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Frank Learning Maths Class 8 Solutions Chapter 5 Algebraic Expressions
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 5 Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions. Here students will find solutions for Exercise 5.1, 5.2 and 5.3. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Algebraic Expressions Exercise 5.1 Solution
Question no – (1)
Solution :
(a) 6x + 4y – 3z
7x – 11y – 9z
14x + 8y – 6z
——————————
27x + y – 18z
or
= (6x + 4y – 3z) + (7x – 11y – 9z) + (14x + 8y – 6z)
= 27x + y – 18z
(b) 7p – 8q + 11r + 13
10p – 13q + 0 + 18
8p – 5q – 14r + 12
5p + 17q + 14r
——————————
30p – 9q + 11r + 43
(c) 30p – 9q + 11r + 43
8x2 + 7x + 12
17x2 – 15x – 21
– 9x2 + 11x + 4
4x2 – 18x + 19
——————————
20x2 – 15x + 14
(d) 6ax – 2by + 3cz
11ax – 16by – 15cz
– 9ax – 3by + 10cz
——————————
8ax – 21by – 2cz
(e) 15m2n – 17mn + 8mn2
13 m2n – 15mn – 9mn2
12m2n + 21mn – 10cz
————————————-
40m2n – 11mn – 15m2n
(f) 13x + 17y – 19z + 2
14x + 12y + 0 – 31
– 15x + 6z + 12
2x + y + 11z + 9
————————————-
14x + 30y – 2z – 8
Question no – (2)
Solution :
(a) 12x + 7y – 15z
17x + 8y – 15z
(-) (-) (+)
—————————-
– 5x – y + 0
(b) 21ax + 15by – 6cz
12ax + 6by + 11cz
(-) (-) (-)
—————————-
9ax + 9by – 17cz
(c) 16x3 + 14x2 – 9x + 15
17x3 – 19x2 + 13x – 21
(-) (+) (-) (+)
—————————-
– x3 + 33x2 – 22x + 36
(d) 7p + 11q – 2r + 9
– 8p – 4q + 6r + 15
(+) (+) (-) (-)
—————————-
15p + 15q – 8r – 6
Question no – (3)
Solution :
9a² – 17ab – 3b²a – 10
7a² + 18ab – 9b²a + 15
(-) (-) (+) (-)
————————————
2a² – 35ab + 6b²a – 25
∴ (2a² – 35ab + 6b²a – 25) must be subtracted for get the given answer.
Question no – (4)
Solution :
We know that, Third side = perimeter – sum of two sides.
= (16x3 + 14x2 – 9x + 15) – (7x3 + 3x2 – 2x + 3 + 4x3 + x2 – 5x + 7)
= (16x3 + 14x2 – 9x + 15) – (11x3 + 4x2 – 7x + 10)
= 5x3 + 10x2 – 2x + 5
Hence, the third side will be 5x3 + 10x2 – 2x + 5
Question no – (5)
Solution :
We know that, Perimeter of a rectangle is 2 (sum of adjacent sides)
= 2(-6p3 + 7p2q2 + 11pq + 7p2q2 + 4p2 + 9pq)
= 2(-6p3 + 14p2q2 + 20pq + 4p2)
= 8p2 + 28p2q2 + 40pq – 12p3
Therefore, its perimeter will be 8p2 + 28p2q2 + 40pq – 12p3
Question no – (6)
Solution :
∴ (13 + 19b+ 12c + 14a – 21b + 11c) – (17a + 13b – 15c + 8a + 12b – 18c)
= 27a – 2b + 23c – 25a – 25b + 33c
= 2a – 27b + 56c …(Answer)
Question no – (7)
Solution :
As per the given question,
A = 5x + 11y – 75z
B = 12x + 13y + 19z
C = 7x – 6y + 21z
(a) A + B + C
= 5x + 11y – 15z + 12x – 13y + 19z + 7x – 6y + 21z
= 24x – 8y + 25z
(b) A – B – C
= 5x + 11y – 15z – 12x + 13y – 19z – 7x + 6y – 21z
= -14x + 30y – 55z
