Hence verified} \\ & \color{red} x + y = y + x \end{align*} \]

4. Verify \(\color{red} x + (y + z) = (x+y) + z \) for following values of \( x ,y \text{ and } z \).

(i) \( \displaystyle x=\frac{3}{4}, y=\frac{5}{6}, z=\frac{-7}{8} \)

Answer

\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{3}{4} + \left(\frac{5}{6} + \frac{-7}{8}\right)\\ \\ &\text{LCM of 6 and 8} = 24 \\ \\ & = \frac{3}{4} + \left(\frac{5 \color{green} \times 4}{6 \color{green} \times 4} + \frac{-7 \color{green} \times 3}{8 \color{green} \times 3}\right)\\ \\ & = \frac{3}{4} + \left(\frac{20}{24} + \frac{-21}{24}\right)\\ \\ & = \frac{3}{4} + \left(\frac{20 - 21}{24}\right)\\ \\ & = \frac{3}{4} + \frac{-1}{24}\\ \\ &\text{LCM} = 24 \\ \\ & = \frac{3 \color{green} \times 6}{4 \color{green} \times 6} + \frac{-1 \color{green} \times 1}{24 \color{green} \times 1}\\ \\ & = \frac{18}{24} + \frac{-1}{24}\\ \\ & = \frac{18 + (-1)}{24}\\ \\ & = \frac{18 - 1}{24}\\ \\ & = \color{green} \boxed{\frac{17}{24}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{3}{4} + \frac{5}{6}\right) + \frac{-7}{8} \\ \\ & \text{LCM of 4 and 6} = 12 \\ \\ & = \left(\frac{3 \color{green} \times 3}{4 \color{green} \times 3} + \frac{5 \color{green} \times 2}{6 \color{green} \times 2}\right) + \frac{-7}{8}\\ \\ & = \left(\frac{9}{12} + \frac{10}{12}\right) + \frac{-7}{8}\\ \\ & = \left(\frac{9 + 10}{12}\right) + \frac{-7}{8}\\ \\ & = \frac{19}{12} + \frac{-7}{8}\\ \\ & \text{LCM} = 24 \\ \\ & = \frac{19 \color{green} \times 2}{12 \color{green} \times 2} + \frac{-7 \color{green} \times 3}{8 \color{green} \times 3}\\ \\ & = \frac{38}{24} + \frac{-21}{24}\\ \\ & = \frac{38 + (-21)}{24}\\ \\ & = \frac{38 - 21}{24}\\ \\ & = \color{green} \boxed{\frac{17}{24}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]

(ii) \( \displaystyle x=\frac{2}{3}, y=\frac{-5}{6}, z=\frac{-7}{9} \)

Answer

\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{2}{3} + \left(\frac{-5}{6} + \frac{-7}{9}\right)\\ \\ &\text{LCM of 6 and 9} = 18 \\ \\ & = \frac{2}{3} + \left(\frac{-5 \color{green} \times 3}{6 \color{green} \times 3} + \frac{-7 \color{green} \times 2}{9 \color{green} \times 2}\right)\\ \\ & = \frac{2}{3} + \left(\frac{-15}{18} + \frac{-14}{18}\right)\\ \\ & = \frac{2}{3} + \left(\frac{-15 - 14}{18}\right)\\ \\ & = \frac{2}{3} + \frac{-29}{18}\\ \\ &\text{LCM} = 18 \\ \\ & = \frac{2 \color{green} \times 6 }{3 \color{green} \times 6 } + \frac{-29 \color{green} \times 1}{18 \color{green} \times 1}\\ \\ & = \frac{12}{18} + \frac{-29}{18}\\ \\ & = \frac{12 - 29}{18}\\ \\ & = \color{green} \boxed{\frac{-17}{18}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{2}{3} + \frac{-5}{6}\right) + \frac{-7}{9} \\ \\ & \text{LCM of 3 and 6} = 6 \\ \\ & = \left(\frac{2 \color{green} \times 2}{3 \color{green} \times 2} + \frac{-5 \color{green} \times 1}{6 \color{green} \times 1}\right) + \frac{-7}{9}\\ \\ & = \left(\frac{4}{6} + \frac{-5}{6}\right) + \frac{-7}{9}\\ \\ & = \left(\frac{4 - 5}{6}\right) + \frac{-7}{9}\\ \\ & = \frac{-1}{6} + \frac{-7}{9}\\ \\ & \text{LCM} = 18 \\ \\ & = \frac{-1 \color{green} \times 3}{6 \color{green} \times 3} + \frac{-7 \color{green} \times 2}{9 \color{green} \times 2}\\ \\ & = \frac{-3}{18} + \frac{-14}{18} \\ \\ & = \frac{-3 - 14}{18}\\ \\ & = \color{green} \boxed{\frac{-17}{18}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]

