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.
Add the followimg:
(i) \(\frac{2}{7}+\frac{5}{7}\)
Answer:
\(\frac{2}{7}+\frac{5}{7}\)
= \(\frac{2+5}{7}\)
= \(\frac{7}{7}\)
= 1
(ii) \(\frac{-5}{9}+\frac{7}{9}\)
Answer:
\(\frac{-5}{9}+\frac{7}{9}\)
= \(\frac{-5+7}{9}\)
= \(\frac{2}{9}\)
(iii) \(2 \frac{3}{11}+\frac{9}{-11}\)
Answer:
\(2 \frac{3}{11}+\frac{9}{-11}\)
= \(\frac{25}{11}+\frac{-9}{11}\)
= \(\frac{25+(-9)}{11}\)
= \(\frac{16}{11}\)
(iv) \(\frac{-5}{4}+\frac{5}{4}\)
Answer:
\(\frac{-5}{4}+\frac{5}{4}\)
= \(\frac{-5+5}{4}\)
= \(\frac{0}{4}\)
= 0
(v) \(-2 \frac{5}{6}+\frac{13}{-6}\)
Answer:
\(-2 \frac{5}{6}+\frac{13}{-6}\)
= \(-\frac{17}{6}+\frac{-13}{6}\)
= \(\frac{-17+(-13)}{6}\)
= \(\frac{-30}{6}\) = – 5
(vi) \(\frac{-21}{3}+\frac{18}{3}\)
Answer:
\(\frac{-21}{3}+\frac{18}{3}\)
= \(\frac{-21+18}{3}\)
= \(\frac{-3}{3}\)
= – 1.
Question 2.
Find the value of:
(i) \(\frac{4}{9}+\frac{7}{4}\)
Answer:
\(\frac{4}{9}+\frac{7}{4}\)
= \(\frac{4 \times 4+7 \times 9}{36}\)
= \(\frac{16+63}{36}\)
= \(\frac{79}{36}\)
(ii) \(\frac{-7}{11}+\frac{1}{4}\)
Answer:
\(\frac{-7}{11}+\frac{1}{4}\)
= \(\frac{-7 \times 4+1 \times 11}{44}\)
= \(\frac{-28+11}{44}\)
= \(\frac{-17}{44}\)
(iii) \(\frac{5}{8}+\frac{-3}{5}\)
Answer:
\(\frac{5}{8}+\frac{-3}{5}\)
= \(\frac{5 \times 5+8 \times(-3)}{40}\)
= \(\frac{25-24}{40}\)
= \(\frac{1}{40}\)
(iv) \(\frac{10}{63}+\frac{6}{7}\)
Answer:
\(\frac{10}{63}+\frac{6}{7}\)
= \(\frac{10 \times 1+6 \times 9}{63}\)
= \(\frac{10+54}{63}\)
= \(\frac{64}{63}\)
(v) \(\frac{-7}{64}+\frac{3}{-16}\)
Answer:
\(\frac{-7}{64}+\frac{3}{-16}\)
= \(\frac{-7}{64}+\frac{-3}{16}\)
= \(\frac{-7 \times 1+(-3) \times 4}{64}\)
= \(\frac{-7-12}{64}\)
= \(\frac{-19}{64}\)
(vi) \(\frac{5}{12}+\frac{-9}{20}\)
Answer:
\(\frac{5}{12}+\frac{-9}{20}\)
= \(\frac{5 \times 5+(-9) \times 3}{60}\)
= \(\frac{25-27}{60}\)
= \(\frac{-2}{60}\)
= \(\frac{-1}{30}\)
Question 3.
Verify x + y = y + x for the following values of x and y:
(i) x = \(\frac{5}{7}\), y = \(\frac{- 3}{2}\)
Answer:
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 = \(\frac{3}{2}\)
Answer:
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 = \(\frac{- 1}{21}\)
Answer:
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 = \(\frac{9}{2}\)
Answer:
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 = \(\frac{-7}{8}\)
Answer:
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 = \(\frac{-7}{9}\)
Answer:
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 = \(\frac{2}{10}\)
Answer:
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 = \(\frac{-8}{15}\)
Answer:
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) \(\frac{-3}{10}+\frac{12}{-10}+\frac{14}{10}\)
Answer:
\(\frac{-3}{10}+\frac{12}{-10}+\frac{14}{10}\) = \(\frac{-3}{10}+\frac{-12}{10}+\frac{14}{10}\)
= \(\frac{-3-12+14}{10}\)
= \(\frac{-15+14}{10}\)
= \(\frac{-1}{10}\)
(ii) \(\frac{-5}{10}+\frac{6}{13}\) + 8
Answer:
\(\frac{-5}{10}+\frac{6}{13}\) + 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) \(\frac{-5}{10}+\frac{9}{7}+\frac{3}{20}+\frac{-11}{14}\)
Answer:
\(\frac{-5}{10}+\frac{9}{7}+\frac{3}{20}+\frac{-11}{14}\) = \(\frac{-70+180+21-110}{140}\)
= \(\frac{201-180}{140}\)
= \(\frac{21}{140}\)
= \(\frac{3}{20}\)
(iv) \(\frac{5}{36}+\frac{-7}{8}+\frac{6}{-72}+\frac{-3}{-12}\)
Answer:
\(\frac{5}{36}+\frac{-7}{8}+\frac{6}{-72}+\frac{-3}{-12}\) = \(\frac{5}{36}+\frac{-7}{8}+\frac{-6}{72}+\frac{3}{12}\)
= \(\frac{5 \times 2+9 \times-7+1 \times-6+3 \times 6}{72}\)
= \(\frac{10-63-6+18}{72}\)
= \(\frac{28-69}{72}\)
= \(\frac{-41}{72}\)
Question 6.
