Rd Sharma Solutions Class 8 Chapter 6 Algebraic Expressions and Identities
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 6, Algebraic Expressions and Identities. Here students can easily find Exercise wise solution for chapter 6, Algebraic Expressions and Identities. Students will find proper solutions for Exercise 6.1, 6.2, 6.3, 6.4, 6.5, 6.6 and 6.7 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.
Algebraic Expressions and Identities Exercise 6.1 Solution :
Question no – (1)
Solution :
(i) 7x²yz – 5xy
∴ Terms = 7xyz, – 5xy
∴ Coefficient = 7, -5
(ii) x²+ x+1
∴ Terms = x², x, 1
∴ Coefficient =1, 1, 1
(iii) 3x² y² – 5x²y²z²+z²
∴ Terms = 3x²y², 5z²y²z², z²
∴ Coefficient = 3, 5, 1, 9
(iv) 9 – ab + bc – ca
∴ Terms = -ab, bc, -ca, a/2, b/2, -ab
∴ Coefficient = 1/2, 1/2, -1
(v) a/2 + b/2 –ab
∴ Terms = a/2 , b/2, -ab
∴ Coefficient = 1/2, 1/2, -1
(vi) 0.2x – 0.3xy + 0.5y
∴ Terms = 0.2x, – 0.3xy, 0.5y
∴ Coefficient = 0.2, -0.3, 0.5
Question no – (2)
Solution :
(i) x + y
= Binomial
(ii) 1000
= Monomial
(iii) x + x² + x³ + x⁴
= None of these
(iv) 7+ a + 5b
= Trinomial
(v) 2b – 3b²
= Binomial
(vi) 2y – 3y² + 4y³
= Trinomial
(vii) 5x- 4y+3x
= Trinomial
(viii) 4a – 15a²
= Binomial
(ix) xy + yz + zt + tx
= None of these
(x) pqr
= Monomial
(xi) p²q + pq²
= Binomial
(xii) 2p + 2q
= Binomial
Algebraic Expressions and Identities Exercise 6.2 Solution :
Question no – (1)
Solution :
(i) 3a²b, -4a²b, 9a²b
= 3a²b – 4a²b + 9a²b
= 8a²b
(ii) 2/3a, 3/5a,- 6/5a
= 2/3a + 3/5a – 6/5a
= 10a + 9a – 18a/15
= 19a – 18a/15
= 1a/15
(iii) 4xy² – 7x²y – 12x²y – 6xy² – 3x²y + 5xy²
= 4xy² – 7x²y + 12x²y – 6xy² – 3x²y + 5xy²
= 3xy² + 2x²y
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
= 3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
= (3/2a + 2/3a + 5/3a) – (5/4b + 7/2b – 5/2b) + (2/5c + 7/2c – 5/4c)
= [(9 + 4 + 10/6)a] + [(5 + 14 – 10/4) b] + [(8 + 70 – 25/20) c]
= 23a/6 – 9b/4 + 53c/20
(vi) (11/2xy + 12/5y + 13/7x) + (- 11/2y – 12/5x – 13/7xy)
= (11/2 – 13/7) xy + (12/5 -11/2)y + (13/7 – 12/5)x
= (77 – 26/14)xy + (24 – 55/10)y + (5 – 84/35)x
= 51/14xy – 31/10y – 19/35x
Question no – (2)
Solution :
(i) -5xy from 12xy
= 12xy – (-5xy)
= 12xy + 5xy
= 17xy
(ii) 2a² from – 7a²
= -7a² – (2a²)
= -7a² – 2a²
= -9a²
(iv) 2x³ – 4x² + 3x+ 5 from 4x³ +x²+ x+6
= (4x³ + x²+ x +6) – (2x³ – 4x² + 3x + 5)
= 4x³ – 2x³ + 4x² + x² + x- 3x + 6 – 5
= 2x³ + 5x² – 2x +1
(v) 2/3y³ -2/7y² – 5 from 1/3y³ + 5/7y² + y – 2
= (1/3y³ + 5/7y² + y – 2) – (2/3y³ – 2/7y² – 5)
= 1/3y³ – 2/3y³ + 5/7y² + y + 3
= 1-2/3y³ + 5/7 + 2y² + y + 3
= -1/3 y³ + 7/7y² + 2/7y² + y + 3
= -1/3y³ + y² + y + 3
(vi) Given, x²y – 4/5xy² + 4/3xy from 2/3x²y + 3/2xy² – 1/3xy
= (2/3x²y + 3/2xy² – 1/3xy) – (x²y – 4/5xy² + 4/3xy)
= (2/3 – 1)x²y + (3/2 + 4/5)xy – (1/3 – 4/3)xy
= (2 – 3/3)x²y + (15 + 8/10)xy² – (1 + 4/3)xy
= -1/3x²y + 23/10xy² – 5/3xy
Question no – (3)
Solution :
(ii) 5a²/2 + 3a³/2 + a/3 – 6/5 from 1/3a³ – 3/4 a² – 5/2
= (1/3a² – 3/4a² – 5/2) – (5a²/2 + 3a²/2 + a/3 – 6/5)
= 1/3a³ – 3/2a³ – 3/4a² – 3/4a² 5/2a² – 5/2 + 6/5 – a/3
= (1/3 – 3/2)a³ – (3/4 + 5/2)a² – (5/2 – 6/5) – a/3
= (2 – 9/6)a³ – (3 + 10/4)a² – (25 -12/10)
= -7/6a³ – 13/4a² – a/3 – 13/10
(iii) 7/4x³ + 3/5x² + 1/2x + 9/2 from 7/2 – x/3 – x²/5
= (7/2 – x/3 – x²/5) – (7/4x³ + 3/5x² + 1/2x + 9/2)
= 7/2 – x/3 – x²/5 – 7/4x³ + 3/5x² + 1/2x + 9/2
= -7/4x³ – (1 + 3/5)x² – (1/3 – 1/2)x + (7/2 – 9/2)
= – 7/4x³ – 4/5x² + 1/6x – 2/2
= – 7/4x³ – 4/5x² + 1/6x – 1
(iv) y³/3 + 7/3y² + 1/2y + 1/2 from 1/3 -5/3y²
= (1/3 – 5/3y²) + (y³/3 + 7/3y² + 1/2y + 1/2)
= 1/3 – 5/3y² – y³/3 + 7/3y² + 1/2y + 1/2 y
= (1/3 – 1/2) – (5/3y² + 7/3y²) – y³/3 + 1/2y
= (2 – 3/6) – (5 + 7/7)b² – y³/3 + 1/2y
= – 1/6 – 12/3 y² – y³/3 + 1/2y
= -4y² – y³/3 – 1/6 + 1/2y
= – 4y² – y³/3 + 1/2y – 1/6
= -y³/3 – 4y² + 1/2 y – 1/6
(v) 2/3ac – 5/7ab +2/3bc from 3/2ab -7/4ac – 5/6bc
= (3/2ab -7/4ac – 5/6bc) – (2/3ac – 5/7ab +2/3bc)
= 3/2ab – 7/4ac – 5/6bc – 2/3ac + 5/7ab – 2/3bc