(c) A + B – C
= 5x +11y – 15z + 12x – 13y + 19z – 7x + 6y – 21z
= 10x + 4y – 17z
(d) 2A + 3B + C
= 10x + 22y – 30z + 36x – 39y + 57z + 7x – 6y + 217
= 53x – 23y + 48z
Question no – (8)
Solution :
Let other side = x
We know that, Perimeter = 2 (one side + other side)
∴ 16a + 8b – 6c = 2 (5a + 3b – 4c + x)
= 8a + 4b – 3c = 5a + 3b – 4c + x
= x = 3 + b + c
Hence, the other side will be 3 + b + c
Question no – (9)
Solution :
As we know that, Perimeter = Sum of 3 side
∴ 18p + 12q + 13r = 34 – 4r + 6p + 5q + 7r+ x
= 18p + 12q + 13r = 11p + 8q +3r + x
= x = 7p + 4q + 10r
Therefore, the third side will be 7p + 4q + 10r
Algebraic Expressions Exercise 5.2 Solution
Question no – (1)
Solution :
(a) (5a + 11b) (8a + 9b)
= (5a × 8a) (11b × 8a) + (5a × 9b) + (11b × 9b)
= 40a2 + 88ab + 45ab + 99b2
= 40a2 + 133ab + 99b2
(b) (11p – 9p) (4p + 17q)
= 44p2 – 36pq + 187pq – 153q2
= 44p2 – 18pq – 153q2
(c) (7x2 + 4y2) (3×2 + 8y2)
= 21x4 + 12x2y2 + 56x2y2 + 32y4
= 214 + 68x2y2 + 32y4
(d) (13xy – 3z) (15xy – 8z)
= 195x2y2 – 45yz – 104xyz + 24z2
= 195x2y2 – 149xyz + 24z2
(e) (7x3y + 5zx) (11x3y + 8xz)
= 77x3 + 55x4yz + 56x4yz + 40x2z2
= 77x3y + 1.11x4yz + 4ox2z2
(f) (2,5a + 0.3b) (1.5c + 0.7d)
= (2,5a + 0.3b/10) (1.5c + 0.7d/10)
= 3a5a + 45bc + 175ad + 21bd/10
= 37.5a + 4.5bc + 17.5ad + 2.1bd
(g) (2x/3 – 5y/4) (6x/5 + 7y/3)
= 3x2/5 – 3xy/2 + 14xy/9- 35y2/12
= 4x2/5 – (27xy – 28xy/18) – 35y2/12
= 4x2/5 + xy/18 – 35y2/12
(h) (2x3 – 3y3) (3x2 – 4y2)
= 6x5 – 9x2yu3 – 8x3y2 + 12y5
Question no – (2)
Solution :
(a) (3a + 7b) (6a + 8b – 11c)
= 18a2 + 24ab – 33ac + 42ab + 56b2 – 77bc
= 18a2 + 6ab – 33ac – 77bc + 56b2
(b) (4x2 + 11) (3x2 + 17x – 16)
= 12x4 + 68x3 =- 64x2 + 33x2 + 187x – 176
(c) (5p2q2 – 7) (7p2q – 13p + 6q)
= 35p4q3 – 65p2q2 + 30p2q3 – 49p2q + 91p – 42q
(d) (xyz) (6x3 + 7cx2 – 12x + 21)
= xyz (6x3 + 7cx2 – 12x + 21)
= 6x4yz + 7x3yz – 12x2yz + 21xyz
(e) (4m2n2) (3m2n + 7mn + 9mn2 – 15)
= 4m2n2 (3m2n + 7mn + 9mn2 – 15)
= 12m4n3 + 28m3n3 + 36m3n4 – 60m2n2
(f) (3x + 4y – 11) (5x – 3y – 8)
= 15x2 – 9xy – 24x + 20xy – 20y2 – 32y – 55x + 33y + 88
= 15x2 + 11xy – 79x + y – 12y2 + 88
(g) (2x2 + 4x + 15) (5x2 – 7x – 12)
= 10x4 + 10x3 – 24x2 + 20x3 – 28x2 – 48x + 75x2 – 105x – 180
= 10x4 – 14x3 + 51x2 – 153x – 180
(h) (5x3 – 3x2 + 6x) (4x2 – 7x + 9)
= 20x5 – 35x4 + 45x3 – 12x4 + 21x3 – 27x2 + 24x3 – 42x2 + 54x
= 20x4 – 47x4 + 90x3 – 69x2 + 54x
(i) (a + 1/a) (a3 + 1/a3)
= a4 + a2 + 1/a2 + 1a4
(j) (a – 1/a (a4 + 1/a4)
= a5 + 1/a3 – a3 – 1/a5
Question no – (3)
Solution :
(a) Given, a = 2
∴ 3a (a2 – 7) + 2 (a + 2)
= 6 (4 – 7) + 2 (2 + 2)
= = 6 (-3) + 2 × 4
= – 18 + 8
= -10
Therefore, the value will be -10.
