# Frank Learning Maths Class 8 Solutions Chapter 6

## 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 = 2x– 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

= 9x– 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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Updated: June 5, 2023 — 7:56 am