= 3/2ab – 5/7ab – 7/4ac – 2/3ac -5/6bc – 2/3bc
= (3/2 + 5/7)ab – (7/4 + 2/3)ac – (5/6 + 2/3)bc
= (21 + 10/14)ab – (21 + 8/12)ac – (15+12/18)bc
= 31/14ab – 29/12ac – 27/18bc
= 31/14ab – 29/12ac – 3/2bc ….. (after cutting)
Question no – (4)
Solution :
Sum = (x – 3y + 3y + 2z) + (-4x + 9y – 11z)
= x – 3y + 2z – 4x + 9y -11z
= -3x + 6y – 9z
∴ The required number,
= (-3x + 6y – 9z) – (3x – 4y – 7z)
= – 3x + 6y – 9z – 4x + 4y + 7z
= – 6x + 10y – 2z
Question no – (5)
Solution :
Sum₁, = (3l – 4m – 7n²) + (2l + 3m – 4n²)
= 3l – 4m – 7n² + 2l + 3m – 4n²
= 5l – m – 11n²
Sum₂, = (9l + 2m – 3n²) + (- 3l + m + 4n²)
= 9l + 2m – 3n² – 3l + m + 4n²
= 6l + 3m + n²
∴ The required number,
= Sum₂ – Sum₁
= (6l + 3m + n²) – (5l – m – 11n²)
= 6l + 3m + n² – 5l – m – 11n²
= l + 4m + 12n²
Question no – (6)
Solution :
Sum = (2x – x² + 5) + (-4x – 3 + 7x²)
= 2x- x² + 5 – 4x – 3 + 7x²
= – 2x – 6x² + 2
∴ The required number,
= 5 – (- 2x + 6x² + 2)
= 5 + 2x – 6x² – 2
= 2x – 6x² + 3
Question no – (7)
Solution :
(i) x²- 3x + 5 – 1/2(3x² – 5x + 7)
= x²- 3x + 5 – 3/2x² + 5/2x – 7/2
= (1 – 3/2)x² – (3 – 5/2) + (5 – 7/2)
= (2 – 3/2)x² – (6 – 5/2)x + (10 – 7/2)
= -1/2x² – 1/2x + 3/2…(Simplified)
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
= 5 – 3x + 2y – 2x + y – 3x + 7y – 9
= – 8x + 10y – 4…(Simplified)
(iv) (1/3y² – 4/7y + 11) – (1/7y – 3 + 2y²) – (2/7y – 2/3y² + 2)
= 1/3y² – 4/7y + 11 – 1/7y – 3 + 2y² – 2/7y – 2/3y² – 2
= (1/3 – 2 + 2/3)y² – (4/7 + 1/7 + 2/7)y + (11 + 3 – 2)
= (1 – 6 + 2/3)y² – (4 + 1 + 2/7)y + 12
= -3/3y² – 7/7y + 12
= -y² – y + 12…(Simplified)
Algebraic Expressions and Identities Exercise 6.3 Solution :
Question no – (1)
Solution :
Given, 5x² × 4x³
= (5 × 4) (x² × x³)
= 20x⁵
Hence, the product is 20x⁵
Question no – (2)
Solution :
Given, -3a² × 4b⁴
=(-3 × 4) (a² × b⁴)
= 12a²b⁴
Thus, the product is 12a²b⁴
Question no – (3)
Solution :
Given, (-5xy) × (-3x²yz)
= (-5 × 3) (xy × x²yz)
= -15x³y²z
Therefore, the product is -15x³y²z
Question no – (4)
Solution :
Given, 1/2xy × 2/3x²yz²
= (1/2 × 2/3) (xy × x²yz²)
= 1/3x³y²z²
Thus, the product is 1/3x³y²z²
Question no – (5)
Solution :
Given, (-7/5xy²z) × (13/3x²yz²)
= (-7/5 × 13/3) (xy²z × x²yz²)
= -91/15x³y³z³
Hence, the product is -91/15x³y³z³
Question no – (7)
Solution :
Given, (-1/27a²b²) × (9/2 a³b²c²)
= (- 1/27 × 9/2) (a²b² ×a³b²c²)
= -1/6a⁵b⁴c²
Thus, the product is -1/6a⁵b⁴c²
Question no – (8)
Solution :
Given, (-7xy) × (1/4x²yz)
= (-7 × 1/4) (xy × x²yz)
= -7/4x³y²z
Therefore, the product is -1/6a⁵b⁴c²
Question no – (9)
Solution :
Given, (7ab) × (-5ab²c) × (6abc²)
= – (7 × 5 × 6) (ab × ab²c × abc²)
= – 210a³b⁴c³
Hence, the product is -210a³b⁴c³
Question no – (11)
Solution :
Given, (-4x²) × (-6xy²) × (-3yz²)
= – (4 × 6 × 3) (x² × xy² × z²)
= -72x³y²z²
Thus, the product is -72x³y²z²
Question no – (13)
Solution :
Given, (7/9ab²) × (15/7ac²b) × (- 3/5a²c)
= -(7/9 × 15/7 × 3/5) (ab² × ac²b × a²c)
= -a⁴b³c³
So, the product is -a⁴b³c³
Question no – (16)
Solution :
Given, (4/3 pq²) × (-1/4p²r) × (16p² q² r²)
= – (4/3 × 1/4 ×16) (pq² × p²r × p²q²r²)
= – 16/3p⁵q⁴r³
Hence, the product is – 16/3p⁵q⁴r³
Question no – (17)
Solution :
Given, (2.3xy) × ( 0.1x) × (0.16)
= (2.3 × 0.1 × 0.16) (xy × x)
= 0.0368x²y
Thus, the product is 0.0368x²y
Question no – (18)
Solution :
Given, (3x) × (4x) × (-5x)
= – 60x³
Question no – (19)
Solution :
Given, (4x²) × (- 3x)× (4/5x³)
= – (4 ×3 × 4/5) (x² × x × x³)
= 16/5x⁶
Question no – (20)
Solution :
Given, (5x⁴) × (x²)³ × (2x²)
= (5 × 2 × 2) (x⁴ × x⁶ × x²)
= 20x^12
Question no – (21)
Solution :
Given, (x²)³ × (2x) × (-4x) × (5)
= – (2 × 4 × 5) (x⁶ × x × x)
= 40x⁸
Question no – (22)
Solution :
= (-8x²y⁶) × (-20xy)
= (8 × 20) (x² × y⁶ × xy)
= 160x³y⁷
= 160 (2.5)³ (1)⁷ [∵ x = 2.5, y = 1]
= 160 × 15.625 × 1
= 2500 ….(Verified)
Question no – (23)
Solution :
Given, (3.2x⁶y³) × (2.1x²y²)
= (3.2 × 2.1) (x⁶y³ × x²y²)
= 6.72x⁸y⁵
= 6.72 × (1)⁸ × (0.5)⁵ [∵ x =1 y = 0.5]
