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Suppose x be the rational number to be added to \(\frac{-5}{11}\) to get \(\frac{26}{33}\)

Then,

\(\frac{-5}{11}\)+x = \(\frac{26}{33}\)

x = \(\frac{26}{33}\)+\(\frac{5}{11}\)

x = \(\frac{26\times 1+5\times 3}{33}\)

x = \(\frac{26+15}{33}\)

x = \(\frac{41}{33}\)

Therefore,

The required number x = \(\frac{41}{33}\)

Given (-5/7)

Let the required number be x

x + (-5/7) = (-2/3)

x = (-2/3) – (-5/7)

x = (-2/3) + (5/7)

LCM of 3 and 7 is 21

Consider (-2/3) = (-2/3) × (7/7) = (-14/21)

Again (5/7) = (5/7) × (3/3) = (15/21)

On substituting

x = (-14/21) + (15/21)

x = (-14 + 15)/21

x = (1/21)

Let the number = x 

Now, 

According to question,

\(\frac{-2}{3}-\) x \(=\frac{-1}{6}\)

\(\Rightarrow\) x = \(\frac{-2}{3}-(\frac{-1}{6})\)

\(\Rightarrow\) x \(=\frac{-4-(-1)}{6}\)

\(\Rightarrow\) x \(\frac{-4+1}{6}\)

\(\Rightarrow\) x \(=\frac{-3}{6}\) 

In lowest terms,

x \(=\frac{-3÷3}{6÷3}= \frac{-1}{2}\)

Therefore, \(\frac{-1}{2}\) should be subtracted from \(\frac{-2}{3}\) so as to get \(\frac{-1}{6}\)

Given (-5/3)

Let the required number be x

(-5/3) – x = (5/6)

– x = (5/6) – (-5/3)

– x = (5/6) + (5/3)

Consider (5/3) = (5/3) × (2/2) = (10/6)

On substituting

– x = (5/6) + (10/6)

– x = (15/6)

x = (-15/6)

= -5 + (-3) + (7/10) + (-4/5) + (3/7) + (5/14)

= 8 + [(7-8)/10] + [(6 + 5)/14]

= – 8 – (1/10) + (11/14)

LCM of 1, 10 and 14 is 70

= (-560 – 7 + 55)/70

= -512/70

= – 256/35

First rearrange the rational numbers and add the numbers with same denominator.

= (4/7) + (3/7) – (4/9) – (13/9)

= ((4 + 3)/7) – ((4 + 13)/9)

= (7/7) – (17/9)

= 1 – (17/9)

= (9 – 17)/9

= -8/9

i) \(\)3/4 from 1/3

\(\Rightarrow \frac13\,-\,\frac34=\frac{4\,-\,9}{12} = \frac{-5}{12}\)

ii) -32/13 from 2

\(\Rightarrow 2\,-\,(\frac{-32}{13})\)

\(\Rightarrow 2\,+\,\frac{32}{13}=\frac{26\,+\,32}{13}=\frac{58}{13}\)

iii) -7 from -4/7

\(\Rightarrow \frac{-4}7\,-\,(-7)=\frac{-4}7\,+\,7\)

\(\frac{-4\,+\,49}{7}=\frac{45}7\)

(-32/13) from (-6/5)

We have:

= (-6/5) – (-32/13)

= (-6/5) + (additive inverse of -32/13)

= (-6/5) + (32/13)

LCM of 5 and 13 is 65

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(-6×13)/ (5×13)] = (-78/65)

= [(32×5)/ (13×5)] = (160/65)

Then,

= (-78/65) + (160/65)

= (-78+160)/65

= (82/65)

(i) 

 We can write, 1 = \(\frac{1}{1}\)

Since the denominators of both the numbers are different 

therefore, 

we will take their LCM 0f 1 and 11 = 11

\(\frac{1}{1} = \frac{1\times11}{1\times11}= \frac{11}{11}\)

And,

\(\frac{18}{11} = \frac{-18\times1}{11\times1}= \frac{-18}{11}\)

Therefore,

\(1-(\frac{-18}{11})\)

= \(\frac{11}{11}-(\frac{-18}{11})\) 

= \(\frac{11-(-18)}{11}\) 

= \(\frac{11+18}{11}\)

= \(\frac{29}{11}\)

(ii) 

\(0-(\frac{-13}{9})\)

= \(0+\frac{13}{9}\)

= \(\frac{13}{9}\)

(iii)

