Exercise 3.3
Question 1
(i)First express $\frac{-5}{11}$ as a rational numbers with positive denominator
It is already with positive denominator
Sum = $\frac{3}{11}+\frac{-5}{11}=\frac{3+(-5)}{11}=\frac{-2}{11}$
ii) We have $\frac{5}{-9}=\frac{5 \times(-1)}{-9 \times(-1)}=\frac{-5}{9}$.
Sum $=\frac{4}{9}+\frac{-5}{9}=\frac{4+(-5)}{9}=\frac{4-5}{9}=-\frac{1}{9} .$
iii) We have $\frac{5}{-7}=\frac{5 \times(-1)}{-7 x-1}=\frac{-5}{7}$
$\frac{-2}{-7}=\frac{-2 \times-1}{-7 \times-1}=\frac{2}{7}$
$\Sum=\frac{-5}{7}+\frac{2}{7}=\frac{-5+2}{7}=\frac{-3}{7}$
iv) $\frac{-2}{5}, \frac{3}{4}$
Given rational number have different denominators
LCM of their denominator 5 and 4 is 20
To write the rational number with this LCM i.e 20as their denominator , we have
$\frac{-2}{5}=\frac{-2 \times 4}{5 \times 4}=-\frac{8}{20}$ and $\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}$.
∴ Sum =$\frac{-8}{20}+\frac{15}{20}=\frac{-8+15}{20}=\frac{7}{20}$
Question 2
i)Given rational numbers have different denominators
LCM of denominators 4 and 8 is 8
In order to have their denominators as LCM i.e 8 we have
$+\frac{27}{-4}=\frac{27 \times(-1)}{-4 \times(-1)}=-\frac{27}{4}$
$\frac{-27}{4}+\frac{-15}{8}=\frac{-27 \times 2}{4 \times 2}+\frac{-15}{8}$
$=\frac{-54}{8}+\frac{-15}{8}$
$=\frac{-54-15}{8}=\frac{-69}{8}$
ii) LCM of denominators 8 and 18 is 72
In order to have their denominators as LCM i.e 72 we have
$\begin{aligned} \frac{-1}{18}+\frac{-3}{8} &=\frac{-1 \times 4}{18 \times 4}+\frac{-3 \times 9}{8 \times 9} \\ &=\frac{-4}{72}+\frac{-27}{72} \end{aligned}$
$=\frac{-4+(-27)}{72}$
$=\frac{-31}{72}$
iii)
$\begin{aligned} &-3 \frac{1}{6}+2 \frac{3}{8} \\ &-3 \frac{1}{6}=\frac{-19}{6} ; 2 \frac{3}{8}=\frac{19}{8} \end{aligned}$
LCM of 6 and 8 is 24
In order to have their denominator is their LCM i.e 24
We have
$-3 \frac{1}{6}+2 \frac{3}{8}=\frac{-19}{6}+\frac{19}{8}=\frac{-19 \times 4}{6 \times 4}+\frac{19 \times 3}{8 \times 3}$
$=\frac{-76}{24}+\frac{57}{24}$
$=\frac{-76+57}{24}$
$=\frac{-19}{24} .$
iv) $-2 \frac{4}{5}+4 \frac{3}{10}$
$-2 \frac{4}{5}=\frac{-14}{5}$ and $4 \frac{3}{10}=\frac{43}{10}$
LCM of 5 and 10 is 10
In order to have their denominators is their LCM i.e 10
We have
$-2 \frac{4}{5}+4 \frac{3}{10}=\frac{-14}{5}+\frac{43}{10}=\frac{-14 \times 2}{5 \times 2}+\frac{43 \times 1}{10 \times 1}$
$=\frac{-28}{10}+\frac{43}{10}$
$=\frac{-28+43}{10}$