= 6.72 × 1 × 0.03125
= 0.21
Question no – (24)
Solution :
(5x⁶) × (- 1.5x²y³) × (- 21xy²)
= (5 × 1.5 × 12) (x⁶ × x²y³ × xy)
= 90x⁹y⁵
= 90(1) (0.5)⁵ [∵ x = 1 y = 0.5]
= 90 × 1 × 0.03125
= 2.8125
Question no – (26)
Solution :
(-8 x²y⁶) × (-20xy)
= (8 × 20) (x²y⁶ × xy)
= 160x³y⁷
= 160 × (2.5) (1)⁷ …[∵ x = 2.5 y = 1]
= 160 × 15.625 × 1
= 2500
Question no – (27)
Solution :
Given, (xy³) × (yx³) × (xy)
= -x⁵ y⁵
Question no – (28)
Solution :
Given, (1/8x²y⁴) × (1/4x⁴y²) ×(xy) × 5
= (1/8 × 1/4 × 5) (x²y⁴ × x⁴y² × xy)
= 5/32x⁷y⁷
Question no – (29)
Solution :
Given, (2/5a²b) × (-15b²ac) × (-1/2c²)
= (2/8 × 15 × 2/1) (a²b × b²ac × c²)
= 3a³b³c³
Question no – (30)
Solution :
Given, (-4/7a²b) × (-2/3b²c) × (-7/6c²a)
= – (4/7 × 2/3 × 7/6) (a²b × b³c × c³a)
= – 4/9a³b³c³
Question no – (31)
Solution :
Given, (4/9abc³) × (- 27/5a³b²) × (-8b³c)
= (4/9 × 27/5³ × 8) (abc² × a³b×b³c)
= 96/5a⁴b⁶c⁴
Question no – (32)
Solution :
Given, (2xy) × (x²y/4) × (x²) ×(y²)
= 1/2x⁵y⁴
= 1/2(2)⁵ (1)² …[∵ x = 2, y = 1]
= 2⁴ × 1
= 16
Question no – (33)
Solution :
Given, (3/5x²y) × (-15/4xy²) × (7/9x²y²)
= – 7/4x⁵y⁵
= -7/4(2)⁵ (1)² [∵ x = 2, y = 1]
= -7/2 × 2 × 2 × 2 × 2 × 2 × 1
= -56
Algebraic Expressions and Identities Exercise 6.4 Solution :
Question no – (1)
Solution :
Given, 2a³(3a + 5b)
= (2a³ × 3a) + (2a³ × 5b)
= 6a⁴ + 10a³b
Question no – (2)
Solution :
-11a(3a + 2b)
= (-11a × 3a) + (-11a × 2b)
= -33a² – 22ab
Question no – (3)
Solution :
-5a(7a – 2b)
= (-5a × 7a) – (5a × 2b)
= 35a² + 10ab
Question no – (4)
Solution :
11y²(3y + 7)
= (- 11y² × 3y) + (-11y² × 7)
= -33y³ – 77y²
Question no – (5)
Solution :
6x/5 (x³ + y³)
= (6x/5 × x³) + (6x/5 × y³)
= 6x⁴/5 + 6xy³/5
Question no – (6)
Solution :
xy(x³ – y³)
= (xy × x³) – (6x/5 × y³)
= x⁴y – xy⁴
Question no – (7)
Solution :
0.1y (0.1x⁵ + 0.1y)
= (0.1x⁵ × 0.1y) + (0.1y × .01y)
= 0.01x⁵y + 0.01y²
Question no – (8)
Solution :
(-7/4ab²c – 6/25a²c²) (- 50a²b²c²)
= (-7/4ab²c × – 50a²b²c²) – (6/25a²c² × 50a²b²c²)
= 175/2a³b⁴c³ – 12a⁴b²c⁴
Question no – (9)
Solution :
-8/27 xyz (3/2xyz² – 9/4xy²z³)
= (- 8/27 xyz × 3/2xyz²) – (- 8/27 xyz× 9/4xy²z³)
= -4/9 x²y²z³ + 2/3 x²y³z⁴
Question no – (10)
Solution :
-4/27 xyz ( 9/2x²y z – 3/4 xyz²)
= (- 4/27xyz × 9/2x²yz) – (- 4/27xyz× 3/4xyz²)
= – 2/3x³y²z² + 1/9x²y²z³
Question no – (11)
Solution :
1.5x (10x²y – 100xy²)
= (1.5x × 10x²y) – (1.5x × 100xy²)
= 15x³y – 150x²y²
Question no – (12)
Solution :
4.1xy (1.x1x-y)
= (4.1xy × 1.1x) – (4.1xy ×y)
= 4.51x²y – 4.1xy²
Question no – (13)
Solution :
Given, 250.5xy(xz + y/10)
= (250.5xy × xz) – (250.5xy × x/10)
= 250.5x²yz + 25.05x²y
Question no – (14)
Solution :
= 7/5x²y (3/5xy² + 2/5x)
= (7/5x²y × 3/5xy²) + (7/5x²y × 2/5x)
= 21/25x³y³ + 14/25x³y
Question no – (16)
Solution :
24x² (1 – 2x)
= 24x² (1 – 2x)
= (24x² × 1) – (24x² × 2x)
= 24x² – 48x³
= 24 × 3² – 48 × 3³ [∵ x = 3]
= 24 × 9 – 48 × 27
= 216 – 1296
= – 1080
Question no – (17)
Solution :
= – 3y (xy + y²)
= (- 3y × xy) + ( -3y × y)
= – 3xy² – 3y³
= – 3× 4 × 5 – 3 × 5³
= – 300 – 375 [∵ x = 4, y = 5]
= – 675
Question no – (18)
Solution :
– 3/2 x²y³ (2x – y)
= (- 3/2 x²y³ × 2x) – (- 3/2 x²y³ × y)
= – 3x³ y³ + 3/2 v² y³ [∵ x = 1, y = 2]
= -3(1) ³ (2) ³ + 3/2 (1)² (2)⁴
= 0
Question no – (19)
Solution :
(i) (15y²(2 – 3x)
= (15y²× 2) – (15y²× 3x)
= 30y² -45xy²
= 30 × (0.25)² – 45 × (-1) × (0.25)² [ ∵ x = 1, y = 0.025]
= 30 × 0.0625+ 45 × 0.0625
= 1.875 + 2.8125
= 4.6875
(ii) – 3x (y²+ z²)
= – 3x y² – 3xz²
= -3 × (-1) × (0.25)² – 3 (-1) (0.05)²
= 3× 0.0625+ 3×0.0075 [∵ x = -1, y = 0.25, z = 0.05]
= 0.195
(iii) z² (x-y)
= z²+yz²
= (-1) (0.05)² – (0.25) (0.50)²
= -0.25-0 .0625 [∵ x = -1, y = 0.25, z = 0.0]
= 0.3125
(iv) xz(x² + y²)
= xz × x² + xz× y²
= x²z + xy²z
= (-1)³ × 0.25 + (-1) × (0.25)² × 0.50
= -1 × 0.25 – 0.03125 [∵ x = -1, y = 0.25, z = 0.50]
= 0.28125
Question no – (20)
Solution :
(i) 2x² (x³ – x) – 3x (x⁴ + 2x) – 2(x⁴ – 3x²)
= 2x⁵ – 2x³ – 3x⁵ – 6x² – 2x⁴ + 6x²