Since the denominators of both the numbers are different 

therefore, 

we will take their LCM 0f 13 and 5 = 65

\(\frac{-6}{5}= \frac{-6\times13}{5\times13}\frac{-78}{65}\)

Therefore,

\(\frac{-6}{5}-(\frac{-32}{13})\)

= \(\frac{-78}{65}-(\frac{-160}{65})\)

= \(\frac{-78-(-160)}{65}\) 

= \(\frac{-78+160}{65}\)

= \(\frac{82}{65}\)

(iv)

We can write, \(-7= \frac{-7}{1}\)

Since the denominators of both the numbers are different 

therefore, 

we will take their LCM 0f 1 and 7 = 7

\(\frac{-7}{1}= \frac{-7\times7}{1\times7}= \frac{-49}{7}\)

And,

\(\frac{-4}{7}= \frac{-4\times1}{7\times1}= \frac{-4}{7}\)

Therefore,

= \(\frac{-4}{7}-(-7)\) 

= \(\frac{-4}{7}-(\frac{-49}{7})\)

= \(\frac{-4-(-49)}{7}\) 

= \(\frac{-4+49}{7}\) 

= \(\frac{45}{7}\)

(-13/9) from (0)

We have:

= (-13/9) – (0)

= (0/1) + (additive inverse of -13/9)

= (0/1) + (13/9)

LCM of 1 and 9 is 9

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(0×9)/ (1×9)] = (0/9)

= [(13×1)/ (9×1)] = (13/9)

Then,

= (0/9) + (13/9)

= (0+13)/9

= (13/9)

(1/5) × [(2/7) + (3/8)] = [(1/5) × (2/7)] + [(1/5) × (3/8)]

∵ From the rule of distributive law of multiplication [a × (b + c) = (a × b) + (a × c)]

It is given that,

The sum of the two numbers = -8

One of the number = \(\frac{-15}{7}\)

Suppose the other number is x

Since,

The sum is -8

Therefore,

x + \(\frac{-15}{7}\) = -8

\(\frac{7x-15}{7}\) = -8

7x - 15 = -56

7x = -56 + 15

7x = -41

x = \(\frac{-41}{7}\)

Suppose x be the rational number to be added to \(\frac{-7}{8}\) to get \(\frac{5}{9}\)

Then,

\(\frac{-7}{8}\)+x = \(\frac{5}{9}\)

x = \(\frac{5}{9}\)+\(\frac{7}{8}\)

x = \(\frac{5\times 8+7\times 9}{72}\)

x = \(\frac{40+63}{72}\)

x = \(\frac{103}{72}\)

Therefore,

The required number x = \(\frac{103}{72}\)

Given cost of 2(1/3) metres of rope = Rs. 75 (1/4)

Cost of cloth per meter = 75 (1/4) ÷ 2 (1/3)

= (301/4) ÷ (7/3)

= (301/4) × (3/7)

= (129/4)

= Rs 32 (1/4)

i) Natural numbers 1, 22 

ii) Whole numbers 0, 1, 22 

iii) Integers 0, 1, 22, -5, -2

iv) Rational numbers 1,1/2,  -2, 0.5, 4\(\frac12\) ,-33/7, 0, 4/7,0.\(\bar3\), 22, -5, 2/19, 0.125 etc.

Would you leave out any of the given numbers from rational numbers?  No

 Is every natural number, whole number and integer is a rational number?  Yes

Let us consider a number as x to be added to (1/2 + 1/3 + 1/5) to get 3

\(x+(\frac{1}{2}+\frac{1}{3}+\frac{1}{5})=3\\\frac{(30x+1\times15+1\times10+1\times6)}{30}=3\\30x+15+10+6=3\times30\\30x+31=90\\30x=90-31\\x=\frac{59}{30}\)

\(\therefore\) the required number is \(\frac{59}{30}\)

Speed of car \(=60\frac{2}{5}\,km/hr\)

Total hours \(=6\frac{1}{4}\,hrs\)

Total Distance = Speed of car × Total hours

\(=60\frac{2}{5}\,km/hr\times6\frac{1}{4}\,hrs\)

\(=\frac{302}{5}\,km/hr\times\frac{25}{4}\,hrs\)

\(=\frac{7550}{20}\,km\)

\(=\frac{755}{2}\,km\)

\(=377\frac{1}{2}\,km\)

Therefore, 

Total Distance \(=377\frac{1}{2}\,km\)

Let the number be x

\(\frac{3}{5}\) of x \(=\frac{3}{5}\text{x}\)

