Question

6. Fill in the blanks by the correct symbol >, = or <.

$\sf (a) \: \frac{3}{4} \: and \: \frac{1}{2} \\ \\ \sf (b) \: \frac{ - 1}{2} \: and \: \frac{ - 3}{4}$


7. Add.

$\sf (a) \: \frac{3}{7} \: and \: \frac{-9}{7} \\ \\ \sf (b) \: \frac{5}{9} \: and \: \frac{7}{-9} \\ \\ \sf (c) \: \frac{2}{5} \: , \frac{5}{-9} \: and \: \frac{ - 6}{15}$


8. Simplify.

$\sf (a) \: -2 + \bigg( \frac{3}{8} \bigg) + \bigg( \frac{ - 1}{5} \bigg) \\ \\ \sf (b) \: \bigg( \frac{2}{3} \bigg) + \bigg( \frac{ - 7}{11} \bigg) + \bigg( \frac{ - 1}{4} \bigg)$


9. Verify that a + b = b + a by taking

$\sf (a) \: a = \frac{ - 7}{5} \: , \: b = \frac{2}{7} \\ \\ \sf (b) \: a = -1 \: , \: b = \frac{ - 2}{3}$


10. Verify that (a + b) + c = a + (b + c) by taking

$\sf (a) \: a = -2 \: , \: b = \frac{ - 2}{3} \: , \: c = \frac{ - 3}{4} \\ \\ \sf (b) \: a = -12 \: , \: b = \frac{ - 9}{11} \: , \: c = \frac{7}{ - 12}$
​

Answer 1

Answer :-

6.

a)

$\implies\sf \dfrac{3}{4}\:and\:\dfrac{1}{2}$

LCM of 2 and 4 = 4

$\implies\sf \dfrac{3}{4}\:\&\:\dfrac{1}{2}\times\dfrac{2}{2}\\\\\implies\sf \dfrac{3}{4}\:\&\:\dfrac{2}{4}$

Comparing the numerator :-

3 > 2

Hence,

$\implies\sf\dfrac{3}{4}>\dfrac{2}{4}$

b)

$\implies\sf \dfrac{-3}{4} \: and \: \dfrac{-1}{2}$

LCM of 2 and 4 = 4

$\implies\sf \dfrac{-3}{4}\:\&\:\dfrac{-1}{2} \times \dfrac{2}{2}\\\\\implies\sf \dfrac{-3}{4}\:\&\:\dfrac{-2}{4}$

Comparing the numerator :-

-2 > -3

Hence,

$\implies\sf\dfrac{-2}{4}>\dfrac{-3}{4}$

7.

a)

$\implies\sf \dfrac{3}{7} + \dfrac{(-9)}{7}$

$\implies\sf \dfrac{3}{7} - \dfrac{9}{7}$

$\implies\sf \dfrac{3-9}{7}$

$\implies\sf \dfrac{-6}{7}$

b)

$\implies\sf \dfrac{5}{9}+\dfrac{7}{-9}\\\\\implies\sf \dfrac{5}{9}-\dfrac{7}{9}\\\\\implies\sf\dfrac{5-7}{9}\\\\\implies\sf\dfrac{-2}{9}$

c)

$\sf \dfrac{2}{5}+\dfrac{5}{(-9)}+\dfrac{(-6)}{15}\\\\\implies\sf\dfrac{2}{5}-\dfrac{5}{9}-\dfrac{6}{15}$

LCM of 5 , 9 and 15 = 45

$\implies\sf\dfrac{2}{5}\times\dfrac{9}{9}-\dfrac{5}{9}\times\dfrac{5}{5}-\dfrac{6}{15}\times\dfrac{3}{3}\\\\\implies\sf \dfrac{18}{45}-\dfrac{25}{45}-\dfrac{-18}{25}\\\\\implies\sf\dfrac{18-25-18}{45}\\\\\implies\sf\dfrac{-25}{45}\\\\\implies\sf \dfrac{-5}{9}$

8.

a)

$\sf-2+\left(\dfrac{3}{8}\right)+\left(\dfrac{-1}{5}\right)\\\\\implies\sf-2+\dfrac{3}{8}-\dfrac{1}{5}$

LCM of 8 and 5 = 40

$\implies\sf-2\times\dfrac{40}{40}+\dfrac{3}{8}\times\dfrac{5}{5}-\dfrac{1}{5}\times\dfrac{8}{8}\\\\\implies\sf\dfrac{-80}{40}+\dfrac{15}{40}-\dfrac{8}{40}\\\\\implies\sf\dfrac{-80+15-8}{40}\\\\\implies\sf \dfrac{-73}{40}$

b)

$\sf\bigg(\dfrac{2}{3}\bigg)+\bigg(\dfrac{-7}{11}\bigg)+\bigg(\dfrac{-1}{4}\bigg)\\\\\implies\sf\dfrac{2}{3}-\dfrac{7}{11}-\dfrac{1}{4}$