= -x⁵ – 2x⁴ – 2x³…(Simplified)
(ii) x³y(x² – 2x) + 2xy(x³ – x⁴)
= x⁵y – 2x⁴y + 2x⁴y – 2x⁵y
= – x⁵y…(Simplified)
(iii) 3a² + 2(a + 2) – 3a (2a + 1)
= 3a² + 2a + 4 – 6a² – 3a
= – 3a² – a + 4…(Simplified)
(vi) a(b – c) + b(c – a) + c(a – b)
= ab – ac + bc – ab + ac – bc
= 0…(Simplified)
(vii) 4ab (a-b) – 6a² (b – b²) – 3b²(2a² – a) + 2ab(b – a)
= 4a²b – 4ab² – 6a²b + 6a²b² – 6a²b² + 3ab² + 2ab² – 2a²b
= (4 – 6 – 2) a²b – (4 – 2 + 3)
= – 4a²b + ab²…(Simplified)
(ix) 2a² + 3a (1 – 2a³) + a(a + 1)
= 2a² + 3a – 6a⁴ + a² + a
= 3a² + 4a – 6a⁴…(Simplified)
(x) a² (2a – 1) + 3a + a³ – 8
= 2a³ – a² + 3a + a³ – 8
= 3a³ + 3a – a² – 8…(Simplified)
(xi) 3/2x² (x² – 1) + 1/4 x² (x² + x) – 3/4x (x³ – 1)
= 3/2x⁴ – 3/2x² + 1/4x⁴ + 1/4x³ – 3/4x⁴ + 3/4x
= (3/2 + 1/4 – 3/4) x⁴ – 3/2x² + 1/4x³ + 3/4x
= x⁴ + 1/4x³ – 3/2x² + 3/4x…(Simplified)
(xii) a²b (a – b²) + ab² (4ab – 2a²) – a³b(1 – 2b)
= a³b – a²b³ + 4a²b³ – 2 a³b² – a³b + 2a³b²
= 3 a²b³…(Simplified)
(xiii) a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b(a³ – a² – 1)
= a⁵b – a³b + a²b – a⁵ b + 2a³b – 2a²b -a³b + a² b + b
= a³b + a³b + a²b – a²b + b
= b…(Simplified)
Algebraic Expressions and Identities Exercise 6.5 Solution :
Question no – (1)
Solution :
Given, (5x + 3) by (7x + 2)
= 5x(7x + 2) + 3 (7x + 2)
= 35x² + 10x + 21x + 6
= 35x² + 31x + 6
Question no – (2)
Solution :
Given, (2x + 8) by (x – 3)
= 2x (x – 3) + 8(x – 3)
= 2x² – 6x + 8x – 24
= 2x² + 2x – 24
Question no – (3)
Solution :
Given, (7x + y) by (x + 5y)
= 7x (x + 5y ) + y(x+5y)
= 7x² + 35y + xy + 5y²
= 7x² + 36xy + 5y²
Question no – (4)
Solution :
Given, (a – 1) by (0.1a² + 3)
= a (0.1a²) – 1 (0.1² + 3)
= 0.1a³ + 3a – 0.1a² – 3
Question no – (5)
Solution :
Given, (3x² + y²) by (2x² + 3y²)
= 3x² (2x² + 3y²) + y²(2x² + 3y²)
= 6x⁴ + 9x²y² + 2x²y² + 3y⁴
= 6x⁴ + 11x²y² + 3y⁴
Question no – (6)
Solution :
Given, (3/5x + 1/2y) by (5/6x + 4y)
= 3x/5 (5/6x +4y) + 1/2y (5/6x + 4y)
= 1/2x² + 12/5xy + 5/12 + 2y²
= 1/2x² + 169/60xy + 2y²
Question no – (7)
Solution :
Given, (x⁶ – y⁶)by (x² + y²)
= x⁶ (x² + y²) – y⁶(x² + y²)
= x⁸ + xy² – y⁶ x² – y⁸
Question no – (8)
Solution :
Given, (x² +y²) by (3a + 2b)
= x²(3a+2b) + y²(3a+2b)
= 3x²a + 2x²b + 3ay² + 2by²
Question no – (9)
Solution :
[-3d + (7f)] by (5d + f)
= -3d (5d + f) + -7f (5d + f)
=-15d² – 3df – 35df + 7f²
= -15d² – 38df + 7f²
Question no – (10)
Solution :
(0.8a – 0.5b) by (1.5a-3b)
= 0.8a (15a-3b) – 0.5b (0.5a -43b)
= 0.2a² – 2.4ab – 0.75ab + 1.5b²
= 1.2a² – 3.15ab + 1.5b²
Question no – (11)
Solution :
As per the question, (2x²y² – 5xy²) by (x² – y²)
= 2x²y²(x² – y²) – 5xy² (x² + y²)
= 2x⁴y² – 2x⁴y² – 5x⁴y² + 5xy⁴
Question no – (12)
Solution :
Given, (x/7 + x²/2) by (2/5 + 9x/4)
= x/7 (2/5 + 9x/4) + x²/2 (2/5 + 9x/4)
= 2x/35 + 9/28x² + 1/5x² + 9/8x³
= 2/35 x + (45 + 28/14)x² + 9/8x³
= 9/8x³ + 73/140x² + 2/35x
Question no – (13)
Solution :
Given in the question, (-a/7+ a²/9) by (b/2 – b²/3)
= -a/47 (b/2 – b²/3) + a²/9 (b/2 – b²/3)
= -ab/14 + ab²/21 + a²b/18 – a²b²/27
Question no – (14)
Solution :
As per the question, (3x²y – 5xy²) by (1/5x² + 1/3y²)
= 3x²y (1/5x² + 1/3y²) – 5xy²(1/5x² + 1/3y²)
= 3/5x⁴y + x²y³ – x³y² – 5/3xy⁴
Question no – (15)
Solution :
Given, (2x² -1) by (4x³ + 5x²)
= 2x² (4x³ + 5x²) -1(4x³ +5x²)
= 8x⁵ + 10x⁴ – 4x³ – 5x²
Question no – (16)
Solution :
Given, (2xy+3y²) (3y²-2)
= 2xy (3y²-2) + 3y² (3y²-2)
= 6xy³ – 4xy + 9y⁴ – 6y²
Question no – (17)
Solution :
Given, (3x – 5y) (x + y)
= 3x (x + y) – 5y (x + y)
= 3x² + 3x – 5xy – 5y²
= 3x² – 2xy – 5y²
Therefore, the product will be 3x² – 2xy – 5y²
Question no – (18)
Solution :
Given, (x²y – 1) (3 – 2x²y)
= x²y(3 – 2x²y) – 1 (3 – 2x²y)
= 3x²y – 2x⁴y² – 3 + 2x²y
= 5x²y – 2x⁴y² – 3
Hence, the product will be 5x²y – 2x⁴y² – 3
Question no – (19)
Solution :
Given, (1/3x – y²/5) (1/3x + y²/5)
= 1/9x²(1/3x + y²/5) – y²/5(1/3x + y²/5)
= 1/9x² – xy² /15 – xy²/15 – y⁴/25
= 1/9x² – y⁴/25
Hence, the product will be 1/9x² – y⁴/25
Simplify.