(b) Given, a = 0
∴ 3a (a2 – 7) + 2 (a + 0)
= 0 + 2 (0 + 3)
= 6
Thus, the value will be 6
(c) Given, a = – 3
∴ 3a (a2 – 7) + 2 (a + -3)
= -9 (9 – 7) + (-3 + 3)
= -9 × 2 + 0
= -18
Therefore, the value will be -18
Question no – (4)
Solution :
(a) x = 2 and y = 3
∴ 2 × 4 (6 – 4) + 9 (2 + 2)
= 8 × 2 + 9 × 4
= 16 + 36
= 52
Thus, the value will be 52
(b) x = -1 and y = 5
∴ 2 ( -5 – 4) + 3 × 5 (- 1 + 2)
= 2 (-9) + 15 (1)
= – 18 + 15
= – 3
Therefore, the value will be -3
Question no – (5)
Solution :
(a) 4(2x + 5y + 8) +3 (5x + 7y – 9)
= 8x + 20y + 32 + 15x + 21y – 27
= 23x + 41y + 5…(Simplified)
(b) (p + q) (p – q) + (p + r) (r – p) – (r2 – q2)
= p2 – q2 + r2 – p2 – r2 + q2
= 0…(Simplified)
(c) (w + x) (y – z) + (w – x) (y +z) + 2 (wy + xz)
= wy + xy – wz – xz + wy – xy + wy – xz + 2 (wy + xz)
= 2 (wy – xz) + 2 (wy + xz)
= 2wy – 2xz + 2wy + 2xz
= 4wy…(Simplified)
(d) (a + b) (a2 – ab + b2)
= a3 + b3…(Simplified)
(e) (x – y) (x2 + xy + y2)
= x3 – y3…(Simplified)
(f) (a + b) (a2 + 2ab + b2)
= (a + b)3
Or,
= a3 + 3a2b + 3ab2 + b3…(Simplified)
(g) ab (cx + y) + bc (a – y) – 2ca (bx + y)
= abcx + aby + abcx – bay – 2abcx – 2acy
= aby – bcy – 2acy…(Simplified)
Question no – (6)
Solution :
| Length (l) | Breadth (b) | Height (h) | Volume (=lbh) | |
| (a) | 4xy | -12xz | -5zy | 24x2y2z2 |
| (b) | 7a2c | 2a2b2 | 2abc2 | 42a5b3c3 |
| (c) | -2mn | -9lmn | 6nl | 108lm2n2 |
| (d) | -x4 | -x3 | -x2 | -x9 |
| (e) | 3ax | 13by | 23cz | 897abcxyz |
| (f) | 15x4 | 10×5 | 25 | 3750x9 |
Algebraic Expressions Exercise 5.2 Solution
Question no – (1)
Solution :
As per the given question,
a = 9,
b = 4
∴ L.H.S = (a – b)2
= (9 – 4)2
= 52
= 25
∴ R.H.S, = a2 – 2ab + b2
= 81 – 72 + 16
= 25
∴ L.H.S = R.H.S …(Verified)
Question no – (2)
Solution :
According to the question,
x = 11
a = 4
b = -7
∴ L.H.S, = (x + a) (x + b)
= (11 + 4) (11 – 7)
= 15 × 4
= 60
∴ R.H.S, = x2 + (a + b) x + ab
= 121 + (-3) × 11 + 4 (- 7)
= 121 – 33 – 28
= 60
Hence, L.H.S = R.H.S …(Verified)
Question no – (3)
Solution :
(a) Given, (4x + 9y) (4x + 9y)
= (4x + 9y)2
= 16x2 + 72 xy + 18y2
Therefore, the product will be 16x2 + 72 xy + 18y2
(b) Given, (3a2 – 8b2) (3a2 – b2)
= (3a2 – 8b2)
= (3a2)2 – 2 × 3a2 × 8b2 + (8b2)2
= 9a4 – 48a2b2 + 64b4
Thus, the product will be, 9a4 – 48a2b2 + 64b4
(c) Given, (3x2/4m+ 4y2/7) (3x2/4 + 4y2/7)
= (3x2/4 + 4y2/7)2
= (3x2/4 + 4y2/7)2
= (3x2/4)2 + 2 × 3x2/4 × 4yz/7 + (4y2/7)2
