# DAV Class 7 Maths Chapter 2 Worksheet 1 Solutions

The DAV Class 7 Maths Book Solutions Pdf and DAV Class 7 Maths Chapter 2 Worksheet 1 Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 2 WS 1 Solutions

Question 1.
(i) $$\frac{2}{7}+\frac{5}{7}$$
$$\frac{2}{7}+\frac{5}{7}$$
= $$\frac{2+5}{7}$$
= $$\frac{7}{7}$$
= 1

(ii) $$\frac{-5}{9}+\frac{7}{9}$$
$$\frac{-5}{9}+\frac{7}{9}$$
= $$\frac{-5+7}{9}$$
= $$\frac{2}{9}$$

(iii) $$2 \frac{3}{11}+\frac{9}{-11}$$
$$2 \frac{3}{11}+\frac{9}{-11}$$
= $$\frac{25}{11}+\frac{-9}{11}$$
= $$\frac{25+(-9)}{11}$$
= $$\frac{16}{11}$$

(iv) $$\frac{-5}{4}+\frac{5}{4}$$
$$\frac{-5}{4}+\frac{5}{4}$$
= $$\frac{-5+5}{4}$$
= $$\frac{0}{4}$$
= 0

(v) $$-2 \frac{5}{6}+\frac{13}{-6}$$
$$-2 \frac{5}{6}+\frac{13}{-6}$$
= $$-\frac{17}{6}+\frac{-13}{6}$$
= $$\frac{-17+(-13)}{6}$$
= $$\frac{-30}{6}$$ = – 5

(vi) $$\frac{-21}{3}+\frac{18}{3}$$
$$\frac{-21}{3}+\frac{18}{3}$$
= $$\frac{-21+18}{3}$$
= $$\frac{-3}{3}$$
= – 1.

Question 2.
Find the value of:
(i) $$\frac{4}{9}+\frac{7}{4}$$
$$\frac{4}{9}+\frac{7}{4}$$
= $$\frac{4 \times 4+7 \times 9}{36}$$
= $$\frac{16+63}{36}$$
= $$\frac{79}{36}$$

(ii) $$\frac{-7}{11}+\frac{1}{4}$$
$$\frac{-7}{11}+\frac{1}{4}$$
= $$\frac{-7 \times 4+1 \times 11}{44}$$
= $$\frac{-28+11}{44}$$
= $$\frac{-17}{44}$$

(iii) $$\frac{5}{8}+\frac{-3}{5}$$
$$\frac{5}{8}+\frac{-3}{5}$$
= $$\frac{5 \times 5+8 \times(-3)}{40}$$
= $$\frac{25-24}{40}$$
= $$\frac{1}{40}$$

(iv) $$\frac{10}{63}+\frac{6}{7}$$
$$\frac{10}{63}+\frac{6}{7}$$
= $$\frac{10 \times 1+6 \times 9}{63}$$
= $$\frac{10+54}{63}$$
= $$\frac{64}{63}$$

(v) $$\frac{-7}{64}+\frac{3}{-16}$$
$$\frac{-7}{64}+\frac{3}{-16}$$
= $$\frac{-7}{64}+\frac{-3}{16}$$
= $$\frac{-7 \times 1+(-3) \times 4}{64}$$
= $$\frac{-7-12}{64}$$
= $$\frac{-19}{64}$$

(vi) $$\frac{5}{12}+\frac{-9}{20}$$
$$\frac{5}{12}+\frac{-9}{20}$$
= $$\frac{5 \times 5+(-9) \times 3}{60}$$
= $$\frac{25-27}{60}$$
= $$\frac{-2}{60}$$
= $$\frac{-1}{30}$$

Question 3.
Verify x + y = y + x for the following values of x and y:
(i) x = $$\frac{5}{7}$$, y = $$\frac{- 3}{2}$$
To verify that x + y = y + x
L.H.S:
$$\frac{5}{7}$$ + $$\frac{- 3}{2}$$ = $$\frac{5 \times 2+7 \times(-3)}{14}$$
= $$\frac{10-21}{14}$$
= $$\frac{-11}{14}$$

R.H.S:
$$\frac{- 3}{2}$$ + $$\frac{5}{7}$$ = $$\frac{-3 \times 7+2 \times 5}{14}$$
= $$\frac{-21+10}{14}$$
= $$\frac{-11}{14}$$
L.H.S = R.H.S
Hence verified.