(iii) \( \displaystyle x=\frac{3}{5}, y=\frac{-6}{9}, z=\frac{2}{10} \)

Answer

\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{3}{5} + \left(\frac{\cancel{-6}^{-2}}{\cancel9_3} + \frac{\cancel2^1}{\cancel{10}_5}\right)\\ \\ & = \frac{3}{5} + \left(\frac{-2}{3} + \frac{1}{5}\right)\\ \\ &\text{LCM of 3 and 5} = 15 \\ \\ & = \frac{3}{5} + \left(\frac{-2 \color{green} \times 5}{3 \color{green} \times 5} + \frac{1 \color{green} \times 3}{5 \color{green} \times 3}\right)\\ \\ & = \frac{3}{5} + \left(\frac{-10}{15} + \frac{3}{15}\right)\\ \\ & = \frac{3}{5} + \left(\frac{-10 + 3}{15}\right)\\ \\ & = \frac{3}{5} + \frac{-7}{15}\\ \\ &\text{LCM} = 15 \\ \\ & = \frac{3 \color{green} \times 3}{5 \color{green} \times 3} + \frac{-7 \color{green} \times 1}{15 \color{green} \times 1}\\ \\ & = \frac{9}{15} + \frac{-7}{15}\\ \\ & = \frac{9 - 7}{15}\\ \\ & = \color{green} \boxed{\frac{2}{15}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{3}{5} + \frac{\cancel{-6}^{-2}}{\cancel9_3}\right) + \frac{\cancel2^1}{\cancel{10}_5} \\ \\ & = \left(\frac{3}{5} + \frac{-2}{3}\right) + \frac{1}{5} \\ \\ & \text{LCM of 5 and 3} = 15 \\ \\ & = \left(\frac{3 \color{green} \times 3}{5 \color{green} \times 3} + \frac{-2 \color{green} \times 5}{3 \color{green} \times 5}\right) + \frac{1}{5}\\ \\ & = \left(\frac{9}{15} + \frac{-10}{15}\right) + \frac{1}{5}\\ \\ & = \left(\frac{9 -10}{15}\right) + \frac{1}{5}\\ \\ & = \frac{-1}{15} + \frac{1}{5}\\ \\ &\text{LCM} = 15 \\ \\ & = \frac{-1 \color{green} \times 1}{15 \color{green} \times 1} + \frac{1 \color{green} \times 3}{5 \color{green} \times 3}\\ \\ & = \frac{-1}{15} + \frac{3}{15}\\ \\ & = \frac{-1 + 3}{15}\\ \\ & = \color{green} \boxed{\frac{2}{15}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]

(iv) \( \displaystyle x=\frac{-3}{5}, y=\frac{-7}{10}, z=\frac{-8}{15} \)

Answer

\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{-3}{5} + \left(\frac{-7}{10} + \frac{-8}{15}\right)\\ \\ &\text{LCM of 10 and 15} = 30 \\ \\ & = \frac{-3}{5} + \left(\frac{-7 \color{green} \times 3}{10 \color{green} \times 3} + \frac{-8 \color{green} \times 2}{15 \color{green} \times 2}\right)\\ \\ & = \frac{-3}{5} + \left(\frac{-21}{30} + \frac{-16}{30}\right)\\ \\ & = \frac{-3}{5} + \left(\frac{-21 - 16}{30}\right)\\ \\ & = \frac{-3}{5} + \frac{-37}{30}\\ \\ &\text{LCM} = 30 \\ \\ & = \frac{-3 \color{green} \times 6}{5 \color{green} \times 6} + \frac{-37 \color{green} \times 1}{30 \color{green} \times 1}\\ \\ & = \frac{-18}{30} + \frac{-37}{30}\\ \\ & = \frac{-18 - 37}{30}\\ \\ & = \frac{\cancel{-55}^{-11}}{\cancel{30}_6}\\ \\ & = \color{green} \boxed{\frac{-11}{6}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{-3}{5} + \frac{-7}{10}\right) + \frac{-8}{15} \\ \\ & \text{LCM of 5 and 10} = 10 \\ \\ & = \left(\frac{-3 \color{green} \times 2}{5 \color{green} \times 2} + \frac{-7 \color{green} \times 1}{10 \color{green} \times 1}\right) + \frac{-8}{15}\\ \\ & = \left(\frac{-6}{10} + \frac{-7}{10}\right) + \frac{-8}{15}\\ \\ & = \left(\frac{-6 - 7}{10}\right) + \frac{-8}{15}\\ \\ & = \frac{-13}{10} + \frac{-8}{15}\\ \\ & \text{LCM} = 30 \\ \\ & = \frac{-13 \color{green} \times 3}{10 \color{green} \times 3} + \frac{-8 \color{green} \times 2}{15 \color{green} \times 2}\\ \\ & = \frac{-39}{30} + \frac{-16}{30}\\ \\ & = \frac{-39 - 16}{30}\\ \\ & = \frac{\cancel{-55}^{-11}}{\cancel{30}_6}\\ \\ & = \color{green} \boxed{\frac{-11}{6}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]