Therefore, the product will be, (3x2/4)2 + 2 × 3x2/4 × 4yz/7 + (4y2/7)2
(d) Given, (13p – 15q) (13p + 15q)
= (13p)2 – (15q)2
= 169p2 – 225q2
(e) Given, (0.5x + 0.9y) (0.5x – 0.9y)
= (0.5x/10)2 – (0.9y/10)2
= 0.5z/10 – 81y2/100
= 25x2 – 81y2
Therefore, the product will be 25x2 – 81y2
(f) Given, (-3p + 7q) (- 3p + 7q)
= (7q – 3p)2
= 49q2 – 42pq + 9p2
Thus, the product will be 49q2 – 42pq + 9p2
(g) Given, (x2y2 + 2) 9x2y2 + 2)
= (x2y2 + 2)2
= x2y2 +2)
= x4y4 + 4x2y2 + 4
Therefore, the product will be x4y4 + 4x2y2 + 4
(h) Given, (11x + 9) (11x + 15)
= (11z + 9) (11x + 9 + 6)
= (11x + 9)2 + 6 (11x + 9)
= 121x2 + 198x + 81 + 66x + 54
= 121x2 + 264x + 135
Thus, the product will be 121x2 + 264x + 135
(i) Given, 9m2 + 7) (9m2 – 5)
= (9m2 + 5 + 2) (9m2 – 5)
= (9m2 + 5) (9m2 – 5) + 2 (9m2 – 5)
= (9m2)2 (5)2 + 18m2 + 10
= 81 m4 – 25 + 18m2 – 10
= 81m4 + 18m2 – 35
Hence, the product will be 81m4 + 18m2 – 35
(j) Given, (7a + 13b) (7a – 18b)
= (7a + 13b) (7 – 13b – 5b)
= (7a + 13b) (7 – 13b) – 5b (7a + 13b)
= (7a)2 – (13b)2 – 35ab2 – 354b + 45b2
= 49a2 – 169b2 – 35ab + 65b2
= 49a2 – 25ab – 104b2
Therefore, the product will be 49a2 – 25ab – 104b2
(k) Given, (x2y2 + 17) (x2y2 – 21)
= (x2y2 + 17) (x2y2 – 17 – 4)
= x4y4 – 289 – 4x2y2 – 68
= x4y4 – 4x2y2 – 357
Therefore, the product will be x4y4 – 4x2y2 – 357
Question no – (4)
Solution :
(a) (a2b2 + c2d2)2
= a4b4 + 2a2b2c2d2 + c4d4
(b) (3/2m – 2/3n)2
= (3/2m – 2/3n)2
= 9m2/4 – 2mn – 4n2/9
(c) (0.6a – 0.7b)2
= 0.36a2– 0 84ab + 0.49b2
(d) (4x2 + 3y2)2
= 16x4 + 24x2y2 + 9y4
Question no – (5)
Solution :
(a) (a2 + b2)2 – (a2 – b2)2
= 4a2b2…(Simplified)
(b) (3.5a – 4.5b)2 – (3.5a + 4.5b)2
= 4 × 3.5a/10 × (- 4.5b/10)
= – 63ab…(Simplified)
(d) (12a + 17b)2 – (12a – 17b)2
= 4×12a × 17b
= 816ab…(Simplified)
(e) 1.98 × 1.98 – 1.02 × 1.02/0.96
= (1.98)2 – 1.02)2/0.96
= (1.98 + 1.02) (198 – 1.02)/096
= 3 × 96/0.96
= 3…(Simplified)
(f) (x + y) (x – y) (x2 + y2)
= (x2 – y2) (x2 + y2)
= x4 – y4…(Simplified)
(g) (r + s)2 – (r + s) (r – s)
= (r + s) – (r + s – r + s)
= (r + s) × 2s
= 2 (sr + s2)…(Simplified)
Question no – (6)
Solution :
(a) Given, (5x + 11)2 – (5x – 11)2 = 220x
= 4 × 5x × 11
= 220x …(Proved)
(b) Given, (4ax + 3y)2 – (4ax – 3y)2 = 48axy
= 4 × 4ax × 3y
= 48axy …(Proved)
(c) Given, (5/6 p – 6/5q) + 2pq = 25/36 = 25/36 p2 + 36/25 q2
= (5/6 p)2 + (6/5 q)2
= (5/6p – 6/5 q) + 2. 5/6.6/5 q
= (5/6 p – 6/5q)2 + 2pq (Proved)
(d) Given, (x + y) (x – y) + (y + z) (y – z) + (z + x) (z – x) = 0
