NCERT Solutions for Class 8 Maths - Algebraic Expressions and Identities - 1 (Exercise 8.1 and 8.2)

Q1. Add the following:
(i) ab - bc, bc - ca, ca - ab
Ans: (ab - bc) + (bc - ca) + (ca - ab)
= ab - bc + bc - ca + ca - ab
= (ab - ab) + (- bc + bc) + (- ca + ca)
= 0

(ii) (a - b + ab) + (b - c + bc) + (c - a + ac)
Ans: a - b + ab + b - c + bc + c - a + ac
= (a - a) + (- b + b) + (- c + c) + ab + bc + ac
= ab + bc + ac

iii) 2p²q² - 3pq + 4, 5 + 7pq - 3p²q²
Ans: (2p²q² - 3pq + 4) + (5 + 7pq - 3p²q²)
= 2p²q² - 3p²q² + (- 3pq + 7pq) + (4 + 5)
= (2 - 3)p²q² + 4pq + 9
= - p²q² + 4pq + 9

(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Ans: (l² + m²) + (m² + n²) + (n² + l²) + (2lm + 2mn + 2nl)
= l² + l² + m² + m² + n² + n² + 2lm + 2mn + 2nl
= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl

Q2. (a) Subtract 4a - 7ab + 3b + 12 from 12a - 9ab + 5b - 3
Ans: (12a - 9ab + 5b - 3) - (4a - 7ab + 3b + 12)
= 12a - 9ab + 5b - 3 - 4a + 7ab - 3b - 12
= (12a - 4a) + (-9ab + 7ab) + (5b - 3b) + (-3 - 12)
= 8a - 2ab + 2b - 15

(b) Subtract 3xy + 5yz - 7zx from 5xy - 2yz - 2zx + 10xyz
Ans: (5xy - 2yz - 2zx + 10xyz) - (3xy + 5yz - 7zx)
= 5xy - 2yz - 2zx + 10xyz - 3xy - 5yz + 7zx
= (5xy - 3xy) + (- 2yz - 5yz) + (- 2zx + 7zx) + 10xyz
= 2xy - 7yz + 5zx + 10xyz

(c) Subtract 4p²q - 3pq + 5pq² - 8p + 7q - 10 from 18 - 3p - 11q + 5pq - 2pq² + 5p²q
Ans: (18 - 3p - 11q + 5pq - 2pq² + 5p²q) - (4p²q - 3pq + 5pq² - 8p + 7q - 10)
= 18 - 3p - 11q + 5pq - 2pq² + 5p²q - 4p²q + 3pq - 5pq² + 8p - 7q + 10
= (18 + 10) + (-3p + 8p) + (-11q - 7q) + (5pq + 3pq) + (- 2pq² - 5pq²) + (5p²q - 4p²q)
= 28 + 5p - 18q + 8pq - 7pq² + p²q

Exercise 8.2

Q1: Find the product of the following pairs of monomials.
(i) 4, 7p
Ans: 4 × 7p = 4 × 7 × p = 28p

(ii) -4p, 7p
Ans: -4p × 7p = (-4 × 7) × (p × p) = -28p²

(iii) -4p, 7pq
Ans: -4p × 7pq = (-4 × 7) × (p × pq) = -28p²q

(iv) 4p³, -3p
Ans: 4p³ × (-3p) = (4 × -3) × (p³ × p) = -12p⁴

(v) 4p, 0
Ans: 4p × 0 = 4 × p × 0 = 0

Q2: Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Ans: We know that,
Area of rectangle = length × breadth
Area of 1st rectangle = p × q = pq
Area of 2nd rectangle = 10m × 5n = 10 × 5 × m × n = 50mn
Area of 3rd rectangle = 20x² × 5y² = 20 × 5 × x² × y² = 100x²y²
Area of 4th rectangle = 4x × 3x² = 4 × 3 × x × x² = 12x³
Area of 5th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn²p

Q3: Complete the table of products.

Q4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a², 7a⁴
Ans: We know that
Volume = length × breadth × height
Volume = 5a × 3a² × 7a⁴
= 5 × 3 × 7 × a × a² × a⁴
= 105a⁷

(ii) 2p, 4q, 8r
Ans: We know that
Volume = length × breadth × height
Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr

(iii) xy, 2x²y, 2xy²
Ans: We know that
Volume = length × breadth × height
Volume = xy × 2x²y × 2xy²
= 2 × 2 × x × x² × x × y × y × y²
= 4x⁴y⁴

(iv) a, 2b, 3c
Ans: We know that
Volume = length × breadth × height
Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc

Q5. Obtain the product of
(i) xy, yz, zx
Ans: xy × yz × zx = x × y × y × z × z × x = x²y²z²

(ii) a, - a², a³
Ans: a × (-a²) × a³ = (-1) × a × a² × a³ = -a⁶

(iii) 2, 4y, 8y², 16y³
Ans: 2 × 4y × 8y² × 16y³ = (2 × 4 × 8 × 16) × y × y² × y³
= 1024 × y⁶ = 1024y⁶

(iv) a, 2b, 3c, 6abc
Ans: a × 2b × 3c × 6abc = (2 × 3 × 6) × a × a × b × b × c × c
= 36a²b²c²

(v) m, - mn, mnp
Ans: m × (-mn) × mnp = (-1) × m × m × m × n × n × p = -m³n²p