Question no – (20)
Solution :
Given, x²(x + 2y) (x – 3y)
= x²[x (x – 3y) + 2y (x – 3y)]
= x²[x² – 3xy + 2y – 6y²]
= x²[x² – xy – 6y²]
= x⁴ – x³y – 6x²y²…(Simplified)
Question no – (21)
Solution :
Given, (x² – 2y²) (x + 4y)x²y²
= [x² (x+4y) -2y² (x+4y)] x²y²
= [x³ + 4x²y – 2xy² – 8y] x²y²
= x⁵y² + 4x⁴y³ – 2x³y⁴ – 8x²y⁵…(Simplified)
Question no – (22)
Solution :
Given, a²b² (a + 2b) (3a + b)
= a²b² [a(3a + b) + 2b (3a + b)]
= a²b² [3a² + ab + 6ab + 2b²]
= a²b² [3a² + 7ab + 2b²]
= 3a⁴b² + 7a³b³ + 2a²b⁴…(Simplified)
Question no – (23)
Solution :
Given, x²(x – y) y²(x + 2y)
= (x³ – x²y) (xy² + 2y³)
= x³ (xy² + 2y³) – x²y (xy² + 2y³)
= x⁴y² + 2x³y³ – x³y³ – 2x²y⁴
= x⁴y² + x³y³ – 2x²y⁴…(Simplified)
Question no – (24)
Solution :
Given, (x³ – 2x² + 5x – 7) (2x – 3)
= 2x(x³ – 2x² + 5x – 7) – 3(x³ – 2x² + 5x – 7)
= 2x – 4x³ + 10xy – 11x – 3x³ + 6x² – 15x + 21
= 2x – 7x³ + 16x² – 29x + 21…(Simplified)
Question no – (25)
Solution :
Given, (5x + 3) (x – 1) (3x – 2)
= (5x + 3) [x (3x – 2) -1 (3x – 2)]
= (5x + 3) (3x² – 2x – 3x + 2)
= (5x + 3) (3x³ – 5x + 2)
= 5x (3x² – 5x + 2) + 3(3x² – 5x + 2)
= 15x³ – 25x² + 10x + 15 + x² – 15x + 6
= 15x³ – 10x² – 5x + 6…(Simplified)
Question no – (27)
Solution :
Given, (2x² + 3x – 5) (3x² – 5x + 4)
= 2x² (3x² – 5x +4) + 3x (3x² – 5x+4) – 5 (3x² – 5x + 4)
=6x⁴ – 10x³ + 8x + 9x³ -15 + 12x – 15x² + 25x – 20
= 6x⁴ – x³ – 22x² + 9x² + 37x – 20…(Simplified)
Question no – (28)
Solution :
Given, (3x – 2) (2x – 3) + (5x – 3) (x + 1)
= 6x² – 9x – 4x + 6 + 5x² + 5x – 3x – 3
= 11x² – 11x + 3…(Simplified)
Question no – (30)
Solution :
Given, (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
= (12x² + 9xy + 8xy + 6y²) – (14x² – 6xy – 7xy + 3y²)
= 12x² + 17xy + 6y² – (14x² – 13xy + 3y²)
= 12x² + 17xy + 6y² – 14x² +13y – 3y²
= – 2x² + 30 xy + 3y²…(Simplified)
Question no – (32)
Solution :
Given, (x³ -2x² + 3x – 4) (x-1) – (2x – 3) (x² – x + 1)
= (x⁴ – 2x³ + 3x² – 4x – x³ – 2x² – 3x + 4) – (2x³ – 2x² + 2x – 3x² + 3x – 3)
= (x⁴ – 3x³ + 5x² – 7x + 4 ) – ( 2x³ – 5x² + 5x – 3)
= x⁴ – 3x³ + 5x² – 7x + 4 – 2x³ + 5x² – 5x + 3
= x⁴ – 5x³ + 10x² – 12x + 7…(Simplified)
Algebraic Expressions and Identities Exercise 6.6 Solution :
Question no – (1)
Solution :
(i) (x + 2)²
= x² + 2.2.x + 2²
= x² + 4x + 4
(ii) (8a + 3b)²
= (8a)² + 2.8a – 3b + (3b)²
= 64a² + 48ab + 9b²
(iv) (9a + 1/6)²
= (9a)² +2.9a.1/6 + (1/6)²
= 81a² + 3a + 1/36
(v) (x + x²/2)²
= x² + 2. x. x²/2 + (x²/2)
= x² + x³ + x⁴/4
(vi) (x/4 – y/3)
= (x/4)² + 2 .x/4. y/3 + (b/3) ²
= x²/16 + xy/6 + y²/9
(vii) (3x – 1/3x)²
= (3x)² + 2. 3x. 1/3x + (1/3x)²
= 9x² + 2 + 1/9x²
(viii) (x/y – x/y)²
= (x/y)² + 2. x/y. y/x + (y/x) ²
= x²/y² + 2 + y²/x²
(ix) (3a/2 – 5b/4)²
= (3a/2) ² +2. 3a/2 . 5b/2 + (5b)²
= 9a²/4 + 15ab/2 + 25b²
(x) (a²b – bc²)²
= (a² b²)² + 2. a²b + bc² + (bc²)²
= a⁴ b² + 2a²b² + b²c⁴
(xi) (2a/3b + 2b/3a)²
= (2a/3b)² + 2. 2a/3b .2b/3a + (2b/3a) ²
= 4a²/9b² + 8/9 + 4b²/9a²
Question no – (2)
Solution :
(i) (2x + y) (2x + y)
= (2xy)²
= (2x)² + 2. 2x y + y²
= 4x² + 4xy + y²
Hence, the product will be 4x² + 4xy + y²
(ii) (a + 2b) (a – 2b)
= a² – (2b)²
= a² – 4b²
Thus, the product will be a² – 4b²
(iii) (a² + bc ) (a² – bc)
= (a²)² – (bc)²
= a⁴ – b²c²
Therefore, the product will be a⁴ – b²c²
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
= (4x/5)² – (3y/4)²
= 16x²/25 – 9y²/16
Therefore, the product will be 16x²/25 – 9y²/16