(ii) x = 5, y = $$\frac{3}{2}$$
To verify that x + y = y + x
L.H.S:
5 + $$\frac{3}{2}$$ = $$\frac{5 \times 2+3 \times 1}{2}$$
= $$\frac{10+3}{2}$$
= $$\frac{13}{2}$$

R.H.S:
$$\frac{3}{2}$$ + 5 = $$\frac{3 \times 1+5 \times 2}{2}$$
= $$\frac{3+10}{2}$$
= $$\frac{13}{2}$$
L.H.S = R.H.S
Hence verified.

(iii) x = $$\frac{- 5}{14}$$, y = $$\frac{- 1}{21}$$
To verify that x + y = y + x
L.H.S:
$$\frac{- 5}{14}$$ – $$\frac{- 1}{21}$$ = $$\frac{-5 \times 3-1 \times 2}{42}$$
= $$\frac{-15-2}{42}$$
= $$\frac{-17}{42}$$

R.H.S:
$$\frac{- 1}{21}$$ – $$\frac{- 5}{14}$$ = $$\frac{-1 \times 2-5 \times 3}{42}$$
= $$\frac{-2-15}{42}$$
= $$\frac{-17}{42}$$
L.H.S = R.H.S
Hence verified.

(iv) x = – 8, y = $$\frac{9}{2}$$
To verify that x + y = y + x
L.H.S:
– 8 + $$\frac{9}{2}$$ = $$\frac{-8 \times 2+9 \times 1}{2}$$
= $$\frac{-16+9}{2}$$
= $$\frac{-7}{2}$$

R.H.S:
$$\frac{9}{2}$$ – 8 = $$\frac{9 \times 1-8 \times 2}{2}$$
= $$\frac{9-16}{2}$$
= $$\frac{-7}{2}$$
L.H.S = R.H.S
Hence verified.

Question 4.
Verify x + (y + z) = (x + y) + z for following values of x, y and z.
(i) x = $$\frac{3}{4}$$, y = $$\frac{5}{6}$$, z = $$\frac{-7}{8}$$
x + (y + z) = (x + y) + z
L.H.S = $$\frac{3}{4}$$ + ($$\frac{5}{6}$$ + $$\frac{-7}{8}$$)
= $$\frac{3}{4}$$ + $$\left(\frac{5 \times 4+(-7) \times 3}{24}\right)$$
= $$\frac{3}{4}$$ + $$\left(\frac{20-21}{24}\right)$$
= $$\frac{3}{4}+\frac{-1}{24}$$
= $$\frac{3 \times 6-1 \times 1}{24}$$
= $$\frac{18-1}{24}$$
= $$\frac{17}{24}$$

R.H.S = ($$\frac{3}{4}$$ + $$\frac{5}{6}$$) + $$\frac{-7}{8}$$
= $$\left(\frac{3 \times 3+5 \times 2}{12}\right)+\left(\frac{-7}{8}\right)$$
= $$\left(\frac{9+10}{12}\right)-\frac{7}{8}$$
= $$\frac{19}{12}-\frac{7}{8}$$
= $$\frac{19 \times 2-7 \times 3}{24}$$
= $$\frac{38-21}{24}$$
= $$\frac{17}{24}$$
L.H.S = R.H.S
Hence verified.

(ii) x = $$\frac{2}{3}$$, y = $$\frac{-5}{6}$$, z = $$\frac{-7}{9}$$
x + (y + z) = (x + y) + z
L.H.S. = $$\frac{2}{3}$$ + ($$\frac{-5}{6}$$ + $$\frac{-7}{9}$$)
= $$\frac{2}{3}$$ + $$\left(\frac{-5 \times 3+2 \times-7}{18}\right)$$
= $$\frac{2}{3}+\left(\frac{-15-14}{18}\right)$$
= $$\frac{2}{3}-\frac{29}{18}$$
= $$\frac{-17}{18}$$

R.H.S = ($$\frac{2}{3}$$ + $$\frac{-5}{6}$$) + $$\frac{-7}{9}$$
= $$\left(\frac{2}{3}+\frac{-5}{6}\right)+\frac{-7}{9}$$
= $$\left(\frac{2 \times 2-5 \times 1}{6}\right)-\frac{7}{9}$$
= $$\left(\frac{4-5}{6}\right)-\frac{7}{9}$$
= $$\frac{-1}{6}-\frac{7}{9}$$
= $$\frac{-3 \times 1-7 \times 2}{18}$$
= $$\frac{-3-14}{18}$$
= $$\frac{-17}{18}$$
L.H.S = R.H.S
Hence verified.