5. Simplify

(i) \( \displaystyle \frac{-3}{10} + \frac{12}{-10} + \frac{14}{10} \)

Answer

\[ \begin{align*} & = \frac{-3}{10} + \frac{-12}{10} + \frac{14}{10} \\ \\ & = \frac{-3 -12 + 14}{10} \\ \\ & = \frac{-15 + 14}{10} \\ \\ & = \boxed{\color{red}\frac{-1}{10}} \\ \\ \end{align*} \]

(ii) \( \displaystyle \frac{-5}{10} + \frac{6}{13} + 8 \)

Answer

\[ \begin{align*} &\color{green} \text{Method - } 1 \\ \\ & = \frac{-5}{10} + \frac{6}{13} + \frac{8}{1} \\ \\ & \text{LCM} = 130 \\ \\ & = \frac{-5 \color{green} \times 13}{10 \color{green} \times 13} + \frac{6 \color{green} \times 10}{13 \color{green} \times 10} + \frac{8 \color{green} \times 130}{1 \color{green} \times 130} \\ \\ & = \frac{-65}{130} + \frac{60}{130} + \frac{1040}{130} \\ \\ & = \frac{-65 + 60 +1040}{130} \\ \\ & = \frac{-65 + 1100}{130} \\ \\ & = \frac{\cancel{1035}^{207}}{\cancel{130}_{26}} \\ \\ & = \boxed{\color{red}\frac{207}{26}} \\ \\ &\color{green} \text{Method - } 2 \\ \\ & = \frac{\cancel{-5}^{-1}}{\cancel{10}_2} + \frac{6}{13} + \frac{8}{1} \\ \\ & = \frac{-1}{2} + \frac{6}{13} + \frac{8}{1} \\ \\ & \text{LCM} = 26 \\ \\ & = \frac{-1 \color{green} \times 13}{2 \color{green} \times 13} + \frac{6 \color{green} \times 1}{13 \color{green} \times 1} + \frac{8 \color{green} \times 26}{1 \color{green} \times 26} \\ \\ & = \frac{-13}{26} + \frac{6}{13} + \frac{208}{13} \\ \\ & = \frac{-13 + 12 + 208}{26} \\ \\ & = \frac{-13 + 220}{26} \\ \\ & = \frac{207}{26} \\ \\ & = \boxed{\color{red}\frac{207}{26}} \\ \\ \end{align*} \]

(iii) \( \displaystyle \frac{-5}{10} + \frac{9}{7} + \frac{3}{20} + \frac{-11}{14} \)

Answer

\[ \begin{align*} &\color{green} \text{Method - } 1 \text{: Grouping } \\ \\ & \left(\frac{-5}{10} + \frac{3}{20}\right)+ \left(\frac{9}{7} + \frac{-11}{14}\right) \\ \\ & = \left(\frac{-10 + 3}{20}\right)+ \left(\frac{18 - 11}{14}\right) \\ \\ & = \frac{-7}{20} + \frac{\cancel7^1}{\cancel{14}_2} \\ \\ & = \frac{-7}{20} + \frac{1}{2} \\ \\ & = \frac{-7 + 10}{20} \\ \\ & = \boxed{\color{red}\frac{3}{20}} \\ \\ &\color{green} \text{Method - } 2 \\ \\ &= \frac{-5}{10} + \frac{9}{7} + \frac{3}{20} + \frac{-11}{14} \\ \\ & \text{LCM} = 140 \\ \\ &= \frac{-5 \color{green} \times 14}{10 \color{green} \times 14} + \frac{9 \color{green} \times 20}{7 \color{green} \times 20} + \frac{3 \color{green} \times 7}{20 \color{green} \times 7} + \frac{-11 \color{green} \times 10}{14 \color{green} \times 10} \\ \\ &= \frac{-70}{140} + \frac{180}{140} + \frac{21}{140} + \frac{-110}{140} \\ \\ &= \frac{-70 + 180 + 21 - 110}{140} \\ \\ &= \frac{\cancel{110} + 21 \cancel{-110}}{140} \\ \\ &= \frac{\cancel{21}^3}{\cancel{140}_{20}} \\ \\ &= \boxed{\color{red}\frac{3}{20}} \\ \\ \end{align*} \]

(iv) \( \displaystyle \frac{5}{36} + \frac{-7}{8} + \frac{6}{-72} + \frac{-3}{-12} \)