$=\frac{15}{10}=\frac{3}{2} .$
Question 3
i) $\frac{4}{13}-\frac{-6}{13}=\frac{4}{13}+$ additive inverse of $\left(\frac{-6}{13}\right)$
$=\frac{4}{13}+\frac{6}{13}=\frac{4+6}{13}=\frac{10}{13} .$
ii) $\frac{-2}{3}-\frac{-1}{2}=\frac{2}{3}+$ additive invesse of $\left(\frac{-1}{2}\right)$
$=\frac{-2}{3}+\frac{1}{2} \quad(\mathrm{LCM}$ of 3 and 2 is 6 $)$
$=\frac{-2 \times 2+1 \times 3}{6}$
$=\frac{-4+3}{6}=\frac{-1}{6}$
iii) $\frac{-2}{3}-\frac{5}{9}=\frac{-2}{3}+$ additive invesse of $\left(\frac{+5}{9}\right)$
$=\frac{-2}{3}+\frac{-5}{9}$ (LCM of 3,9 is 9)
$=\frac{-2 \times 3+(-5 \times 1)}{9}$
$=\frac{-6-5}{9}=\frac{-11}{9} .$
Question 4
i) $\frac{5}{63}-\left(\frac{-6}{21}\right)$
$=\frac{5}{63}+\frac{6}{21}$
LCM of 21,63 is 63
$=\frac{5 \times 1+6 \times 3}{63}=\frac{5+18}{63}=\frac{23}{63}$
ii) $\frac{-6}{3}-\left(\frac{-7}{15}\right)$
$=\frac{-6}{3}+\frac{7}{15}$
LCM of 3,15 is 15
$=\frac{-6 \times 5+7 \times 1}{15}=\frac{-30+7}{15}=\frac{-23}{15}$
iii) $3 \frac{1}{8}-\left(-1 \frac{5}{6}\right)$
$\begin{aligned}=\frac{25}{8}+\frac{11}{6} \\ & \mathrm{LCM} \text { of } 8,6 \text { is } 24 \end{aligned}$
$=\frac{25 \times 3+1.1 \times 4}{24}$
$=\frac{75+44}{24}$
$=\frac{119}{24}$
Question 5
Let the other rational number be 'x'
Given Sum = $\frac{2}{5}$
1.e $x+\frac{-4}{9}=\frac{2}{5}$
$x=\frac{2}{5}-\left(-\frac{4}{9}\right)$
$x=\frac{2}{5}+\frac{4}{9}$
LCM of 5,9 is 45
$x=\frac{2 \times 9+4 \times 5}{45}$
$x=\frac{18+20}{45}$
$x=\frac{38}{45}$
∴ The other rational number is $\frac{38}{45}$
Question 6
Let the rational added to $\frac{-5}{2}$ is " $x^{\prime}$
Given the result of sum is $-frac{-7}{8} .$
i.e $x+\left(\frac{-5}{12}\right)=\frac{-7}{8}$
$x=\frac{-7}{8}-\left(\frac{-5}{12}\right)$
$x=\frac{-7}{8}+\frac{5}{12}$
$\mathrm{Lcm}$ of 8,12 is 24
$x=\frac{-7 \times 3+5 \times 2}{24}$
$x=\frac{-21+10}{24}$
$x=\frac{-11}{24}$
∴ Therefore $\frac{-11}{24}$ is to be added to $\frac{-5}{12}$ to get $\frac{-7}{8}$
Question 7
Let the rational number to be subtracted from $\frac{-2}{3}$ be $x^{\prime}$
The result is $\frac{-5}{6} .$
i.e $x-\left(\frac{-2}{3}\right)=-\frac{5}{6}$
$x=\frac{=5}{6}+\left(-\frac{2}{3}\right) .$
LCM of 3 and 6 is 6
$x=\frac{-5 \times 1+-2 \times 2}{6}$
$x=\frac{-5+(-4)}{6}$
$x=\frac{-5-4}{6}=\frac{-9}{6}$
$x=-\frac{3}{2}$
∴ $\frac{-3}{2}$ should be substracted from $\frac{-2}{3}$ to get $\frac{-5}{6}$.