(iii) x = $$\frac{3}{5}$$, y = $$\frac{-6}{9}$$, z = $$\frac{2}{10}$$
x + (y + z) = (x + y) + z
L.H.S. = $$\frac{3}{5}$$ + ($$\frac{-6}{9}$$ + $$\frac{2}{10}$$)
= $$\frac{3}{5}$$ + $$\left(\frac{-6 \times 10+2 \times 9}{90}\right)$$
= $$\frac{3}{5}$$ + $$\left(\frac{-60+18}{90}\right)$$
= $$\frac{3}{5}+\left(\frac{-42}{90}\right)$$
= $$\frac{3}{5}-\frac{42}{90}$$
= $$\frac{3 \times 18-42 \times 1}{90}$$
= $$\frac{54-42}{90}$$
= $$\frac{12}{90}$$

R.H.S. = ($$\frac{3}{5}$$ + $$\frac{-6}{9}$$) + $$\frac{2}{10}$$
= $$\left(\frac{3 \times 9+5 \times-6}{45}\right)$$ + $$\frac{2}{10}$$
= $$\left(\frac{27-30}{45}\right)$$ + $$\frac{2}{10}$$
= $$\frac{-3}{45}+\frac{2}{10}$$
= $$\frac{-3 \times 2+2 \times 9}{90}$$
= $$\frac{-6+18}{90}$$
= $$\frac{12}{90}$$
L.H.S = R.H.S
Hence verified.

(iv) x = $$\frac{-3}{5}$$, y = $$\frac{-7}{10}$$, z = $$\frac{-8}{15}$$
x + (y + z) = (x + y) + z
L.H.S. = $$\frac{-3}{5}$$ + ($$\frac{-7}{10}$$ + $$\frac{-8}{15}$$)
= $$\frac{-3}{5}$$ + $$\left(\frac{-7 \times 3+2 \times-8}{30}\right)$$
= $$\frac{-3}{5}$$ + $$\left(\frac{-21-16}{30}\right)$$
= $$\frac{-3}{5}+\left(\frac{-37}{30}\right)$$
= $$\frac{-3}{5}-\frac{37}{30}$$
= $$\frac{-3 \times 6-37 \times 1}{30}$$
= $$\frac{-18-37}{30}$$
= $$\frac{-55}{30}$$

R.H.S. = ($$\frac{-3}{5}$$ + $$\frac{-7}{10}$$) + $$\frac{-8}{15}$$
= $$\left(\frac{-3 \times 2+1 \times-7}{10}\right)-\frac{8}{15}$$
= $$\left(\frac{-6-7}{10}\right)-\frac{8}{15}$$
= $$\frac{-13}{10}-\frac{8}{15}$$
= $$\frac{-39-16}{30}$$
= $$\frac{-55}{30}$$

Question 5.
Simplify:
(i) $$\frac{-3}{10}+\frac{12}{-10}+\frac{14}{10}$$
$$\frac{-3}{10}+\frac{12}{-10}+\frac{14}{10}$$ = $$\frac{-3}{10}+\frac{-12}{10}+\frac{14}{10}$$
= $$\frac{-3-12+14}{10}$$
= $$\frac{-15+14}{10}$$
= $$\frac{-1}{10}$$

(ii) $$\frac{-5}{10}+\frac{6}{13}$$ + 8
$$\frac{-5}{10}+\frac{6}{13}$$ + 8 = $$\frac{-5}{10}+\frac{6}{13}+\frac{8}{1}$$
= $$\frac{-65+60+1040}{130}$$
= $$\frac{-65+1100}{130}$$
= $$\frac{1035}{130}$$
= $$\frac{207}{26}$$
= $$7 \frac{25}{26}$$

