NCERT Class 8 Maths Chapter 8 – Algebraic Expressions and Identities
Exercise 8.1
1. Add the following.
(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) (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)
= a – b + ab + b – c + bc + c – a + ac
= a – a +b – b +c – c + ab + bc + ca
= 0 + 0 + 0 + ab + bc + ca
= ab + bc + ca
iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
= (2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)
= 2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5
= – p2q2 + 4pq + 9
iv)(l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
2. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) 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) (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) (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) (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 + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q
= 28 + 5p – 18q + 8pq – 7pq2 + p2q
Exercise 8.2
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 , 7 p = 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 × 0 = 0
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:
Area of rectangle = Length x breadth. So, it is multiplication of two monomials.
The results can be written in square units.
(i) p × q = pq
(ii)10m × 5n = 50mn
(iii) 20x2 × 5y2 = 100x2y2
(iv) 4x × 3x2 = 12x3
(v) 3mn × 4np = 12mn2p
3. Complete the following table of products:
Solution:
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:
Volume of rectangle = length x breadth x height. To evaluate volume of rectangular boxes, multiply all the monomials.
(i) 5a x 3a2 x 7a4 = (5 × 3 × 7) (a × a2 × a4 ) = 105a7
(ii) 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr
(iii) y × 2x2y × 2xy2 =(1 × 2 × 2 )( x × x2 × x × y × y × y2 ) = 4x4y4
(iv) a x 2b x 3c = (1 × 2 × 3 ) (a × b × c) = 6abc
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 = x2 y2 z2
(ii) a × – a2 × a3 = – a6
(iii) 2 × 4y × 8y2 × 16y3 = 1024 y6
(iv) a × 2b × 3c × 6abc = 36a2 b2 c2
(v) m × – mn × mnp = –m3 n2 p
Exercise 8.3
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) = 4pq + 4pr
(ii)ab(a – b) = a2 b – a b2
(iii)(a + b) (7a2b2) = 7a3b2 + 7a2b3
(iv) (a2 – 9)(4a) = 4a3 – 36a
(v) (pq + qr + rp) × 0 = 0 ( Anything multiplied by zero is zero )
2. Complete the table.
Solution:
3. Find the product.
i) a2 x (2a22) x (4a26)
ii) (2/3 xy) ×(-9/10 x2y2)
(iii) (-10/3 pq3/) × (6/5 p3q)
(iv) (x) × (x2) × (x3) × (x4)
Solution:
i) a2 x (2a22) x (4a26)
= (2 × 4) ( a2 × a22 × a26 )
= 8 × a2 + 22 + 26
= 8a50
ii) (2xy/3) ×(-9x2y2/10)
=(2/3 × -9/10 ) ( x × x2 × y × y2 )
= (-3/5 x3y3)
iii) (-10pq3/3) ×(6p3q/5)
= ( -10/3 × 6/5 ) (p × p3× q3 × q)
= (-4p4q4)
iv) ( x) x (x2) x (x3) x (x4)
= x 1 + 2 + 3 + 4
= x10
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =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) 3x (4x – 5) + 3
=3x ( 4x) – 3x( 5) +3
=12x2 – 15x + 3
(i) Putting x=3 in the equation we gets 12x2 – 15x + 3 =12(32) – 15 (3) +3
= 108 – 45 + 3
= 66
(ii) Putting x=1/2 in the equation we get
12x2 – 15x + 3 = 12 (1/2)2 – 15 (1/2) + 3
= 12 (1/4) – 15/2 +3
= 3 – 15/2 + 3
= 6- 15/2
= (12- 15 ) /2
= -3/2
b) a(a2 +a +1)+5
= a x a2 + a x a + a x 1 + 5 =a3+a2+a+ 5
(i) putting a=0 in the equation we get 03+02+0+5=5
(ii) putting a=1 in the equation we get 13 + 12 + 1+5 = 1 + 1 + 1+5 = 8
(iii) Putting a = -1 in the equation we get (-1)3+(-1)2 + (-1)+5 = -1 + 1 – 1+5 = 4
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 – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )
Solution:
a) p ( p – q) + q ( q – r) + r ( r – p)
= (p2 – pq) + (q2 – qr) + (r2 – pr)
= p2 + q2 + r2 – pq – qr – pr
b) 2x (z – x – y) + 2y (z – y – x)
= (2xz – 2x2 – 2xy) + (2yz – 2y2 – 2xy)
= 2xz – 4xy + 2yz – 2x2 – 2y2
c) 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)
= (40ln – 12lm + 8l2) – (3l2 – 12lm + 15ln)
= 40ln – 12lm + 8l2 – 3l2 +12lm -15 ln
= 25 ln + 5l2
d) 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))
= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – ( 2ab – 2b2 + 2bc ))
=-4ac + 4bc + 4c2 – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)
= -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac +2ab – 2b2 + 2bc
= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2
Exercise 8.4
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) (3/4 a2 + 3b2) and 4( a2 – 2/3 b2)
Solution :
(i) (2x + 5)(4x – 3)
= 2x x 4x – 2x x 3 + 5 x 4x – 5 x 3
= 8x² – 6x + 20x -15
= 8x² + 14x -15
ii) ( y – 8)(3y – 4)
= y x 3y – 4y – 8 x 3y + 32
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32
(iii) (2.5l – 0.5m)(2.5l + 0.5m)
= 2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 0.5m
= 6.25l2 + 1.25 lm – 1.25 lm – 0.25 m2
= 6.25l2– 0.25 m2
iv) (a + 3b) (x + 5)
= ax + 5a + 3bx + 15b
v) (2pq + 3q2) (3pq – 2q2)
= 2pq x 3pq – 2pq x 2q2 + 3q2 x 3pq – 3q2 x 2q2
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4
(vi) (3/4 a² + 3b² ) and 4( a² – 2/3 b² )
=(3/4 a² + 3b² ) x 4( a² – 2/3 b² )
=(3/4 a² + 3b² ) x (4a² – 8/3 b² )
=3/4 a² x (4a² – 8/3 b² ) + 3b² x (4a² – 8/3 b² )
=3/4 a² x 4a² -3/4 a² x 8/3 b² + 3b² x 4a² – 3b² x 8/3 b²
=3a4 – 2a² b² + 12 a² b² – 8b4
= 3a4 + 10a² b² – 8b4
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)
=15 + 5x – 6x – 2x2
= 15 – x -2 x 2
(ii) (x + 7y) (7x – y)
= x(7x-y) + 7y ( 7x-y)
=7x2 – xy + 49xy – 7y2
= 7x2 – 7y2 + 48xy
iii) (a2+ b) (a + b2)
= a2 (a + b2) + b(a + b2)
= a3 + a2b2 + ab + b3
= a3 + b3 + a2b2 + ab
iv) (p2– q2) (2p + q)
= p2 (2p + q) – q2 (2p + q)
=2p3 + p2q – 2pq2 – q3
= 2p3 – q3 + p2q – 2pq2
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
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x
ii) (a2+ 5) (b3+ 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 5b3 + 3a2 + 20
iii) (t + s2)(t2 – s)
= t (t2 – s) + s2(t2 – s)
= t3 – st + s2t2 – s3
= t3 – s3 – st + s2t2
iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
=(ac – ad + bc – bd) + (ac + ad – bc – bd) + (2ac + 2bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= 4ac
v) (x + y)(2x + y) + (x + 2y)(x – y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 3x2 + 4xy – y2
vi) (x + y)(x2– xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y = 2.25x2 – 16y2
viii) (a + b + c)(a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + b2 – c2 + 2ab
No comments:
Post a Comment