(v) (2x + 3/y) (2x – 3/y)
= (2x) ² – (3/y) ²
= 4x² – 9/y²
Thus, the product will be 4x² – 9/y²
(vi) (2a³ + b³) (2a³ – b³)
= (2a³)² – (b³)²
= 4a⁶ – b⁶
Hence, the product will be 4a⁶ – b⁶
(vii) (x⁴ + 2/x²) (x⁴ – 2/x²)
= (x⁴)² – (2/x²)
= x⁸ – 4/x⁴
Thus, the product will be x⁸ – 4/x⁴
(viii) (x³+ 1/x³) (x³ – 1/x³)
= (x³)² – (1/x³)²
= x⁶ – 1/x⁶
Therefore, the product will be x⁶ – 1/x⁶
Question no – (3)
Solution :
(i) (100)²
= (100 + 10²
= (100) ² + 2.100. 1 + 1²
= 10000 + 200 + 1
= 10201
(ii) (99)²
= (100 – 1)²
= (100)² – 2.1000.1 + 1²
= 10000 – 200 + 1
= 9801
(iii) (1001)²
= (100 + 1)
= (1000)² + 2.100 + 1 + 1²
= 1000000 + 2000 + 1
= 10020001
(iv) (999)²
= (100-1)²
= (1000)² – 2.1000 + 1 + 1²
= 1000000 – 2000 + 1
= 998001
(v) (700 + 3)²
= (700 + 3)²
= (700)² + 2.700 + 3 + 3²
= 490000 + 4200 + 9
= 532009
Question no – (4)
Solution :
(i) (82)² – (18) ²
= (82 + 18) (82 – 18)
= 100 × 64
= 6400…(Simplified)
(ii) (467)² – (33)²
= (467 + 33) (467 – 33)
= 500 × 434
= 217000…(Simplified)
(iii) (79) ² – (69) ²
= (79 + 69) (79 – 69)
= 148 × 10
= 1480…(Simplified)
(iv) 197 × 203
= (200 – 3) (200 + 3)
= (200)² – 3²
= 40000 – 9
= 39991…(Simplified)
(v) 113 × 87
= (100 + 13) (100 – 13)
= (100)² – (13)²
= 1000 – 169
= 9831…(Simplified)
(vi) 95 × 105
= (100 – 5) (100 + 5)
= (100)² – 5² – 1000 – 25
= 9975…(Simplified)
(viii) 9.8 × 10.2
= (10.0 – 0.2) (10.0 + 0.2)
= (10.0)² – (0.2)²
= 100.00 – 0.04
= 99.96…(Simplified)
Question no – (5)
Solution :
(i) 58² – 42²/16
= (58 + 42) (58 – 42)/16
= 100 × 16/16
= 100…(Simplified)
(ii) 178 × 178 – 22 × 22
= (178)² – (22)²
= (178 × 22) (178 – 22)
= 200 × 156
= 31200…(Simplified)
(iii) 198 × 198 – 102 × 102/96
= (198)² – (102)²/96
= (198 + 102) (198 – 102)/96
= 300 × 96/96
= 300…(Simplified)
(iv) 1.73 × 1.73 – 0.27 × 0.27
= (1.73)² – (0.27)²
= (1.73 + 0.27) (1.73 – 0.27)
= 2.00 × 1.46
= 2.92 …(Simplified)
(v) 8.63 × 8.63- 1.37 × 1.37/0.726
= (8.63)² – (1.37)² /0.726
= (8.63 + 1.37) (8.63 – 1.37)/0.726
= 10.00 × 7.26/0.26
= 100.0 × 7.26/10 × 0.726
= 100 × 7.26/7.26
= 100 …(Simplified)
Question no – (6)
Solution :
(i) 4x = (52) ² – (48) ²
or 4x = (52 + 48) (52 – 48)
or 4x = 100 × 4
or x = 100 × 4/4
∴ x = 100
Therefore, the value of x will be 100
(ii) 14x = (47)² – (33)²
or, 14x = (47 + 33) (47 – 33) = 80 × 14
or x = – 80 × 14/14 = 80
Thus, the value of x will be 80
(iii) 5x = (50)² – (40)²
Or, 5x = (50 + 40) (50 – 40) = 90 × 10
Or, x = 90 × 10/5 = 180
Hence, the value of x will be 180
Question no – (7)
Solution :
Given, x + 1/x = 20
or (x + 1/x)² = (20)²
or x² + 2x 1/x + 1/x² = 400
or x² + 1/x² + 2 = 400
or x² + 1/x² = 400 – 2 = 398
or x² + 1/x² = 398
Hence, the value will be 398
Question no – (8)
Solution :
Give, x – 1/x = 3
or, (x- 1/x)² = 32
or, x²-2 x. 1/x + (1/x)² = 9
or, x² + 1/x² – 2 = 9
or, x² + 1/x² = 9 + 2 = 11
(x² + 1/x²)² = (11)² …… (Squaring both the term)
= x² + 2 . x . 1/x² + 1/x⁴ = 121
= x⁴ + 1/x⁴ = 121 – 2
= x⁴ + 1/x⁴ = 119
Therefore, the values are 11 and 119
Question no – (9)
Solution :
As per the question, x² + 1/x² = 18
or, x² + 1/x² + 2 – 2= 18
or, x² + 2. x. 1/x + 1/x² – 2 = 18
or, (x + 1/x)² = 18 + 2 = 20
or, (x + 1/x) = √20
Again,
x² + 1/x² = 10
or, x² + 1/x² + 2 – 2= 18
or, x² – 2. x 1/x + 1/x² + 2 = 18
or, (x – 1/x)² = 18 – 2 = 16
or, x – 1/x = √16 = 4
∴ x + 1/x = √20
x – 1/x = 4
Therefore, the values are √20 and 4.