(iii) $$\frac{-5}{10}+\frac{9}{7}+\frac{3}{20}+\frac{-11}{14}$$
$$\frac{-5}{10}+\frac{9}{7}+\frac{3}{20}+\frac{-11}{14}$$ = $$\frac{-70+180+21-110}{140}$$
= $$\frac{201-180}{140}$$
= $$\frac{21}{140}$$
= $$\frac{3}{20}$$

(iv) $$\frac{5}{36}+\frac{-7}{8}+\frac{6}{-72}+\frac{-3}{-12}$$
$$\frac{5}{36}+\frac{-7}{8}+\frac{6}{-72}+\frac{-3}{-12}$$ = $$\frac{5}{36}+\frac{-7}{8}+\frac{-6}{72}+\frac{3}{12}$$
= $$\frac{5 \times 2+9 \times-7+1 \times-6+3 \times 6}{72}$$
= $$\frac{10-63-6+18}{72}$$
= $$\frac{28-69}{72}$$
= $$\frac{-41}{72}$$

Question 6.
For x = $$\frac{1}{5}$$ and y = $$\frac{3}{7}$$, verify that – (x + y) = (- x) + (- y).
– (x + y) = (- x) + (- y)
= – $$\left(\frac{1}{5}+\frac{3}{7}\right)$$ = $$\left(\frac{-1}{5}\right)+\left(-\frac{3}{7}\right)$$
= – $$\left(\frac{1 \times 7+3 \times 5}{35}\right)$$ = $$\left(\frac{-1 \times 7+5 \times-3}{35}\right)$$
= – $$\left(\frac{7+15}{35}\right)$$ = $$\left(\frac{-7-15}{35}\right)$$
= – $$\frac{22}{35}$$ = $$\frac{-22}{35}$$
Hence verified.

Question 7.
Write True (T) or False (F) for the following statements:

(i) $$\frac{-2}{-3}$$ is the additive inverse of $$\frac{2}{3}$$. ________
False

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

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

(iv) 1 is the identity element of addition. ________
False

(v) $$\frac{-9}{7}$$ + 0 = 0 ________
False

(vi) Additive inverse of $$\frac{-3}{5}$$ is $$\frac{3}{5}$$. ________
True

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

### DAV Class 7 Maths Chapter 2 Worksheet 1 Notes

(i) If the denominators of the given rational numbers are same then add their numerators.
e.g. $$\frac{5}{8}+\frac{4}{8}$$ = $$\frac{5+4}{8}$$
⇒ $$\frac{9}{8}$$

(ii) If the denominators of the given rational numbers are not same then their denominators are made equal by taking their L.C.M.
e.g. $$\frac{2}{3}+\frac{5}{4}$$ = $$\frac{2 \times 4}{3 \times 4}+\frac{5 \times 3}{4 \times 3}$$
⇒ $$\frac{8}{12}+\frac{15}{12}$$
⇒ $$\frac{8+15}{12}$$
⇒ $$\frac{23}{12}$$

The sum of the given rational numbers is same even if their orders are changed.
e.g. x + y = y + x
where x and y are rational numbers.

The sum of any three rational numbers remains same if their groups are changed.
e.g. x + (y + z) (x + y) + z where x, y, z are rational numbers.

When zero is added to any rational number, the sum is the rational number itself, i.e., if x is a rational number, then 0 + x = x + 0 = x.
Zero is called the identity element of addition.

Every rational number has its additive inverse such that their sum is zero.
e.g. Let x be a rational number
∴ Its additive inverse is – x
so, x + (- x) = 0

Subtraction of Rational Numbers:

• Commutative property for subtraction of rational numbers does not hold true.
e.g. x – y ≠ y – x if x and y are two rational numbers.
but x – y = – (y – x)
• Associative property for the subtraction of rational numbers also does not hold true
e.g. (x – y) – z ≠ x – (y – z)
• Identity element for subtraction of rational numbers does not exist.
e.g. x – 0 ≠ 0 – x

Multiplication of Rational Numbers:

• The product of two rational numbers is a rational number.
e.g. $$\frac{-6}{5} \times \frac{3}{4}$$ = $$\frac{(-6) \times(3)}{(5) \times(4)}$$ = $$\frac{-18}{20}$$ a rational number
• Commutative property for multiplication of rational numbers holds true.
e.g. x × y = y × x if x and y are rational numbers.
• Associative property for multiplication of rational numbers also holds true:
e.g. (x × y) × z = x × (y × z) where x, y and z are all rational numbers.
Product of any rational number with zero is always zero.
e.g. x × 0 = 0 × x =0 where x is a rational number.