LCM of 3 , 11 and 4 = 132

$\implies\sf\dfrac{2}{3}\times\dfrac{44}{44}-\dfrac{7}{11}\times\dfrac{12}{12}-\dfrac{1}{4}\times\dfrac{33}{33}\\\\\implies\sf\dfrac{88}{132}-\dfrac{77}{132}-\dfrac{33}{132}\\\\\implies\sf\dfrac{88-77-33}{132}\\\\\implies\sf\dfrac{-22}{132}\\\\\implies\sf\dfrac{-1}{6}$

9.

a)

$\sf a=\dfrac{-7}{5},\:b=\dfrac{2}{7}$

LHS = a + b

$\implies\sf\dfrac{-7}{5}+\dfrac{2}{7}$

LCM of 5 and 7 = 35

$\implies\sf\dfrac{-7}{5}\times\dfrac{7}{7}+\dfrac{2}{7}\times\dfrac{5}{5}\\\\\implies\sf\dfrac{-49}{35}+\dfrac{10}{35}\\\\\implies\sf\dfrac{-39}{35}$

RHS = b + a

$\implies\sf\dfrac{2}{7}+\dfrac{(-7)}{5}\\\\\implies\sf\dfrac{2}{7}-\dfrac{7}{5}$

LCM of 7 and 5 = 35

$\implies\sf\dfrac{2}{7}\times\dfrac{5}{5}-\dfrac{7}{5}\times\dfrac{7}{7}\\\\\implies\sf\dfrac{10}{35}-\dfrac{49}{35}\\\\\implies\sf \dfrac{-39}{35}$

LHS = RHS

Hence verified.

b)

$\sf a=-1,\:b=\dfrac{-2}{3}$

LHS = a + b

$\implies\sf-1+\dfrac{(-2)}{3}\\\\\implies\sf-1-\dfrac{2}{3}\\\\\implies\sf\dfrac{-3-2}{3}\\\\\implies\sf\dfrac{-5}{3}$

RHS = b + a

$\implies\sf \dfrac{-2}{3}+(-1)\\\\\implies\sf\dfrac{-2-3}{3}\\\\\implies\sf\dfrac{-5}{3}$

LHS = RHS

Hence verified.

10.

a)

$\sf a=-2,\:b=\dfrac{-2}{3},\:c=\dfrac{-3}{4}$

LHS = (a + b) + c

$\implies\sf\left(-2+\dfrac{(-2)}{3}\right)+\dfrac{(-3)}{4}\\\\\implies\sf\left(-2-\dfrac{2}{3}\right)-\dfrac{3}{4}\\\\\implies\sf\left(\dfrac{-6-2}{3}\right)-\dfrac{3}{4}\\\\\implies\sf\left(\dfrac{-8}{3}\right)-\dfrac{3}{4}$

LCM of 3 and 4 = 12

$\implies\sf\dfrac{-8}{3}\times\dfrac{4}{4}-\dfrac{3}{4}\times\dfrac{3}{3}\\\\\implies\sf\dfrac{-32}{12}-\dfrac{9}{12}\\\\\implies\sf\dfrac{-41}{12}$

RHS = a + (b + c)

$\implies\sf-2+\left(\dfrac{(-2)}{3}+\dfrac{(-3)}{4}\right)$

LCM of 3 and 4 = 12

$\implies\sf-2\left(\dfrac{-2}{3}\times\dfrac{4}{4}-\dfrac{3}{4}\times\dfrac{3}{3}\right)\\\\\implies\sf-2\left(\dfrac{-8}{12}-\dfrac{9}{12}\right)\\\\\implies\sf-2\left(\dfrac{-17}{12}\right)\\\\\implies\sf\dfrac{-24-17}{12}\\\\\implies\sf\dfrac{-41}{12}$

LHS = RHS

Hence verified.

b)

$\sf a=-12,\:b=\dfrac{-9}{11},\:c=\dfrac{7}{-12}$

LHS = (a + b) + c

$\implies\sf\left(-12+\dfrac{(-9)}{11}\right)+\dfrac{7}{-12}\\\\\implies\sf\left(\dfrac{-12\times11-9}{11}\right)-\dfrac{7}{12}\\\\\implies\sf\left(\dfrac{-132-9}{11}\right)-\dfrac{7}{12}\\\\\implies\sf\left(\dfrac{-141}{11}\right)-\dfrac{7}{12}$

$\implies\sf\dfrac{-141}{11}\times\dfrac{12}{12}-\dfrac{7}{12}\times\dfrac{11}{11}\\\\\implies\sf\dfrac{-1692}{132}-\dfrac{77}{132}\\\\\implies\sf\dfrac{-1692-77}{132}\\\\\implies\sf\dfrac{-1769}{132}$

RHS = a + (b + c)

$\implies\sf-12+\left(\dfrac{(-9)}{11}+\dfrac{7}{-12}\right)\\\\\implies\sf-12+\left(\dfrac{-9}{11}-\dfrac{7}{12}\right)\\\\\implies\sf-12+\left(\dfrac{-9}{11}\times\dfrac{12}{12}-\dfrac{7}{12}\times\dfrac{11}{11}\right)\\\\\implies\sf-12+\left(\dfrac{-108-77}{132}\right)\\\\\implies\sf-12-\dfrac{185}{132}\\\\\implies\sf\dfrac{-1584-185}{132}\\\\\implies\sf\dfrac{-1769}{132}$

LHS = RHS

Hence verified.