Question no – (10)
Solution :
Given, x + y = 4; xy = 2
or, (x + y)² = 4
or x² + 2x + 2xy +y² = 16
or, x² + y² + 2.2 = 16
or, x² + y² +4 = 16
or, x² + y²
= 16 – 4
= 12
Therefore, the value of x² + y² will be 12.
Question no – (11)
Solution :
Given, x – y = 7
Squaring both side,
= (x – y)² = 7²
= x² + y² – 2xy = 49
= x² + y² = 49 + 2xy
Now,
= x² + y² = 49 + 2 × 9
= (49 + 18)
= 67
Hence, the value of (x² + y²) will be 67
Question no – (12)
Solution :
As per the question, 3x + 5y = 11; xy = 2
or, (3x + 5y)² = 11²
or, (3x) ² + 2. 3x. 5y + (5y)² = 121
or, 9x² + 30xy + 25 y² + 30.2 = 121
or, 9x² + 25y² = 121 – 60 = 61
or, 9x² + 25y² = 61
Therefore, the value will be 61
Question no – (13)
Solution :
(i) 16x² + 24x + 9
= (4x)² + 2. 4x .3 + 3²
= (4x + 3)² = (4. 7/4 + 3) ….[∵ x = 7/4]
= (7 + 3)²
= (10)² = 100
Thus, the value will be 100
(ii) 64x² + 81y² + 144xy
= (8x) ² + (9y) + 2. 8x. 9y
= (8x + 9y) (8.11 + 9.4/3) ….[∵ x = 11, y = 4/3]
= 988 + 12)² = (100)²
= 10000
Therefore, the value will be 10000
(iii) x = 2/3; y = 3/2
∴ 81x² + 16y² – 72xy
= 81x² – 72xy + 16y²
= (9x)² – 2. 9x. 4y + (4y)²
= (9x – 4y)² = (9 .2/3 – 4 . 3/4)²
= (6 – 3)² = 3² = 9
Therefore, the value will be 9
Question no – (14)
Solution :
Given value, x + 1/x = 9
∴ x + 1/x = 9
or (x + 1/x) = 9²
or, x² + 2.x 1/x + (1/x)²= 81
or x² +1/x² + 2 = 81
or, x²1/x² = 81 – 2 = 79
or, (x² + 1/x²)² = (79)²
or (x²) + 2. x²1/x² + (x²)² = 6241
or, x⁴ + 1/x ⁴ + 2 = 6241
or, x⁴ + 1/x⁴
= 6241 – 2
= 6239
Therefore, the value will be 6239
Question no – (16)
Solution :
As per the question,
2x + 3y = 14; 2x – 3y = 2; xy = ?
∴ (2x + 3y) ² – (2x-3y)² = 24xy
or (14)² – 92)² = 24xy
or, 196 – 4 = 24xy
or, xy = 192/24 = 8
Therefore, the value of xy will be 8
Question no – (17)
Solution :
(i) x² + y² = 29; xy = 2
∴ x² + y² = 29
or, (x – y)² – 2xy = 29
or, (x + y)² 2. 2 = 29
or, (x + y) = 29 + 4 = 33
or x + y = ±√33
(ii) x² + y² = 29; xy = 2
∴ x² + y² = 29
or, (x – y) ² – 2xy = 29
or, (x – y)² + 2.2 = 29
or, (x – y)² = 29 – 4 = 33
or, x – y = √ = ±5
Therefore, the value of x – y will be ±5
(iii) x² + y² = 29; xy = 2
∴ x² + y² = 29
or (x² + y²)² = (29)²
or (x²)² + 2. x²y² + y⁴ = 841
or, x⁴ + y⁴ + 2.4 = 841
or, x⁴ + y⁴ = 841 – 8 = 833
Hence, the value of x⁴ + y⁴ = 833
Question no – (18)
Solution :
(i) 4x² – 12x + 7
= (2x)²- 12 + 12x + 7 + 2
= (2x)² – 2.2x .3 + (3)²
= (2x) – 2. 2x 3 + (3)²
= (2x + 3)²
∴ Here need to add 2 (answer)
(ii) 4x² – 20x + 20
= (2x)² – 20x + 20 + 5
= (2x)² – 20x + 25
= (2x)² 2 . 2x .5 + 5²
= (2x + 5)²
∴ Here need to add 5 (answer)
Question no – (19)
Solution :
(i) (x – y) (x + y) (x² + y²) (x⁴ + y⁴)
= (x² – y²) (x² + y²) (x⁴+ y⁴)
= {(x²)² – (y²)²} (x⁴ + y⁴)
= (x⁴- y⁴) (x⁴ + y⁴)
= (x⁴)² – (x⁴)²
= x⁸ – y⁸…(Simplified)
(ii) (2x – 1) (2x + 1) (4x² + 1) (16x⁴ + 1)
= [(2x)² – 1²] (4x² + 1) (16x + 1)
= (4x² – 1) (4x² + 1) (16x⁴ + 1)
= [(4x²)² – (1²)² ] [16x⁴ + 1]
= (16x⁴ – 1) (16x⁴ + 1) = (16x⁴) – 1²
= 256x⁸ – 1…(Simplified)
(iii) (7m – 8n)² + (7m + 8n)²
= (7m)² – 2.7m 8.n + (8n)² + (7m)² + 2.7m .8n + (8n)²
= 49m² + 49m² + 64n² + 64n²
= 98m² + 128n²…(Simplified)
(iv) (2.5p – 1.5q)² – (1.5 – 2.5q)²
= [(2.5p)² – 2 × 2. 5p × 1. 5q +(1.5q)²] – [(1.5p) – 2. 1. 5p. 2. 5q + (2.5q)²]
= 6.25p² – 7.5 pq + 2.25q² + 2. 25p² + 7.5pq – 6.25q²
= 4p² – 4q² = 4(p² – q²)
= 4(p² – q²)…(Simplified)
(v) Given, (m² – n²m) ² + 2m³n²
= (m²)² – 2.m². n².m + (n²m)² + 2m³n²
= m⁴ – 2m³n² + n⁴m² + 2m³n²
= m⁴ + n⁴m²…(Simplified)
Question no – (20)
Solution :
(i) L.H.S, (3x + 7)² – 84x
= (3x + 7) – 84x
= (3x)² + 2. 3x .7 + 7² – 84x
= (3x)² + 42x + (7)² – 84x
= (3x)² – 42x + (7)²
= (3x – 7)²
∴ L.H.S = R.H.S …[proved]
(ii) L.H.S (9a-5b)² + 180ab
= (9a)² – 2. 9a. 5b + (5b)² + 180ab
= (9a)² – 90ab + (5b) ² + 180 ab
= (9a)² + 90ab + (5b)² = (9a + 5b)²
∴ L.H.S = R.H.S …[proved]
(iii) L.H.S (4m/3 – 3n/4)² + 2mn