\(\frac{2}{7}\) of x \(=\frac{2}{7}\text{x}\)

According to the question,

\(\frac{3}{5}\text{x}-\frac{2}{7}\text{x}=44\)

\(\Rightarrow\) \(\frac{3\text{x}\times7-2\text{x}\times5}{35}=44\) 

\(\Rightarrow\) \(\frac{21\text{x}-10\text{x}}{35}=44\)

\(\Rightarrow\) \(\frac{11}{35}\times\text{x}=44\)

\(\Rightarrow\) \(\text{x}=44\div\frac{11}{35}\)

\(\Rightarrow\) \(\text{x}=44\times\frac{35}{11}\)

\(\Rightarrow\) \(\text{x}=\frac{1540}{11}\)

\(\Rightarrow\) \(\text{x}=140\)

The number is 140

Suppose x be the rational number to be subtracted to \(\frac{3}{7}\) to get \(\frac{5}{4}\)

Then,

\(\frac{3}{7}\) - x = \(\frac{5}{4}\)

-x = \(\frac{5}{4}\) - \(\frac{3}{7}\)

-x = \(\frac{5\times 7-3\times 4}{28}\)

-x = \(\frac{35-12}{28}\)

-x = - \(\frac{23}{28}\)

Therefore,

The required number x = - \(\frac{23}{28}\)

Let us consider a number as x to be added to (2/3 + 3/5) to get -2/15
\(x+(\frac{2}{3}+\frac{3}{5})=\frac{-2}{15}\\=\frac{(15x+2\times5+3\times3)}{15}\\=\frac{-2}{15}\\15x+10+9=-2\\15x=-2-19\\x=\frac{-21}{15}\\=\frac{-7}{5}\)

\(\therefore\)the required number is \(\frac{-7}{5}\)

In the question is given to verify the property = x + y = y + x

Where, x = -2/3, y = -5/6

Then, -2/3 + (-5/6) = -5/6 + (-2/3)

LHS = -2/3 + (-5/6)

= -2/3 – 5/6

The LCM of the denominators 3 and 6 is 6

(-2/3) = [(-2×2)/ (3×2)] = (-4/6)

and (-5/6) = [(-5×1)/ (6×1)] = (-5/6)

Then,

= – 4/6 – 5/6

= (- 4 – 5)/ 6

= – 9/6

RHS = -5/6 + (-2/3)

= -5/6 – 2/3

The LCM of the denominators 6 and 3 is 6

(-5/6) = [(-5×1)/ (6×1)] = (-5/6)

and (-2/3) = [(-2×2)/ (3×2)] = (-4/6)

Then,

= – 5/6 – 4/6

= (- 5 – 4)/ 6

= – 9/6

By comparing LHS and RHS

LHS = RHS

∴ -9/6 = -9/6

Hence x + y = y + x

In the question is given to verify the property = x + y = y + x

Where, x = -2/5, y = -9/10

Then, -2/5 + (-9/10) = -9/10 + (-2/5)

LHS = -2/5 + (-9/10)

= -2/5 – 9/10

The LCM of the denominators 5 and 10 is 10

(-2/5) = [(-2×2)/ (5×2)] = (-4/10)

and (-9/10) = [(-9×1)/ (10×1)] = (-9/10)

Then,

= – 4/10 – 9/10

= (- 4 – 9)/ 10

= – 13/10

RHS = -9/10 + (-2/5)

= -9/10 – 2/5

The LCM of the denominators 10 and 5 is 10

(-9/10) = [(-9×1)/ (10×1)] = (-9/10)

and (-2/5) = [(-2×2)/ (5×2)] = (-4/10)

Then,

= – 9/10 – 4/10

= (- 9 – 4)/ 10

= – 13/10

By comparing LHS and RHS

LHS = RHS

∴ -9/6 = -9/6

Hence x + y = y + x

Length of park \(=36\frac{3}{5}\,m\)

Breadth of park \(=16\frac{2}{3}\,m\)

\(=36\frac{3}{5}\,m\times16\frac{2}{3}\,m\)

\(=\frac{183}{5}\,m\times\frac{50}{3}\,m\)

\(=\frac{183\times50}{5\times3}\,m\)

\(=\frac{9150}{15}\,m^2\)

\(=610\,m^2\)

Hence, 

Area of park \(=610\,m^2\)

