HomeNCERT SOLUTIONS8th CLASSClass 8 Maths Chapter 9 Algebraic Expressions and Identities

## NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Solutions

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1

Ex 9.1 Class 8 Maths Question 1.
Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy
(ii) 1 + x + x2
(iii) 4x2y2 – 4x2y2z2 + z2
(iv) 3 – pq + qr – rp
(v) $\frac { x }{ 2 }$ + $\frac { y }{ 2 }$ – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution:  Ex 9.1 Class 8 Maths Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y – 3y2, 2y – 3y2 + 4y3, 5x – 4y + 3xy, 4z – 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q
Solution: Ex 9.1 Class 8 Maths Question 3.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Solution:
(i) Given: ab – bc, bc – ca, ca – ab
We have
(ab – bc) + (bc – ca) + (ca – ab) (Adding all the terms)
= ab – bc + bc – ca + ca – ab
= (ab – ab) + (bc – bc) + (ca – ca) (Collecting the like terms together)
= 0 + 0 + 0
= 0

(ii) Given:
a – b + ab, b – c + bc, c – a + ac
We have (a – b + ab) + (b – c + bc) + (c – a + ac) (Adding all the terms)
= a – b + ab + b – c + bc + c – a + ac
= (a – a) + (b – b) + (c – c) + ab + bc + ac (Collecting all the like terms together)
= 0 + 0 + 0 + ab + bc + ac
= ab + bc + ac

(iii) Given:
2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
By arranging the like terms in the same column, we have (iv) Given: l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + nl
By arranging the like terms in the same column, we have Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl)

Ex 9.1 Class 8 Maths Question 4.
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(6) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
Solution:
(a) Arranging the like terms column-wise, we have [Change the signs of all the terms of lower expressions and then add]
(b) Arranging the like terms column-wise, we have [Change the signs of all the terms of lower expressions and then add]
(c) Arranging the like terms column-wise, we have [Change the signs of all the terms of lower expressions and then add]
The terms are p2q – 7pq2 + 8pq – 18q + 5p + 20

# Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

## NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Solutions

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Ex 9.2 Class 8 Maths Question 1.
Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) -4p, 7p
(iii) -4p, 7pq
(iv) 4p3, -3p
(v) 4p, 0
Solution:
(i) 4 × 7p = (4 × 7) × p = 28p
(ii) -4p × 7p = (-4 × 7) × p × p = -28p2
(iii) -4p × 7pq = (-4 × 7) × p × pq = -28p2q
(iv) 4p3 × -3p = (4 × -3) × p3 × p = -12p4
(v) 4p x 0 = (4 × 0) × p = 0 × p = 0

Ex 9.2 Class 8 Maths Question 2.
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)
Solution:
(i) Length = p units and breadth = q units
Area of the rectangle = length × breadth = p × q = pq sq units
(ii) Length = 10 m units, breadth = 5n units
Area of the rectangle = length × breadth = 10 m × 5 n = (10 × 5) × m × n = 50 mn sq units
(iii) Length = 20x2 units, breadth = 5y2 units
Area of the rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq units
(iv) Length = 4x units, breadth = 3x2 units
Area of the rectangle = length × breadth = 4x × 3x2 = (4 × 3) × x × x2 = 12x3 sq units
(v) Length = 3mn units, breadth = 4np units
Area of the rectangle = length × breadth = 3mn × 4np = (3 × 4) × mn × np = 12mn2p sq units

Ex 9.2 Class 8 Maths Question 3.
Complete the table of Products. Solution:
Completed Table Ex 9.2 Class 8 Maths Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a2, 7a4
(ii) 2p, 4q, 8r
(iii) xy, 2x2y, 2xy2
(iv) a, 2b, 3c
Solution:
(i) Here, length = 5a, breadth = 3a2, height = 7a4
Volume of the box = l × b × h = 5a × 3a2 × 7a4 = 105 a7 cu. units
(ii) Here, length = 2p, breadth = 4q, height = 8r
Volume of the box = l × b × h = 2p × 4q × 8r = 64pqr cu. units
(iii) Here, length = xy, breadth = 2x2y, height = 2xy2
Volume of the box = l × b × h = xy × 2x2y × 2xy2 = (1 × 2 × 2) × xy × x2y × xy2 = 4x4y4 cu. units
(iv) Here, length = a, breadth = 2b, height = 3c
Volume of the box = length × breadth × height = a × 2b × 3c = (1 × 2 × 3)abc = 6 abc cu. units

