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Frank Learning Maths Class 8 Solutions Chapter 6 Linear Equations In One Variable
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 6 Linear Equations In One Variable. Here students can easily find step by step solutions of all the problems for Linear Equations In One Variable. Here students will find solutions for Exercise 6.1, 6.2 and 6.3. 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.
Linear Equations In One Variable Exercise 6.1 Solution
Question no – (1)
Solution :
(a) Given, 5x – 13 = 2x + 8
∴ 5x – 2x = 8 + 13
= 3x = 21
= x = 21/3
∴ x = 7
(b) Given, 17x + 12 = 13x + 24
= 4x = 12
= x = 12/4
∴ x = 3
(c) Given, 7 (x + 3) = 5 (2x – 3)
∴ 7x + 21 = 10x – 15
= 3x = 36
= x = 36/3
∴ x = 12
(d) Given, 0.5x + 1.5 = 0.7x – 0.9
= 0.5x + 15/10 = 7x – 9/10
= 2x = 24
= x = 24/12
∴ x = 12
(e) Given, 7y/3 + 11 = 9y/4 + 15
= 7y + 33/3 = 9y + 60/4
= 28y + 132 = 27y + 180
∴ y = 48
(f) Given, 2m – 2/9 = 4m – 4/3
= 2m – 2/9 = 4m – 4/3
= 2m = 4/3 – 2/9
= 2m = 12 – 2/9
= 2m = 10/9
∴ m = 5/9
(g) Given, 6(4y – 2) = 3 (5 + 3y)
= 8y – 4 = 5 + 3y
= 5y = 9
∴ y = 9/5
(h) Given, 9(x – 1) = 4 (x – 3)
= 9x – 9 = 4x – 3
= 5x = 12
∴ x = 12/5
(i) As per the question,
2x -3/3x + 2 = – 2/3
= 6x – 9 = – 6x – 4
= 12x = 5
∴ x = 5/12
(j) Given, 6/3x – (3 – 4x) = 3/2
= 3x – 3 + 4x = 4
= 7x = 7
∴ x = 1
(k) Given, (2x + 3) – (5x + 4)/5x + 9 = -7/3
= (2x + 3) – (5x + 4) = -7/3
= 2x + 3 – 5x – 4/5x + 9
= – 3x – 1/3x + 9 = -7/3
= 3x + 1/5x + 9 = 7/3
= 35x + 63 = 9x + 3
= 26x = – 60
= x = 60/26
∴ x = 30/13
(l) Given, 2x – (7 – 5x)/9x – (3x + 4x) = 7/6
= 2x – 7 + 5x/9x – 3 – 4x = 7/6
= 7x – 7/x – 3 = 7/6
= x – 1/5x – 3 = 1/6
= 6x – 6 = 5x – 3
∴ x = 3
Question no – (2)
Solution :
(a) Given, 4x + 1/3 + 2x+1/2 = 3x – 7/5 + 6
= 8x + 2 + 6x + 3/6 = 3x – 7 + 30/5
= 14x + 5/6 = 3x – 23/5
= 70x + 25 = 18x + 138
= 52x = 138 – 25
∴ x = 113/52
(b) Given, (x – 5)/4 = (3 + 4x)/3 – (x – 3)/6
x/4 – 5/4 = 3/3 + 4x/3 – x/6 + 3/6
= x/4 – 4x/3 + x/6 = 1 + 1/2 + 5/4
= 3x – 16x + 2x/12 = 4 + 2 + 5/4
= – 11x = 3 × 11
= x = – 3
L.H.S, = x/4 – 4x/3 + x/
= -3/4 – 4×(-3)/3 + -3/6
= 4 – 1/2 – 3/4
= 16 – 2 – 3/4
= 11/4
R.H.S, = 1 + 1/2 + 5/4
= 4 + 2 + 5/4
= 11/4
Therefore, L.H.S = R.H.S …[Verified]
(c) Given, 2x =/5 – 6x = 9/7
= 5 – 6x/2x = 7/9
= 5 – 6x/2x = 7/9
= 5/2 – x/2 = 7x/9
= 7x/9 + 27x = 45/2
= 34x = 2 × 34
= 45/68