In the question is given to verify the property = x + y = y + x

Where, x = ½, y = ½

Then, ½ + ½ = ½ + ½

LHS = ½ + ½

= (1 + 1)/2

= 2/2

= 1

RHS = ½ + ½

= (1 + 1)/2

= 2/2

= 1

By comparing LHS and RHS

LHS = RHS

∴ 1 = 1

Hence x + y = y + x

The sum is = 65/12 + 12/7

The difference is =  65/12 – 12/7

When we divide, 

= (65/12 + 12/7) / (65/12 – 12/7)

= ((65×7 + 12×12)/84) / ((65×7 – 12×12)/84)

= ((455+144)/84) / ((455 – 144)/84)

= (599/84) / (311/84)

= 599/84 × 84/311

= 599/311

Let us consider a number as x to be subtracted from (3/4 – 2/3) to get -1/6
So, 

\((\frac{3}{4}-\frac{2}3)-x=\frac{-1}6\\x=\frac{3}{4}-\frac{2}{3}+\frac{1}{6}\\x=\frac{(3\times3-2\times4)}{12}+\frac{1}{6}\\=\frac{(9-8)}{12}+\frac{1}{6}\\=\frac{(1\times1+1\times2)}{12}\\=\frac{3}{12}\\=\frac{1}{4}\)

\(\therefore\) the required number is \(\frac{1}{4}\)

Let the number be x to subtracted from 3/7 to get 5/4
So,

\(\frac{3}{7}-x=\frac{5}{4}\\x=\frac{3}{7}-\frac{5}{4}\\x=\frac{(3\times4-5\times7)}{28}\\=\frac{(12-35)}{28}\\=\frac{-23}{28}\)

\(\therefore\)the required number is \(\frac{-23}{28}\)

Side of plot \(=8\frac{1}{2}\,m\)

Area of plot = Side of plot × Side of plot

\(=8\frac{1}{2}\,m\times8\frac{1}{2}\,m\)

\(=\frac{17}{2}\,m\times\frac{17}{2}\,m\)

\(=\frac{17\times17}{2\times2}\,m\)

\(=\frac{289}{4}\,m^2\)

\(=72\frac{1}{4}\,m^2\)

Hence, 

Area of plot \(=72\frac{1}{4}\,m^2\)

Let the other number be x 

Then,

\(\frac{-3}{10}\times\text{x}=\frac{-1}{4}\)

\(\Rightarrow\) \(\text{x}=\frac{-1}{4}\div\frac{-3}{10}\)

\(\Rightarrow\) \(\text{x}=\frac{-1}{4}\times\frac{10}{-3}\) 

\(\Rightarrow\) \(\text{x}=\frac{-1\times10}{4\times-3}\)

\(\Rightarrow\) \(\text{x}=\frac{-10}{-12}=\frac{-10\times-1}{-12\times-1}=\frac{10}{12}\) 

\(\Rightarrow\) \(\text{x}=\frac{10}{12}=\frac{10\div2}{12\div2}=\frac{5}{6}\)

(-12) ÷ (…. ) = (-6/5)

Let the required number be (a/b). Then,

(-12/1) ÷ (a/b) = (-6/5)

⇒ (a/b) = (-12/1) × (5/-6)

⇒ (a/b) = (-12 × 5) / (1 × -6)

⇒ (a/b) = (-2× 5) / (1×-1)

⇒ (a/b) = (-10/-1)

⇒ (a/b) = 10

As we know that  7/3 meters of cloth = Rs 301/4

Let a number be  = x

So,

\(x\times\frac{7}{3}=\frac{301}{4}\)

x = (301/4) / (7/3)

x = (301/4) × (3/7)

= (301×3) / (4×7)

= (43×3) / (4×1)

= 129/4

= 32.25

According to question,

\(\frac{-13}{5}+\frac{12}{7}\)

= \(\frac{-91+60}{35}\)

= \(\frac{-30}{35}\)

= \(\frac{-6}{7}\)

And,

\(\frac{-31}{7}\times \frac{1}{2}\)

 = \(\frac{-31}{14}\)

Now,

We have to divide \(\frac{-6}{7}\) by \(\frac{-31}{14}\)

Therefore

= \(\frac{-6}{7}\times \frac{14}{31}\)

= \(\frac{-12}{31}\)

 Sum of -13/5 and 12/7

= \(\frac{-13}{5}\,+\,\frac{12}{7} = \frac{-13\,\times\,7\,+\,12\,\times\,5}{35}\)

\(\frac{-91\,+\,60}{35}=\frac{-31}{35}\)...............(1)

the product of -13/7 and -1/2

= \((\frac{-13}{7})\,\times\,(\frac{-1}{2})= \frac{13}{14}\).................(2)