Reciprocal of a Rational Number:
Identity Element under Multiplication:
One multiplied by any rational number is the rational number itself, i.e., if x is a rational number, then
x × 1 = 1 × x = x
i.e., 1 is the identity element under multiplication.

Distributive Law of Multiplication over Addition:

If x, y and z are rational numbers, then
(i) x × (y + z) = x × y + x × z
(ii) x × (y – z) = x × y – x × z

Reciprocal of any non-zero rational number exists.
e.g. Reciprocal of $$\frac{a}{b}$$ is $$\frac{b}{a}$$where $$\frac{a}{b}$$ ≠ 0
Two non-zero numbers $$\frac{a}{b}$$ and $$\frac{b}{a}$$ are reciprocals of each other.

Division of Rational Numbers:

• Division of a rational number by any non-zero rational number is a rational number.
e.g. x ÷ y = $$\frac{x}{y}$$ (y ≠ 0) is a rational number.
• When a rational number (non-zero) is divided by the same rational number, the quotient is one.
• When a rational number is divided by 1, the quotient is the same rational number.
• If x, y and z are rational numbers then
(x + y) ÷ z = x + z + y ÷ z
and (x – y) + z = x ÷ z – y ÷ z
• If x and y are two non-zero rational numbers then
x + y ≠ y + x so, commutative property for division does not hold true.
• Associative property for the division does not hold true.
e.g. x + (y + z) ≠ (x ÷ y) + z where x, y, z are non-zero rational numbers.
• Distributive property for division of rational numbers does not hold true.
e.g. x ÷ (y + z) ≠ x ÷ y + x ÷ z where x, y, z are non-zero rational numbers.
• If x and y are two rational numbers then the number lying between them are
$$\frac{x+y}{2}, \frac{x+y}{3}, \frac{x+y}{4} \ldots$$
• Between any two rational numbers, there are infinitely many rational numbers.

Example 1:

(i) $$\frac{2}{3}$$ and $$\frac{5}{6}$$
$$\frac{2}{3}$$ + $$\frac{5}{6}$$
⇒ $$\frac{2 \times 2+5 \times 1}{6}$$
⇒ $$\frac{4+5}{6}$$
⇒ $$\frac{9}{6}$$
⇒ $$\frac{3}{2}$$

(ii) $$\frac{1}{5}$$ and $$\frac{1}{6}$$
$$\frac{1}{5}$$ + $$\frac{1}{6}$$
⇒ $$\frac{6 \times 1+5 \times 1}{5 \times 6}$$
⇒ $$\frac{6+5}{30}$$
⇒ $$\frac{11}{30}$$

(iii) $$\frac{2}{5}$$ and $$\frac{-4}{10}$$
$$\frac{2}{5}$$ + $$\frac{-4}{10}$$
⇒ $$\frac{2}{5}+\left(-\frac{4}{10}\right)$$
⇒ $$\frac{2}{5}-\frac{4}{10}$$
⇒ $$\frac{4-4}{10}$$
⇒ $$\frac{0}{10}$$
⇒ 0

Example 2:
Verify that: $$\frac{4}{7}+\frac{2}{3}=\frac{2}{3}+\frac{4}{7}$$.
L.H.S.
$$\frac{4}{7}+\frac{2}{3}$$
= $$\frac{4 \times 3+2 \times 7}{7 \times 3}$$
= $$\frac{12+14}{21}$$
= $$\frac{26}{21}$$

R.H.S.
$$\frac{2}{3}+\frac{4}{7}$$
= $$\frac{2 \times 7+4 \times 3}{3 \times 7}$$
= $$\frac{14+12}{21}$$
= $$\frac{26}{21}$$

Example 3:
Verify that: (x + y) + z = x + (y + z) if
(i) x = $$\frac{1}{2}$$, y = $$\frac{5}{6}$$ and z = $$\frac{3}{4}$$
L.H.S. = (x + y) + z
= ($$\frac{1}{2}$$ + $$\frac{5}{6}$$) + $$\frac{3}{4}$$
= $$\left(\frac{3+5}{6}\right)+\frac{3}{4}$$
= $$\frac{8}{6}+\frac{3}{4}$$
= $$\frac{8 \times 2+3 \times 3}{12}$$
= $$\frac{16+9}{12}$$
= $$\frac{25}{12}$$