(d) Given, (3y – 1)/4 + (2y + 3)/3 = (1 – 7y)/6
= 3y/4 – 1/4 + 2y/3 + 1 = 1/6 – 7y/6
= 3y/4 + 2y/3 + 7y/6 = 1/6 + 1/4 – 1
= 9y + 8y + 14y/12 = 2 + 3 – 12/12
= 31y = -7
= y – 7/3
L.H.S = 31y
= 31 × -7/31
= -7 = R.H.S
(e) Given, 0.6 (1.5n + 1.2) = 0.8 (0.3n – 0.35
= 4.5n + 3.6 = 1.2n – 1.40
= 4.5n – 1.2n = 1.4 – 3.6
= 3.3n = – 5.0
= x = 50/3.3
= – 50/33
L.H.S = 3.3
= 3.3/10 × -50/3.3
= – 5 = R.H.S
(f) Given, 5-8x/7x + 1 = 5/4
= 35x + 5= 20 – 32x
= 67x =15
= x = 15/67
∴ L.H.S = 67 × 15/67
= 15 = R.H.S
(g) Given, 2/3y + 1 = 5/3 (y + 1/4)
= 2y + 3/3 = 5/3 94y + 1/4)
= 8y + 12 = 20y + 5
= 12y = 7
= y = 7/12
∴ L.H.S = 12 × 7/12
= 7 = R.H.S
(h) Given, 5(y+12)-17(2-y)/7y-1 = 8
= 5y + 60 34 + 17y = 56y – 8
= 22y + 26 = 56y – 8
= 34y = 3y
= y = 1
L.H.S = (591 + 12) – 17 (2 – 1)/7.1 – 1
= 65 – 17/6
= 48/6
= 8 R.H.S
(i) Given, x + 2/6 + x – 3/3 = x
= x + 2- 2x – 6/6 = x
= – x – 4 = 6x
= x = – 4
= x – 4/7
∴ L.H.S = 7 × (-4/7)
= – 4 = R.H.S
(j) Given, 2/5(n – 4)/3/4(n + 7) = 4/5
= 2/5 (n – 4) × 5 – 4x = × 3/4 (n + 7)
= 2n – 8 = 3n + 21
= n = – 29
∴ L.H.S, 2 (-29) -8
= -58 – 8
= -68
∴ R.H.S, 3 × (-29) + 21
= -87 + 21
= -66
∴ L.H.S = R.H.S …(Solved)
Linear Equations In One Variable Exercise 6.2 Solution
Question no – (1)
Solution :
(a) Given, 6x + 7/3x + 2 = 4x + 5/2x + 3
= (6x + 7) (2x + 3) (4x + 5) (3x + 2)
= 12x² + 14x + 18x + 21 = 12x² + 15x + 8x + 10
= 32x + 21 = 23x + 10
= 9x = – 11
∴ x = – 11/9
(b) Given, 6 (5x + 1) + x² = (x + 1) (x + 2)
= 30x + 6 + x² = x² + 3x + 2
= -2 + x = -4
∴ x = -4/27
(c) Given, 5x + 4/7x + 3 = 5x – 1/7x – 2
= 35x² + 28x – 10x – 8 = 35x² – 7x + 75x – 3
= 18x – 8 = 8x – 3
= 10x = 5
∴ x = 5/10
(d) Given, 9x -7/3x + 5 = 3x – 4/x + 6
= 9x² – 7x + 54x – 42x = 9x² – 12x + 15x – 20
= 47x – 42 = 3x – 20
= 44x = 22
= x = 22/442
∴ x = 1/2
(e) Given, x – 3/x + 3 = x – 2/x + 2
= x² – 3x + 2x – 6 = x² – 3x – 23x – 6
= – x = x
= 2x = 0
∴ x = 0
(f) Given, 1/x – 1 + 3/x + 1 = 4/x
= x + 1 + 3x – 3/x2 = 4/x
= 4x – 2/x2 – 1 = 4/x
= 2x – 1/x2 = 2/x
= 2x2 – 2 = 2x2 – x
∴ x = 2
(g) Given, (x + 2) (2x – 3) = (2x + 5) (x + 4)
= 2x² – 3x + 4x – 6
= 2x² + 5x + 8x + 20
= x – 6 = 13x + 20
= 12x = – 26
= x – 26/12
∴ x = – 13/6
(h) Given, (6x + 5) (2x + 3) = (4x + 7) (3x + 2)
= 12x2 + 18x + 10x + 15 = 12x2 + 8x + 21x + 14
= 28x + 15 = 29x + 14
∴ x = 1
Question no – (2)
Solution :
(a) x2 – 9/7 x2 = -7/9
= 9x2 – 81 = – 49 + 7×2
= 2x2 = 32
= x2 = 16
= x = ± 4
∴ x = + 4 are x = – 4