Answer

\[ \begin{align*} &= \frac{5}{36} + \frac{-7}{8} + \frac{-6}{72} + \frac{3}{12} \\ \\ \end{align*} \] \[ \begin{array}{c|c} 2 & 36, 8, 72, 12 \\ \hline 2 & 18, 4, 36, 6 \\ \hline 2 & 9, 2, 18, 3 \\ \hline 3 & 9, 1, 9, 3 \\ \hline 3 & 3, 1, 3, 1 \\ \hline & 1, 1, 1, 1 \\ \end{array} \] \[ \begin{align*} & \text{LCM} = 72 \\ \\ & = \frac{5 \color{green} \times 2}{36 \color{green} \times 2} + \frac{-7 \color{green} \times 9}{8 \color{green} \times 9} + \frac{-6 \color{green} \times 1}{72 \color{green} \times 1} + \frac{3 \color{green} \times 6}{12 \color{green} \times 6} \\ \\ &= \frac{10}{72} + \frac{-63}{72} + \frac{-6}{72} + \frac{18}{72} \\ \\ &= \frac{10 +(-63) + (-6) + 18}{72} \\ \\ &= \frac{10 - 63 - 6 + 18}{72} \\ \\ &= \frac{28 - 69}{72} \\ \\ &= \boxed{\color{red} \frac{-41}{72}} \end{align*} \]

6. For \( \displaystyle x = \frac{1}{5} \) and \( \displaystyle y = \frac{3}{7}\) , verify that \( \color{red} -(x + y) = (-x) + (-y) \)

Answer

\[ \begin{align*} \text{LHS} &= -(x + y) \\ \\ & = -\left(\frac{1}{5} + \frac{3}{7} \right) \\ \\ \text{LCM} & = 35 \\ \\ & = -\left(\frac{1 \color{green} \times 7}{5 \color{green} \times 7} + \frac{3 \color{green} \times 5}{7 \color{green} \times 5} \right) \\ \\ & = -\left(\frac{7}{35} + \frac{15}{35} \right) \\ \\ & = -\left(\frac{7 + 15}{35} \right) \\ \\ & = -\left(\frac{22}{35} \right) \\ \\ & = \color{green} \boxed{\frac{-22}{35}} \\ \\ \text{RHS} &= (-x) + (-y) \\ \\ & = \left(-\frac{1}{5} \right) + \left(-\frac{3}{7} \right) \\ \\ & = \frac{-1}{5} + \frac{-3}{7} \\ \\ \text{LCM} & = 35 \\ \\ & = \frac{-1 \color{green} \times 7}{5 \color{green} \times 7} + \frac{-3 \color{green} \times 5}{7 \color{green} \times 5} \\ \\ & = \frac{-7}{35} + \frac{-15}{35} \\ \\ & = \frac{-7 + (-15)}{35} \\ \\ & = \frac{-7 -15}{35} \\ \\ & = \color{green} \boxed{\frac{-22}{35}} \\ \\ & \text{Hence verified} \\ & \color{red} -(x + y) = (-x) + (-y) \end{align*} \]

7. Write True or False for the following statements

(i) \( \displaystyle \frac{-2}{-3} \) is the additive inverse of \( \displaystyle \frac{2}{3} \)

Answer False

\[ \begin{align*} \frac{-2}{3} \text{ is the the additive inverse of } \frac{2}{3} \\ \\ \end{align*} \]

(ii) \( \displaystyle \frac{2}{3} + \frac{4}{5} \) is a rational number.

Answer True

\[ \begin{align*} & = \frac{2}{3} + \frac{4}{5} \\ \\ & = \frac{10 + 12}{15} \\ \\ & = \frac{22}{15} \text{ is a rational number}\\ \\ \end{align*} \]

(iii) \( \displaystyle \frac{-5}{3} + \frac{5}{-3} \) is equal to zero.

Answer False

\[ \begin{align*} & = \frac{-5}{3} + \frac{-5}{3} \\ \\ & = \frac{-5 -5}{3} \\ \\ & \frac{-10}{3} \neq 0 \\ \\ \end{align*} \]

(iv) 1 is the identity element of addition.

Answer False

\( \color{red} 0 \) is the identity element of addition.

(v) \(\displaystyle \frac{-9}{7} + 0 = 0 \)

Answer False

\[ \begin{align*} \frac{-9}{7} + 0 & = \frac{-9}{7} \\ \\ \end{align*} \]

(vi) Additive inverse of \( \displaystyle \frac{-3}{5} \) is \( \displaystyle \frac{3}{5} \)

Answer True

\[ \begin{align*} & = \frac{-3}{5} + \frac{3}{5} \\ \\ & = \frac{-3 + 3}{5} \\ \\ & = 0 \end{align*} \]

(vii) Negative of a negative rational number is negative.

Answer False

\[ \begin{align*} & = -\left(\frac{-1}{5} \right) \\ \\ & = +\frac{1}{5} \\ \\ \end{align*} \]