= (4m/3)² – 2 . 4m/3 . 3n/4 + (3n/4)² + 2mn
= (4m/3)² – 2mn + 2mn + (3n/4)²
= (4m/3)² + (3n/4)² = 16m²/9 + 9n²/16
∴ L.H.S = R.H.S …[proved]
(iv) L.H.S (4pq + 3q)² – (4pq – 3p)
= (4pq)² + 2.4 pq. 3a + (3q)² – [(4pq)² – 2. 4 pq. 3q + (3q)²
= (4pq)² + 24pq + (3q)² – (4pq)² + 24pq – (3q)² = 48 pq²
∴ L.H.S = R.H.S …..[proved]
Algebraic Expressions and Identities Exercise 6.7 Solution :
Question no – (1)
Solution :
(i) (x + 4) (x + 7)
= x² + (4 + 7) x + 4.7
= x² + 11x + 28
Thus, the product will be x² + 11x + 28
(ii) (x – 11) (x + 4)
= [x + (-11)] [x + 4]
= x² + (-11 + 4) x + (- 11). 4
= x² + (-7) x + (- 44)
= x² – 7x – 44
Hence, the product will be x² – 7x – 44
(iii) Given, (x + 7) (x – 5)
= [(x + 7)] [x + (-5)]
= x²[7 + (-5)] x + 7.(-5)
= x² + (2) – 35
= x² + 2x – 35
So, the product will be x² + 2x – 35
(iv) (x – 3) (x – 2)
= x² – (3 + 2) x + 3.2
= x² – 5x + 6
Thus, the product will be x² – 5x + 6
(v) (y² – 4) (y² – 3)
= (y²)² – (4 + 3) y² + 4.3
= 4² – 7y² + 12
Therefore, the product will be 4² – 7y² + 12
(vi) (x + 4/3) (x + 3/4)
= x² + (4/3 + 3/4) x + 4/3 . 3/4
= x² + (16 + 9/12) x + 1
= x² + 25/12x + 1
Hence, the product will be x² + 25/12x + 1
(vii) Let, 3x = y
∴ (y + 5) (y + 11) = y² + (5 + 11) y + 5. 11
= y² + 16y + 55
= (3x)² + 16. 3x + 55 …[Putting value of y = 3x]
= 9x² + 48x + 55
Thus, the product will be 9x² + 48x + 55
(viii) Let, 2x² = y
∴ (y – 3) (y + 5) = (y + 5) (y – 3)
= [y + 5] [y + (-3)]
= y² + [5 + (-3)] y 5 × (-3)
= y² + 2y – 15
= (2x²)² + 2. 2x² – 15 ….[∵ y = 2x²]
= 4x⁴ + 4x⁴ – 15
So, the product will be 4 x⁴ + 4x⁴ – 15
(ix) Let , z²= x
∴ (x + 2) (x – 3) = [x + 2] [x + (- 3)]
= x² + [2 + (-3)] x + 2. (-3)
= x² – x – 6 = (z²) – z² – 6
= x² – z² – 6
Thus, the product will be x² – z² – 6
(x) (3x – 4y) (2x – 4y)
= 3x. 2px – 3x. 4y – 4y – 2x + 4y. 4y
= 6x² – 12xy – 8xy + 16y²
= 6x² – 20xy + 16y²
Hence, the product will be 6x² – 20xy + 16y²
(xi) (3x² – 4xy)(3x² – 3xy)
= (3x²) – (4xy + 3xy) 3x² + (4xy) . (3xy)
= 9x⁴ – 7xy. 3x² + 12x²y²
= 9x⁴ – 21x³y + 12x²y²
Therefore, the product will be 9x⁴ – 21x³y + 12x²y²
(xii) (x + 1/5) (x + 5)
= x² + (1/5 + 5) x + 1/8. 5
= x² + (1 + 25) x + 1
= x² + 26/5 x + 1
Hence, the product will be x² + 26/5 x + 1
(xiii) (z + 3/4) (z + 4/3)
= z² + (3/4 + 4/3) z + 3/4 . 4/3
= z² + (9 + 16/12) z + 1
= z² + 25/12 z + 1
Thus, the product will be z² + 25/12 z + 1
(xiv) (x² + 4) (x² + 9)
= (x²)² + (4 + 9) x² +4 × 9
= x⁴ + 13x² + 36
So, the product will be x⁴ + 13x² + 36
(xv) (y² + 12) (y² + 6)
= (y²)² + (12 + 6) y² + 12 × 6
= y⁴ + 18y² + 72
Thus, the product will be y⁴ + 18y² + 72
(xvi) (y² + 5/7) (y² – 14/5)
= (y² + 5/7) [ y² + (-14/5)]
= (y²)² – [5/7 + (-14/5)] y² + 5/7 × (-14/8)²
= y⁴ – [25 – 98/35] y² – 2
= y⁴ – 73/35 y² – 2
Therefore, the product will be y⁴ – 73/35 y² – 2
(xvii) (p² + 16) (p² – 1/4 )
= [p² + 16] [p² + (-1/4)]
= (p²)² + [16 + (-1/4)] p² + 16 × (-1/4)
=p⁴ + [ 64 – 1/4] p² – 4
= p⁴ + 63/4 p² – 4
Therefore, the product will be p⁴ + 63/4 p² – 4
Question no – (2)
Solution :
(i) 102 × 106
= (100 + 2) (100 + 6)
= (100)² + (2 + 6) 100 + 2.6
= 10000 + 800 + 12
= 10000 + 800 + 12
= 10812 …(answer)
(ii) 109 × 107
= (100 + 9) (100 + 7)
= (100)² + (9 + 7) 100 + 9.7
= 1000 + 16 × 100 + 63
= 10000 + 1600 + 63
= 11663 …(answer)
(iii) (30 + 5) (30 + 7)
= (30)² + (5 + 7) 30 + 5.7
= 900 + 12 × 30 + 35
= 900 + 360 + 35
= 1295 …(answer)
(iv) 53 × 55
= (50 + 3) (50 + 5)
= (50)² + (3 + 5) 50 + (3 × 5)
= 2500 + 8 × 50 + 15
= 2500 + 400 + 15
= 2915 …(answer)
(v) 103 × 96
= (100 + 3) (100-4)
= [100 + 3] [100 + (4)]
= (100)² – ( 3 + (- 4)] 100 + 3. (-4)
= 1000 + 100 – 12
= 9888 …(answer)
(vi) 34 × 36
= (40 – 6) (40 – 4)
= (40)² – (6 + 4) 40 + 6.4
= 1600 – (10 × 40) + 24
= 1600 – 400 + 24
= 1224 …(answer)
(vii) 994 × 1006
= (100 – 06) (100 + 6)
= (1000)² – 6²
= 100000000 – 36
= 999964 …(answer)
Next Chapter Solution :
👉 Chapter 7 👈