R.H.S = x + (y + z)
= $$\frac{1}{2}$$ + ($$\frac{5}{6}$$ + $$\frac{3}{4}$$)
= $$\frac{1}{2}$$ + $$\left(\frac{2 \times 5+3 \times 3}{12}\right)$$
= $$\frac{1}{2}$$ + $$\left(\frac{10+9}{12}\right)$$
= $$\frac{1}{2}+\frac{19}{12}$$
= $$\frac{6+19}{12}=\frac{25}{12}$$
L.H.S. = R.H.S
Hence Verified.

(ii) x = $$\frac{- 2}{3}$$, y = $$\frac{1}{4}$$ and z = $$\frac{-3}{2}$$
L.H.S. = (x + y) + z
= $$\left(\frac{-2}{3}+\frac{1}{4}\right)+\left(\frac{-3}{2}\right)$$
= $$\left(\frac{-8+3}{12}\right)-\frac{3}{2}$$
= $$\frac{-5}{12}-\frac{3}{2}$$
= $$\frac{-5-18}{12}$$
= $$\frac{-23}{12}$$

R.H.S. = x + (y + z)
= $$\frac{-2}{3}+\left(\frac{1}{4}-\frac{3}{2}\right)$$
= $$\frac{-2}{3}+\left(\frac{1-6}{4}\right)$$
= $$\frac{-2}{3}-\frac{5}{4}$$
= $$\frac{-8-15}{12}$$
= $$\frac{-23}{12}$$
L.H.S = R.HS.
Hence verified.

Example 4:
Simplify: $$\frac{4}{25}+\frac{-3}{5}+\frac{-6}{15}+\frac{1}{10}$$
$$\frac{4}{25}+\frac{-3}{5}+\frac{-6}{15}+\frac{1}{10}$$
L.C.M. of 25, 5, 15 and 10 = 150
Making the denominators same, we get
= $$\frac{4 \times 6}{25 \times 6}+\frac{-3 \times 30}{5 \times 30}+\frac{-6 \times 10}{15 \times 10}+\frac{1 \times 15}{10 \times 15}$$
= $$\frac{24}{150}+\frac{-90}{150}+\frac{-60}{150}+\frac{15}{150}$$
= $$\frac{24-90-60+15}{150}$$
= $$\frac{-111}{150} \text { or } \frac{-37}{50}$$

Example 5:
State whether the given operation is true or false.
$$\left(\frac{5}{8}-\frac{3}{2}\right)-\frac{7}{8}=\frac{5}{8}-\left(\frac{3}{2}-\frac{7}{8}\right)$$
Solution:
L.H.S. = $$\left(\frac{5}{8}-\frac{3}{2}\right)-\frac{7}{8}$$
= $$\left(\frac{5-3 \times 4}{8}\right)-\frac{7}{8}$$
= $$\left(\frac{5-12}{8}\right)-\frac{7}{8}$$
= $$-\frac{7}{8}-\frac{7}{8}$$
= $$-\frac{14}{8}$$

R.H.S. = $$\frac{5}{8}-\left(\frac{3}{2}-\frac{7}{8}\right)$$
= $$\frac{5}{8}-\left(\frac{3 \times 4-7}{8}\right)$$
= $$\frac{5}{8}-\left(\frac{12-7}{8}\right)$$
= $$\frac{5}{8}-\frac{5}{8}$$
= 0
Hence the given option is false.