Ex 9.2 Class 8 Maths Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, -a2, a3
(iii) 2, 4y, 8y2, 16y3
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp
Solution:
(i) xy × yz × zx = x2y2z2
(ii) a × (-a2) × a3 = -a6
(iii) 2 × 4y × 8y2 × 16y3 = (2 × 4 × 8 × 16) × y × y2 × y3 = 1024y6
(iv) a × 2b × 3c × 6abc = (1 × 2 × 3 × 6) × a × b × c × abc = 36 a2b2c2
(v) m × (-mn) × mnp = [1 × (-1) × 1 ]m × mn × mnp = -m3n2p

# Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

## NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Solutions

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Ex 9.3 Class 8 Maths Question 1.
Carry out the multiplication of the expressions in each of the following pairs:
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a2b2
(iv) a2 – 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) 4p × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
(ii) ab, a – b = ab × (a – b) = (ab × a) – (ab × b) = a2b – ab2
(iii) (a + b) × 7a2b2 = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3
(iv) (a2 – 9) × 4a = (a2 × 4a) – (9 × 4a) = 4a3 – 36a
(v) (pq + qr + rp) × 0 = 0
[∵ Any number multiplied by 0 is = 0]

Ex 9.3 Class 8 Maths Question 2.
Complete the table.

Solution:
(i) a × (b + c + d) = (a × b) + (a × c) + (a × d) = ab + ac + ad
(ii) (x + y – 5) (5xy) = (x × 5xy) + (y × 5xy) – (5 × 5xy) = 5x2y + 5xy2 – 25xy
(iii) p × (6p2 – 7p + 5) = (p × 6p2) – (p × 7p) + (p × 5) = 6p3 – 7p2 + 5p
(iv) 4p2q2 × (p2 – q2) = 4p2q2 × p2 – 4p2q2 × q2 = 4p4q2 – 4p2q4
(v) (a + b + c) × (abc) = (a × abc) + (b × abc) + (c × abc) = a2bc + ab2c + abc2

Completed Table:

Ex 9.3 Class 8 Maths Question 3.
Find the products. Solution: Ex 9.3 Class 8 Maths Question 4.
(a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = $\frac { 1 }{ 2 }$.
(b) Simplify: a(a2 + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1
Solution:
(a) We have 3x(4x – 5) + 3 = 4x × 3x – 5 × 3x + 3 = 12x2 – 15x + 3
(i) For x = 3, we have
12 × (3)2 – 15 × 3 + 3 = 12 × 9 – 45 + 3 = 108 – 42 = 66 (b) We have a(a2 + a + 1) + 5
= (a2 × a) + (a × a) + (1 × a) + 5
= a3 + a2 + a + 5
(i) For a = 0, we have
= (0)3 + (0)2 + (0) + 5 = 5
(ii) For a = 1, we have
= (1)3 + (1)2 + (1) + 5 = 1 + 1 + 1 + 5 = 8
(iii) For a = -1, we have
= (-1)3 + (-1)2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4

Ex 9.3 Class 8 Maths Question 5.
(a) Add: p(p – q), q(q – r) and r(r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
Solution:
(a) p(p – q) + q(q – r) + r(r – p)
= (p × p) – (p × q) + (q × q) – (q × r) + (r × r) – (r × p)
= p2 – pq + q2 – qr + r2 – rp
= p2 + q2 + r2 – pq – qr – rp
(b) 2x(z – x – y) + 2y(z – y – x)
= (2x × z) – (2x × x) – (2x × y) + (2y × z) – (2y × y) – (2y × x)
= 2xz – 2x2 – 2xy + 2yz – 2y2 – 2xy
= -2x2 – 2y2 + 2xz + 2yz – 4xy
= -2x2 – 2y2 – 4xy + 2yz + 2xz
(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l × 10n) – (4l × 3m) + (4l × 2l) – (3l × l) – (3l × -4m) – (3l × 5n)
= 40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln
= (40ln – 15ln) + (-12lm + 12lm) + (8l2 – 3l2)
= 25ln + 0 + 5l2
= 25ln + 5l2
= 5l2 + 25ln
(d) [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]
= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)
= -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac + 2ab – 2b2 + 2bc
= -3a2 – 2b2 + 4c2 – ab + 6bc – 7ac

# Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

## NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Solutions

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Ex 9.4 Class 8 Maths Question 1.
Multiply the binomials:
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
(vi) ( $\frac { 3 }{ 4 }$a2 + 3b2) and 4(a2 – $\frac { 2 }{ 3 }$ b2)
Solution:
(i) (2x + 5) × (4x – 3)
= 2x × (4x – 3) + 5 × (4x – 3)
= (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3)
= 8x2 – 6x + 20x – 15
= 8x2 + 14x – 15

(ii) (y – 8) × (3y – 4)
= y × (3y – 4) – 8 × (3y – 4)
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32

(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l2 + 1.25ml – 1.25ml – 0.25m2
= 6.25l2 + 0 – 0.25m2
= 6.25l2 – 0.25m2

(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 36 × (x + 5)
= (a × x) + (a × 5) + (36 × x) + (36 × 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) × (3pq – 2q2)
= 2pq × (3pq – 2q2) + 3q2 (3pq – 2q2)
= (2pq × 3pq) – (2pq × 2q2) + (3q2 × 3pq) – (3q2 × 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4

Ex 9.4 Class 8 Maths Question 2.
Find the product:
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2 + b) (a + b2)
(iv) (p2 – q2)(2p + q)
Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
= 15 + 5x – 6x – 2x2

(ii) (x + 7y) (7x – y)
= x(7x – y) + 7y(7x – y)
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
= 7x2 – xy + 49xy – 7y2
= 7x2 + 48xy – 7y2

(iii) (a2 + b) (a + b2)
= a2 (a + b2) + b(a + b2)
= (a2 × a) + (a2 × b2) + (b × a) + (b × b2)
= a3 + a2b2 + ab + b3

(iv) (p2 – q2)(2p + q)
= p2(2p + q) – q2(2p + q)
= (p2 × 2p) + (p2 × q) – (q2 × 2p) – (q2 × q)
= 2p3 + p2q – 2pq2 – q3

Ex 9.4 Class 8 Maths Question 3.
Simplify:
(i) (x2 – 5) (x + 5) + 25
(ii) (a2 + 5)(b3 + 3) + 5
(iii) (t + s2) (t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y)(x2 – xy + y2)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c) (a + b – c)
Solution:
(i) (x2 – 5) (x + 5) + 25
= x2(x + 5) + 5(x + 5) + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x + 0
= x3 + 5x2 – 5x

(ii) (a2 + 5)(b3 + 3) + 5
= a2(b3 + 3) + 5(b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20

(iii) (t + s2) (t2 – s)
= t(t2 – s) + s2(t2 – s)
= t3 – st + s2t2 – s3
= t3 + s2t2 – st – s3

(iv) (a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 2x2 + x2 + xy + 2xy – xy + 2xy + y2 – 2y2
= 3x2 + 4xy – y2

(vi) (x + y)(x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + x2y + xy2 – xy2 + y3
= x3 – 0 + 0 + y3
= x3 + y3

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x.+ 12y
= 1.5x (1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
= 2.25x2 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y2
= 2.25x2 + 0 + 0 + 0 – 16y2
= 2.25x2 – 16y2

(viii) (a + b + c) (a + b – c)
= a(a + b – c) + b(a + b – c) + c(a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + ab + ab – bc + bc – ac + ac + b2 – c2
= a2 + 2ab + b2 – c2 + 0 + 0
= a2 + 2ab + b2 – c2

# Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

## NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Solutions

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Ex 9.5 Class 8 Maths Question 1.
Use a suitable identity to get each of the following products:
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – $\frac { 1 }{ 2 }$) (3a – $\frac { 1 }{ 2 }$)
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2 + b2) (-a2 + b2)
(vii) (6x – 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix) ( $\frac { x }{ 2 }$ + $\frac { 3y }{ 4 }$) ( $\frac { x }{ 2 }$ + $\frac { 3y }{ 4 }$)
(x) (7a – 9b) (7a – 9b)
Solution:    Ex 9.5 Class 8 Maths Question 2.
Use the identity (x + a)(x + b) = x2 + (a + b)x + ab to find the following products.
(i) (x + 3) (x + 7)
(ii) (4x + 5)(4x + 1)
(iii) (4x – 5) (4x – 1)
(iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)
Solution: Ex 9.5 Class 8 Maths Question 3.
Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv) ( $\frac { 2 }{ 3 }$ m + $\frac { 3 }{ 2 }$ n)2
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
Solution:  Ex 9.5 Class 8 Maths Question 4.
Simplify:
(i) (a2 – b2)2
(ii) (2x + 5)2 – (2x – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2 – 2ab2c
(vii) (m2 – n2m)2 + 2m3n2
Solution:   Ex 9.5 Class 8 Maths Question 5.
Show that:
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2 + 180pq = (9p + 5q)2
(iii) ( $\frac { 4 }{ 3 }$ m – $\frac { 3 }{ 4 }$ n)2 + 2mn = $\frac { 16 }{ 9 }$ m2 + $\frac { 9 }{ 16 }$ n2
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:   Ex 9.5 Class 8 Maths Question 6.
Using identities, evaluate:
(i) 712
(ii) 992
(iii) 1022
(iv) 9982
(v) 5.22
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.92
(ix) 1.05 × 9.5
Solution:   Ex 9.5 Class 8 Maths Question 7.
Using a2 – b2 = (a + b) (a – b), find
(i) 512 – 492
(ii) (1.02)2 – (0.98)2
(iii) 1532 – 1472
(iv) 12.12 – 7.92
Solution:
(i) 512 – 492 = (51 + 49) (51 – 49) = 100 × 2 = 200
(ii) (1.02)2 – (0.98)2 = (1.02 + 0.98) (1.02 – 0.98) = 2.00 × 0.04 = 0.08
(iii) 1532 – 1472 = (153 + 147) (153 – 147) = 300 × 6 = 1800
(iv) 12.12 – 7.92 = (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84

Ex 9.5 Class 8 Maths Question 8.
Using (x + a) (x + b) = x2 + (a + b)x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
Solution:
(i) 103 × 104 = (100 + 3)(100 + 4) = (100)2 + (3 + 4) (100) + 3 × 4 = 10000 + 700 + 12 = 10712
(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5)2 + (0.1 + 0.2) (5) + 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52
(iii) 103 × 98 = (100 + 3) (100 – 2) = (100)2 + (3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)2 – (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06

# Extra Questions

## Algebraic Expressions and Identities Class 8 Extra Questions Maths Chapter 9

Extra Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

### Algebraic Expressions and Identities Class 8 Extra Questions Very Short Answer Type

Question 1.
Write two examples of each of
(i) Monomials
(ii) Binomials
(iii) Trinomials
Solution:
(i) Monomials:
(a) 3x
(b) 5xy2
(ii) Binomials:
(a) p + q
(b) -5a + 2b
(iii) Trinomials:
(a) a + b + c
(b) x2 + x + 2

Question 2.
Identify the like expressions.
5x, -14x, 3x2 + 1, x2, -9x2, xy, -3xy
Solution:
Like terms: 5x and -14x, x2 and -9x2, xy and -3xy

Question 3.
Identify the terms and their coefficients for each of the following expressions:
(i) 3x2y – 5x
(ii) xyz – 2y
(iii) -x – x2
Solution: Question 4.
Add: -3a2b2, – $\frac { 5 }{ 2 }$ a2b2, 4a2b2 $\frac { 2 }{ 3 }$ a2b2
Solution:  Question 5.
Add: 8x2 + 7xy – 6y2, 4x2 – 3xy + 2y2 and -4x2 + xy – y2
Solution: Question 6.
Subtract: (4x + 5) from (-3x + 7)
Solution:
(-3x + 7) – (4x + 5) = -3x + 7 – 4x – 5 = -3x – 4x + 7 – 5 = -7x + 2

Question 7.
Subtract: 3x2 – 5x + 7 from 5x2 – 7x + 9
Solution:
(5x2 – 7x + 9) – (3x2 – 5x + 7)
= 5x2 – 7x + 9 – 3x2 + 5x – 7
= 5x2 – 3x2 + 5x – 7x + 9 – 7
= 2x2 – 2x + 2