Question 8
i) $\frac{2}{3} \times-\frac{7}{8}=\frac{2 x-7}{3 \times 8}=\frac{-7}{12}$
ii) $\frac{-6}{7} \times \frac{5}{7}=\frac{-6 \times 5}{7 \times 7}=-\frac{30}{49}$
iii) $\frac{-2}{9} \times(-5)=\frac{-2 \times-5}{9}=\frac{10}{9}$.
iv) $\frac{-5}{11} \times \frac{11}{-5}=+\frac{8 \times 11}{4 \times+5}=1$
v) $\frac{8}{35} \times \frac{21}{-32}=\frac{18 \times 21}{35 \times-32}=\frac{3}{-20}$
Vi) $\frac{-105}{128} \times\left(-1 \frac{29}{35}\right)=-\frac{105}{128} \times\frac{-64}{35}$
$=\frac{3}{2}$
Question 9
i) $(-6) \div \frac{2}{5}$
$=\frac{(-6)}{(2 / 5)}=\frac{-6 \times 5}{21}=-15$
ii) $\frac{-1}{10} \div \frac{-8}{5}$
$=\frac{(-1 / 10)}{(-8 / 5)}=+\frac{1}{10} \times \frac{5}{+8}$
$=\frac{1}{2 \times 8}=\frac{1}{16}$
iii) $\frac{-65}{14} \div \frac{13}{-7}$
$\begin{aligned}=\frac{-65 / 14}{13 /-7} &=\frac{-65}{14} \times \frac{-7}{13} \\ &=\frac{5}{2} \end{aligned}$
iv)
$\begin{aligned} &(-6) \div 3 \frac{3}{5} \\ &=(-6) \div \frac{18}{5} \end{aligned}$
$=\frac{-6}{(18 / 5)}=\frac{-6 \times 5}{18}=\frac{-5}{3} .$
V) $\frac{-48}{49} \div \frac{72}{-35}$
$=-\frac{48 / 49}{72 /-35}$
$=\frac{-48}{49} \times \frac{-35}{72}$ = $\frac{10}{21}$
vi) $3 \frac{1}{7} \div\left(\frac{-33}{34}\right)$
$=\frac{22}{7} \div\left(\frac{-33}{34}\right)$
$=\frac{22 / 7}{-33 / 34}=\frac{22}{7} \times \frac{34}{-33}$
$=\frac{68}{-21}$
Question 10
Let the other rational number be 'x '
product of them is $\frac{18}{35}$
i.e $\frac{-2}{5} \times x=\frac{18}{35}$
x=$\frac{18 \times 5}{35 \times -2}$
$x=\frac{9}{-7}=\frac{-9}{7} .$
∴ The other rational number is $\frac{-9}{7}$
Question 11
i) $\left(\frac{13}{21} \div \frac{39}{42}\right) \times\left(\frac{-3}{5}\right)$
$=\frac{13 / 21}{39 / 42} \times \frac{-3}{5}$
$=\frac{13}{21} \times \frac{42}{39} \times \frac{-3}{5}=\frac{-2}{5}$
(ii) $\left(-5 \frac{5}{21}\right) \div\left(\frac{7}{11} \times \frac{5}{12}\right)$
$=\frac{-110}{21} \div \frac{35}{132}$
$=\frac{-110 / 21}{35 / 132}$
$\frac{-110}{21} \times \frac{132}{35}$
$=\frac{-22 \times 44}{7 \times 7}=\frac{-768}{49}$
Question 12
i) $\frac{3}{13} \div \frac{-4}{65}$
$=\frac{(3 / 13)}{(-4 / 65)}$ = $\frac{3}{13} \times \frac{65}{-4}$
$=\frac{15}{-4}=\frac{-15}{4}$
∴ The reciprocal is $\frac{-4}{15} .$
ii)$\left(-8 \times \frac{12}{15}\right)-\left(-3 \times \frac{2}{9}\right)$
$=(-4)-\left(-frac{-2}{3}\right)$
$\begin{aligned}=-4+\frac{2}{3} &=\frac{-12+2}{3} \\ &=\frac{-10}{3} . \end{aligned}$
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