= x2 – y2 + y2 – x2 + x2 – x2
= 0 …(Proved)
Question no – (7)
Solution :
(a) (93)2
= (90 + 3)2
= 8100 + 2 × 90 × 3 + 9
= 8100 + 540 + 9
= 8649
(b) (103)2
= (100 + 3)2
= 10000 + 600 + 9
= 10609
(c) (54)2
= (50 + 4)2
= 2500 + 400 + 6
= 2916
(d) (997)2
= (1000 – 3)2
= 1000000 – 6000 + 9
= 994,009
(e) (10.1)2
= (10 + 1)2
= 100 + 2 + 01
= 102.01
(f) (8.6)2
= 1/100 [(86)2]
= 1/100 [(90 – 4)2]
= 1/100 [8100 – 120 + 16]
= 7396/100 = 73.96
(g) 103 × 97
= (100 + 3) (100 – 3)
= (100)2 – (3)2
= 1000000 – 9
= 999,991
(h) 53 × 47
= (50 + 3) (50 – 3)
= (50)2 – (3)2
= 2500 – 9
= 2491
(i) (12.5)2 – (7.5)2
= (12.5 + 7.5) (12.5 – 7.5)
= (20.0) × (5)
= 100
(j) (10.2)2 – (9.8)2
= (10.2 + 9.8) (10.2 – 9.8)
= 20 + 4
= 20.4
(k) 93 × 109
= (100 – 7) (100 + 7 + 2)
= (100 – 7) (100 + 7) + 2 × 93
= (100)2 – (7)2 + 186
= 10000 – 49 + 186
= 10,137
(l) 54 × 59
= (50 + 4) × (50 + 4 + 5)
= (50 + 4)2+5 × 54
= 2500 + 400 + 16 + 270
= 3186
(m) 12.5 × 10.5
= 12.5/10 × 12.5/10
= 1/100 [(10 + 5 + 20) × (100 + 5)]
= 1/100 [(100 + 5)2 + 20 × 105)]
= 1/100 910000 + 1000 + 25 + 2100)
= 1/100 × 12125
(n) 997 × 1007
= (100 – 3) (100 + 3 + 4)
= (100)2 (3)2 + 4 × 997
= 100000 – 9 3988
= 1003979
(o) 82 × 85
= (80 + 2) (80 + 2 + 3)
= (80 + 2)2 × 3 × 82
= 6400 + 320 + 4 + 246
= 6970
Question no – (8)
Solution :
In the question we get,
(x + y) = 13
xy = 22
∴ x2 + y2 = (x + y)2 – 2xy
= (13)2 – 2 × 22
= 169 – 44
= 125
Therefore, the value of x2 + y is2 125
Question no – (9)
Solution :
As per the given question,
(x – y) = 5
xy = 36
∴ x2 + y2 = (x – y)2 + 2xy
= 25 + 72
= 97
Therefore, the value of x2 + y2 will be 97
Question no – (10)
Solution :
Given in the question,
(x2 + y2) = 74
xy = 35
∴ x2 + y2 = (x + y)2 – 2xy
= 74 = (x + y)2 – 2 × 35
= (x + y) = 74 + 70
= x + y = √144
= 12
Now, x2 + y2 = (x – y)2 – 2xy
= 7x (x – y)2 + 2 × 35
= (x – y)2 = 74 – 70
= x – y = √4
= 2
Question no – (11)
Solution :
(a) Given, x2 + 1/x2
= x2 + 1/x2
= (x + 1/x)2 2.x 1/x
= 52 – 2
= 25 – 2
= 23
(b) Given, x + 1/x = 5,
∴ x4 + 1/x4
= (x2 + 1/x2)2 – 2.x4 1/x4
= (23)2 – 2
= 529 – 2
= 527
Question no – (12)
Solution :
According to the given question,
x2 + 1/x2 = 3
(a) x – 1/x
= x2 × 1/x2 = 3
= x2 + 1/x2 = (x – 1/x)2 + 2.x 1/4
= 3 = (x – 1/x)2 + 2
= (x – 1/x)2 = 1
= x – 1/x = 1
(b) x + 1/x
= x2 + 1/x2 = (x + 1/x)2 – 2.x.1/x
= 3 = (x + 1/x)2 = 2
= x + 1/x = √5
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