∴ (1) % (2) 

= \(\frac{-31}{35}\,\div \,\frac{13}{14}= \frac{-31}{35}\,\times\,\frac{14}{13}= \frac{-62}{65}\) 

In the question is given to verify the property = x + y = y + x

Where, x = -3/7, y = 20/21

Then, -3/7 + 20/21 = 20/21 + (-3/7)

LHS = -3/7 + 20/21

The LCM of the denominators 7 and 21 is 21

(-3/7) = [(-3×3)/ (7×3)] = (-9/21)

and (20/21) = [(20×1)/ (21×1)] = (20/21)

Then,

= – 9/21 + 20/21

= (- 9 + 20)/ 21

= 11/21

RHS = 20/21 + (-3/7)

The LCM of the denominators 21 and 7 is 21

(20/21) = [(20×1)/ (21×1)] = (20/21)

and (-3/7) = [(-3×3)/ (7×3)] = (-9/21)

Then,

= 20/21 – 9/21

= (20 – 9)/ 21

= 11/21

By comparing LHS and RHS

LHS = RHS

∴ 11/21 = 11/21

Hence x + y = y + x

Suppose x be the rational number to be added to \(\frac{-5}{7}\) to get \(\frac{-2}{3}\)

Then,

\(\frac{-5}{7}\)+ x = \(\frac{-2}{3}\)

x - \(\frac{5}{7}-\frac{2}{3}\)

x = \(\frac{5\times 3-2\times 7}{21}\)

x = \(\frac{15-14}{21}\)

x = \(\frac{1}{21}\)

Therefore,

The required number x = \(\frac{1}{21}\)

The cost of 7\(\frac23\) mts ( \(\frac{23}3\) mts ) of cloth = ₹12\(\frac34\) = ₹\(\frac{51}4\)

∴ The cost of 1m cloth

= \(\frac{51}{4}\,\div\,\frac{23}3=\frac{51}4\,\times\,\frac3{23}\)

= \(\frac{153}{92}\) = ₹ 1.66

Cost of one litre petrol \(=Rs\,63\frac{3}{4}=Rs\,\frac{255}{4}\)

Cost of 34 litre petrol = 34 × Cost of one litre petrol

\(=34\times\,Rs\,\frac{255}{4}\)

\(=Rs\,\frac{34\times255}{4}\)

\(=Rs\,\frac{8670}{4}\)

\(=Rs\,\frac{4335}{2}\)

\(=Rs\,2167\frac{1}{2}\)

Cost of 34 litre petrol \(=Rs\,2167\frac{1}{2}\)

((65/12) + (8/3)) ÷ ((65/12) – (8/3))

= ((65/12) + (32/12)) ÷ ((65/12) – (32/12))

= (65 + 32)/12 ÷ (65 -32)/12

= (65 + 32)/12 × (12/ (65 – 32)

= (65 + 32)/ (65 – 32)

= (97/33)

First we have to find the sum of (65/12) + (8/3)

LCM of 12 and 3 is 12

Then,

= (65 × 1) / (12×1) = (65/12)

= (8× 4) / (3× 4) = (32/12)

Now,

= (65 + 32)/12

= (97/12)

Now find the difference of (65/12) – (8/3)

LCM of 12 and 3 is 12

Then,

= (65 × 1) / (12×1) = (65/12)

= (8× 4) / (3× 4) = (32/12)

Now,

= (65 – 32)/12

= (33/12)

Now divide (97/12) ÷ (33/12)

= (97/12) × (12/33)

= (97 × 12) / (12× 33)

= (97 × 1) / (1× 33)

= (97/33)

\(\frac{-5}{6}\div\frac{-2}{3}\)

\(=\frac{-5}{6}\times\frac{3}{-2}\)

\(=\frac{-5\times3}{6\times-2}\)

\(=\frac{-15}{-12}=\frac{-15\times-1}{-12\times-1}=\frac{15}{12}\)

\(=\frac{15}{12}=\frac{15\div3}{12\div3}=\frac{5}{4}\)

(-5/-8) – (-3/4)

First we write each of the given numbers with a positive denominator.