Example 6.
Simplify : $$\frac{-3}{5}-\left(\frac{-4}{15}\right)-\left(\frac{7}{-10}\right)$$
Solution:
$$\frac{-3}{5}-\left(\frac{-4}{15}\right)-\left(\frac{7}{-10}\right)$$ = $$\frac{-3}{5}-\left(\frac{-4}{15}\right)-\left(\frac{-7}{10}\right)$$
L.C.M. of 5, 15 and 10 = 30.
= $$\frac{-3}{5} \times \frac{6}{6}-\left(\frac{-4 \times 2}{15 \times 2}\right)-\left(\frac{-7 \times 3}{10 \times 3}\right)$$
= $$\frac{-18}{30}-\left(\frac{-8}{30}\right)-\left(\frac{-21}{30}\right)$$
= $$\frac{-18}{30}+\frac{8}{30}+\frac{21}{30}=\frac{11}{30}$$

Example 7.
If x = $$\frac{-3}{4}$$, y = $$\frac{5}{7}$$ and z = $$\frac{-2}{11}$$
Verify that (x × y) × z = x × (y × z)
L.H.S. = (x × y) × z
= $$\left(\frac{-3}{4} \times \frac{5}{7}\right) \times \frac{-2}{11}$$
= $$\left(\frac{-15}{28}\right) \times\left(\frac{-2}{11}\right)$$
= $$\frac{-15 \times-2}{28 \times 11}=\frac{30}{308}$$
= $$\frac{15}{154}$$ (Standard form)

R.H.S. = x × (y × z)
= $$\frac{-3}{4} \times\left(\frac{5}{7} \times \frac{-2}{11}\right)$$
= $$\left(\frac{-3}{4}\right) \times\left(\frac{-10}{77}\right)$$
= $$\frac{(-3) \times(-10)}{4 \times 77}$$
= $$\frac{30}{308}$$
= $$\frac{15}{154}$$ (Standard form)
L.H.S = R.HS.
Hence verified.

Example 8.
If x = $$\frac{-4}{7}$$, y = $$\frac{-3}{4}$$ and z = $$\frac{5}{6}$$.
Verify that x × (y + z) = x × y + x × z
L.H.S. = x × (y + z)
= $$\frac{-4}{7} \times\left[\left(\frac{-3}{4}\right)+\left(\frac{5}{6}\right)\right]$$
= $$\frac{-4}{7} \times\left[\frac{-3 \times 3+5 \times 2}{12}\right]$$
= $$\frac{-4}{7} \times\left[\frac{-9+10}{12}\right]$$
= $$\frac{-4}{7} \times \frac{1}{12}$$
= $$\frac{-4}{84}=\frac{-1}{21}$$ (Standard form)

R.H.S. = x × y + x × z
= $$\left(\frac{-4}{7}\right) \times\left(\frac{-3}{4}\right)+\left(\frac{-4}{7}\right) \times\left(\frac{5}{6}\right)$$
= $$\frac{12}{28}+\left(\frac{-20}{42}\right)$$
= $$\frac{3 \times 12-20 \times 2}{84}$$
= $$\frac{36-40}{84}$$
= $$\frac{-4}{84}=\frac{-1}{21}$$ (Standard form)
L.H.S = R.HS.
Hence verified.

Example 9:
Find the value of x, if
$$\frac{5}{6} \times\left(\frac{-7}{8} \times \frac{8}{3}\right)$$ = $$\left(\frac{5}{6} \times x\right) \times \frac{8}{3}$$
$$\frac{5}{6} \times\left(\frac{-7}{8} \times \frac{8}{3}\right)$$ = $$\left(\frac{5}{6} \times x\right) \times \frac{8}{3}$$
$$\frac{5}{6} \times \frac{-7}{3}=\frac{5 x}{6} \times \frac{8}{3}$$
$$\frac{-35}{18}=\frac{40 x}{18}$$
– 35 = 40 x (multiplying both sides by 18)
x = $$\frac{-35}{40}=\frac{-7}{8}$$ (Standard form)

Example 10:
Verify that (x × y)-1 = x-1 × y-1 if x = $$\frac{2}{5}$$, y = $$-\frac{3}{10}$$
Solution:
L.H.S. = (x × y)-1
= $$\left[\frac{2}{5} \times\left(-\frac{3}{10}\right)\right]^1$$
= $$\left[\frac{-6}{50}\right]^{-1}$$
= $$\frac{50}{-6}=\frac{25}{-3}$$ (Standard form)

R.H.S. = x-1 × y-1
= $$\left(\frac{2}{5}\right)^{-1} \times\left(-\frac{3}{10}\right)^{-1}$$
= $$\frac{5}{2} \times \frac{-10}{3}$$
= $$\frac{-50}{6}=\frac{-25}{3}$$ (Standard form)
L.H.S. = R.H.S.
Hence verified.