Therefore, Positive value = + 4
(b) x2 + 7/3x2 + 10 = 1/2
= 3x2 + 10 = 2×2 + 1y
= x2 = 4
= x = ± 4
∴ x = +4 are x = – 4
Therefore, Positive value = +4
Linear Equations In One Variable Exercise 6.3 Solution
Question no – (1)
Solution :
Let, the integers are = x and 4x
∴ 4x – x = 75
= 3x = 75
= x = 25
Therefore, the integers will be 25 and 100
Question no – (3)
Solution :
Let, the numbers be = 3x, 4x, 5x
∴ (3x + 5x) = 4x = 64
= x – 4x = 64
= 4x = 64
= x = 16
Therefore, the required numbers are 48, 64 and 80
Question no – (6)
Solution :
Let, the units place number = x
∴ Tens digit = 4x
New, number after reversed,
= (x × 10) + 4x × 1
= 10x + 4x
= 14x
∴ 4x × 10 × x × 1 – 14x = 54
= 41x – 14x = 54
= 27x = 54
= x = 2
∴ The number,
= (4 × 2)²
= 82
Therefore, the required number will be 82.
Question no – (7)
Solution :
Let, the number be – x
= x – 2/3 = 3x – 2/3
= 3x – 2/3 × 9 = 9x – 6
= 9x – 6 = 7x
= 2x = 6
= x = 3
Hence, 3 is the number.
Question no – (8)
Solution :
Let, present of son = x
Present age of father = (53 – x)
4 years ago, (x – 4) son’s age
= (53 – x – 4) father’s age
= (49 – x) – 11
Now, (49 -x) = 4 (x – 4)
= 49 – x = 4x – 16
= 5x = 49 + 16
= x 65/5
= 13 years
∴ Son’s present age = 13 years
∴ Father’s present age,
= (53 – 13)
= 40 years
Question no – (9)
Solution :
Let, son’s present age – x
Ankit present age = 2x
Son’s age 9 years ago = (x – 9)
∴ In four year = (2x + 4)
= (2x + 4) = 4 (x – 9)
= 2x + 4 = 4 x – 36
= 2x = 40
= x = 20
∴ Son’s present age = 20
∴ Ankit’s present age = 40
Question no – (10)
Solution :
Let the total amount x
Deposited = x/2 in bank
∴ Gives her daughter,
= x/2 × 1/2 + 6000
= x/4 + 6000
∴ Gives her son,
= x/4 + 6000/2
= x/8 + 3000
Thus, x/4 + 6000 + x/8 + 3000 = x/2
= 2x + x/8 + 9000 = x/2
= x/2 – 3x/8 = 9000
= 4x – 3x/8 = 9000
= x = 72000
Therefore, Srividya will received Rs. 72000 on retirement.
Question no – (15)
Solution :
Let the number at coins = x
∴ 5x + 2 × 5x/8 = 1000
40x + 10x/8 = 1000
50x = 8000
x = 8000/50
x = 160
∴ 5 rupees coins = 160
∴ 2 rupees coins,
= 160 × 5/8
= 100
Question no – (17)
Solution :
Let A’s share is x
B’s share 5/6x and C’s share
= 4/5 × 5/6
= 2/3x
∴ x + 5/6x + 2/3x = 1500
= 6x + 5x + 4x/6 = 1500
= 15x = 1500 × 6
= x = 1500 × 6/15
= x = 600
∴ A’s share = 600
∴ B’s share = 5/6 × 600 = 500
∴ C’s share = 2/3 × 600 = 400
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