For x = \(\frac{1}{5}\) and y = \(\frac{3}{7}\), verify that – (x + y) = (- x) + (- y).
Answer:
– (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 \(\frac{2}{3}\). ________
Answer:
False
(ii) \(\frac{2}{3}\) + \(\frac{4}{5}\) is a rational number. ________
Answer:
True
(iii) \(\frac{-5}{3}\) + \(\frac{5}{-3}\) is equal to zero. ________
Answer:
False
(iv) 1 is the identity element of addition. ________
Answer:
False
(v) \(\frac{-9}{7}\) + 0 = 0 ________
Answer:
False
(vi) Additive inverse of \(\frac{-3}{5}\) is \(\frac{3}{5}\). ________
Answer:
True
(vii) Negative of a negative rational number is negative. ________
Answer:
False
DAV Class 7 Maths Chapter 2 Worksheet 1 Notes
Addition of Rational Numbers:
(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. \(\frac{2}{3}+\frac{5}{4}\) = \(\frac{2 \times 4}{3 \times 4}+\frac{5 \times 3}{4 \times 3}\)
⇒ \(\frac{8}{12}+\frac{15}{12}\)
⇒ \(\frac{8+15}{12}\)
⇒ \(\frac{23}{12}\)
Commutative Property of Addition:
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.
Associative Property of Addition:
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.
Identity Element of Addition:
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.
Additive Inverse:
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:
Add the following rational numbers.
(i) \(\frac{2}{3}\) and \(\frac{5}{6}\)
Answer:
\(\frac{2}{3}\) + \(\frac{5}{6}\)
⇒ \(\frac{2 \times 2+5 \times 1}{6}\)
⇒ \(\frac{4+5}{6}\)
⇒ \(\frac{9}{6}\)
⇒ \(\frac{3}{2}\)
(ii) \(\frac{1}{5}\) and \(\frac{1}{6}\)
Answer:
\(\frac{1}{5}\) + \(\frac{1}{6}\)
⇒ \(\frac{6 \times 1+5 \times 1}{5 \times 6}\)
⇒ \(\frac{6+5}{30}\)
⇒ \(\frac{11}{30}\)
(iii) \(\frac{2}{5}\) and \(\frac{-4}{10}\)
Answer:
\(\frac{2}{5}\) + \(\frac{-4}{10}\)
⇒ \(\frac{2}{5}+\left(-\frac{4}{10}\right)\)
⇒ \(\frac{2}{5}-\frac{4}{10}\)
⇒ \(\frac{4-4}{10}\)
⇒ \(\frac{0}{10}\)
⇒ 0
Example 2:
Verify that: \(\frac{4}{7}+\frac{2}{3}=\frac{2}{3}+\frac{4}{7}\).
Answer:
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 = \(\frac{3}{4}\)
Answer:
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 = \(\frac{-3}{2}\)
Answer:
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: \(\frac{4}{25}+\frac{-3}{5}+\frac{-6}{15}+\frac{1}{10}\)
Answer:
\(\frac{4}{25}+\frac{-3}{5}+\frac{-6}{15}+\frac{1}{10}\)
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)
Answer:
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
Answer:
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
\(\frac{5}{6} \times\left(\frac{-7}{8} \times \frac{8}{3}\right)\) = \(\left(\frac{5}{6} \times x\right) \times \frac{8}{3}\)
Answer:
\(\frac{5}{6} \times\left(\frac{-7}{8} \times \frac{8}{3}\right)\) = \(\left(\frac{5}{6} \times x\right) \times \frac{8}{3}\)
\(\frac{5}{6} \times \frac{-7}{3}=\frac{5 x}{6} \times \frac{8}{3}\)
\(\frac{-35}{18}=\frac{40 x}{18}\)
– 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.