Question 8.
Multiply the following expressions:
(a) 3xy2 × (-5x2y)
(b) $\frac { 1 }{ 2 }$ x2yz × $\frac { 2 }{ 3 }$ xy2z × $\frac { 1 }{ 5 }$ x2yz
Solution: Question 9.
Find the area of the rectangle whose length and breadths are 3x2y m and 5xy2 m respectively.
Solution:
Length = 3x2y m, breadth = 5xy2 m
Area of rectangle = Length × Breadth = (3x2y × 5xy2) sq m = (3 × 5) × x2y × xy2 sq m = 15x3y3 sq m

Question 10.
Multiply x2 + 7x – 8 by -2y.
Solution: ### Algebraic Expressions and Identities Class 8 Extra Questions Short Answer Type

Question 11.
Simplify the following:
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
Solution:
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
= a2b2 – a2c2) + b2c2 – b2a2) + c2a2 – c2b2)
= 0
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
= x3 – 3x2y2 – xy3 + 2x2y2 – xy3 + 5x3
= x3 + 5x3 – 3x2y2 + 2x2y2 – xy3 – xy3
= 6x3 – x2y2 – 2xy3

Question 12.
Multiply (3x2 + 5y2) by (5x2 – 3y2)
Solution:
(3x2 + 5y2) × (5x2 – 3y2)
= 3x2(5x2 – 3y2) + 5y2(5x2 – 3y2)
= 15x4 – 9x2y2 + 25x2y2 – 15y4
= 15x4 + 16x2y2 – 15y4

Question 13.
Multiply (6x2 – 5x + 3) by (3x2 + 7x – 3)
Solution:
(6x2 – 5x + 3) × (3x2 + 7x – 3)
= 6x2(3x2 + 7x – 3) – 5x(3x2 + 7x – 3) + 3(3x2 + 7x – 3)
= 18x4 + 42x3 – 18x2 – 15x3 – 35x2 + 15x + 9x2 + 21x – 9
= 18x4 + 42x3 – 15x3 – 18x2 – 35x2 + 9x2 + 15x + 21x – 9
= 18x4 + 27x3 – 44x2 + 36x – 9

Question 14.
Simplify:
2x2(x + 2) – 3x (x2 – 3) – 5x(x + 5)
Solution:
2x2(x + 2) – 3x(x2 – 3) – 5x(x + 5)
= 2x3 + 4x2 – 3x3 + 9x – 5x2 – 25x
= 2x3 – 3x3 – 5x2 + 4x2 + 9x – 25x
= -x3 – x2 – 16x

Question 15.
Multiply x2 + 2y by x3 – 2xy + y3 and find the value of the product for x = 1 and y = -1.
Solution:
(x2 + 2y) × (x3 – 2xy + y3)
= x2(x3 – 2xy + y3) + 2y(x3 – 2xy + y3)
= x5 – 2x3y + x2y3 + 2x3y – 4xy2 + 2y4
= x5 + x2y3 – 4xy2 + 2y4
Put x = 1 and y = -1
= (1)5 + (1)2 (-1)3 – 4(1)(-1)2 + 2(-1)4
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2

Question 16.
Using suitable identity find:
(i) 482 (NCERT Exemplar)
(ii) 962
(iii) 2312 – 1312
(iv) 97 × 103
(v) 1812 – 192 = 162 × 200 (NCERT Exemplar)
Solution: Question 17. Solution:  Question 18.
Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2 (NCERT Exemplar)
Solution:
LHS = (11pq + 4q)2 – (11pq – 4q)2 = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a2 -b2 = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq2
= RHS.
Hence Verified.

Question 19.
Find the value of $\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }$, using a suitable identity. (NCERT Exemplar)
Solution: Question 20.
Find the value of x, if 10000x = (9982)2 – (18)2 (NCERT Exemplar)
Solution:
RHS = (9982)2 – (18)2 = (9982 + 18)(9982 – 18)
[Since a2 -b2 = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964

### G20 సదస్సు..పోటీ పరీక్షలకు నోట్స్​

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