(-5/-8) = [(-5× (-1))/ (-8×-1)]

= (5/8)

We have:

= (5/8) – (-3/4)

= (5/8) + (additive inverse of -3/4)

= (5/8) + (3/4)

LCM of 8 and 4 is 8

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(5×1)/ (8×1)] = (5/8)

= [(3×2)/ (4×2)] = (6/8)

Then,

= (5/8) + (6/8)

= (5+6)/8

= (11/8)

Let the required number be x. Then,

= (5/7) + x = (4/21)

By sending (5/7) from left hand side to the right hand side it changes to – (5/7)

x = (4/21) – (5/7)

LCM of 21 and 7 is 21

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(4×1)/ (21×1)] = (4/21)

= [(5×3)/ (7×3)] = (15/21)

Then,

= (4/21) – (15/21)

= (4-15)/21

= (-11/21)

Hence the required number is (-11/21)

\(\frac{4}{3}\div\text{x}=\frac{-5}{2}\)

\(\Rightarrow\) \(\text{x}=\frac{4}{3}\div\frac{-5}{2}\) 

\(\Rightarrow\) \(\text{x}=\frac{4}{3}\times\frac{2}{-5}\) 

\(\Rightarrow\) \(\text{x}=\frac{4\times2}{3\times-5}\) 

\(\Rightarrow\) \(\text{x}=\frac{8}{-15}=\frac{8\times-1}{-15\times-1}=\frac{-8}{15}\)

Rational number between \(\frac{-1}{3}\) and \(\frac{1}{2}\) 

\(=\frac{1}{2}(\frac{-2}{3}+\frac{1}{2})\)

\(=\frac{1}{2}(\frac{-2\times2+1\times3}{6})\)

\(=\frac{1}{2}(\frac{-4+3}{6})\) 

\(=\frac{1}{2}\times\frac{-1}{6}\)

\(=\frac{-1}{12}\)

Given sum of two numbers is (5/9)

And one them is (1/3)

Let the unknown number be x

x + (1/3) = (5/9)

x = (5/9) – (1/3)

LCM of 3 and 9 is 9

Consider (1/3) = (1/3) × (3/3) = (3/9)

On substituting we get

x = (5/9) – (3/9)

x = (5 – 3)/9

x = (2/9)

According to the question,

\((\frac{65}{12}+\frac{8}{3})\div(\frac{65}{12}-\frac{8}{3})\)

= \((\frac{65\times1+8\times4}{12})\div(\frac{65\times1-8\times4}{12})\)

= \((\frac{65+32}{12})\div(\frac{65-32}{12})\)

= \((\frac{97}{12})\div(\frac{33}{12})\)

= \(\frac{97}{12}\times\frac{12}{33}\)

= \(\frac{97}{33}\)

Distance covered in one hour = 1020 km

Distance covered in \(4\frac{1}{6}\) hours \(=4\frac{1}{6}\times\) Distance covered in one hour

\(=4\frac{1}{6}\times1020\,km\)

\(=\frac{25}{6}\times1020\,km\)

\(=\frac{25\times1020}{6}\,km\)

\(=\frac{25500}{6}\,km\)

\(=4250\,km\)

Distance covered in \(4\frac{1}{6}\) hours \(=4250\,km\)

The reciprocal of \(\frac{-5}{14} \,is\,\frac{-14}{5}\)

( \(\because\) \((\frac{-5}{14})\,\times\,(\frac{-14}{5})=1)\)

∴ The product of \(\frac{2}{14} \,and\,\frac{-14}{5} \,is\,\frac{2}{11}\,\times\,(\frac{-14}{5}) = \frac{-28}{55}\)

Reciprocal of \(\frac{-7}{9}=\frac{9}{-7}\)

\(\frac{9}{-7}=\frac{9\times-1}{-7\times-1}=\frac{-9}{7}\)

(i) Given (2/3) – (3/5)

The LCM of 3 and 5 is 15

Consider (2/3) = (2/3) × (5/5) = (10/15)

Now again (3/5) = (3/5) × (3/3) = (9/15)

(2/3) – (3/5) = (10/15) – (9/15)

= (1/15)

(ii) Given (-4/7) – (2/-3)

The LCM of 7 and 3 is 21

Consider (-4/7) = (-4/7) × (3/3) = (-12/21)

Again (2/-3) = (-2/3) × (7/7) = (-14/21)

(-4/7) – (2/-3) = (-12/21) – (-14/21)

= (-12 + 14)/21

= (2/21)

(iii) Given (4/7) – (-5/-7)

(4/7) – (5/7) = (4 -5)/7

= (-1/7)

(iv) Given -2 – (5/9)

Consider (-2/1) = (-2/1) × (9/9) = (-18/9)

-2 – (5/9) = (-18/9) – (5/9)

= (-18 -5)/9

= (-23/9)