ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.1 Class 8 ICSE Ch-12 Maths Solutions. We Provide Step by Step Answer of  Exe-12.1 Questions for Linear Equations and Inequalities in One Variable as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

## ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.1 Class 8 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-12 Linear Equations and Inequalities in One Variable Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-12.1 Questions Edition 2023-2024

### Linear Equations and Inequalities in One Variable Exe-12.1

ML Aggarwal Class 8 ICSE Maths Solutions

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Solve the following equations (1 to 12):

#### Question 1.

(i) 5x – 3 = 3x – 5
(ii) 3x – 7 = 3(5 – x)

(i) 5x – 3 = 3x – 5

5x – 3x = – 5 + 3

2x = – 2

We get,

x = – 2 / 2

x = – 1

(ii) 3x – 7 = 3(5 – x)

3x – 7 = 15 – 3x

3x + 3x = 15 + 7

We get,

6x = 22

x = 22 / 6

x = 11 / 3

#### Question 2.

(i) 4(2x + 1) = 3(x – 1) + 7
(ii) 3(2p – 1) = 5 – (3p – 2)

(i) 4(2x + 1) = 3(x – 1) + 7

⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 0
⇒ x = 0

(ii) 3(2p – 1) = 5 – (3p – 2)

6p – 3 = 5 – 3p + 2

6p + 3p = 5 + 3 + 2

We get,

9p = 10

p = 10 / 9

p = 1(1/9)

#### Question 3.

(i) 5y – 2[y – 3(y – 5)] = 6
(ii) 0.3(6 – x) = 0.4(x + 8)

(i) 5y – 2[y – 3(y – 5)] = 6
⇒ 5y – 2[y – 3y + 15] = 6
⇒ 5y – 2[-2y + 15] = 6
⇒ 5y + 4y – 30 = 6
⇒ 9y = 6 + 30
⇒ 9y = 36
⇒ y = 36/9
⇒ y = 4

(ii) 0.3 (6 – x) = 0.4 (x + 8)
⇒ 1.8 – 0.3x = 0.4x + 3.2
⇒ 0.3x – 0.4x = 3.2 – 1.8
⇒ -0.7x = 1.4
⇒ x = -1.4/0.7
⇒ x = -14/7
⇒ x = -2

Linear Equations and Inequalities in One Variable Exe-12.1

### ML Aggarwal Class 8 ICSE Maths Solutions

Page-207

#### Question 4.

(i) (x – 1) / 3 = {(x + 2) / 6} + 3

(ii) (x + 7) / 3 = 1 + {(3x – 2) / 5}

(i) {(x – 1) / 3} = {(x + 2) / 6} + 3

{(x – 1) / 3} – {(x + 2) / 6} = 3

On further calculation, we get,

{2 (x – 1) – 1(x + 2)} / 6 = 3

(2x – 2 – 1x – 2) / 6 = 3

(x – 4) / 6 = 3

We get,

x – 4 = 6 × 3

x – 4 = 18

x = 18 + 4

We get,

x = 22

(ii) (x + 7) / 3 = 1 + {(3x – 2) / 5}

{(x + 7) / 3} – {(3x – 2) / 5} = 1

{5 (x + 7) – 3 (3x – 2)} / 15 = 1

{(5x + 35) – (9x – 6)} / 15 = 1

(5x – 9x + 35 + 6) / 15 = 1

(- 4x + 41) / 15 = 1

– 4x + 41 = 15

– 4x = 15 – 41

– 4x = – 26

x = – 26 / – 4

x = 13 / 2

x = 6(1/2)

#### Question 5.

(i) {(y + 1) / 3} – {(y – 1) / 2} = (1 + 2y) / 3

(ii) (p / 3) + (p / 4) = 55 – {(p + 40) / 5}

(i) {(y + 1) / 3} – {(y – 1) / 2} = (1 + 2y) / 3

{2 (y + 1) – 3 (y – 1)} / 6 = (1 + 2y) / 3

(2y + 2 – 3y + 3) / 6 = (1 + 2y) / 3

(- y + 5) / 6 = (1 + 2y) / 3

3 (- y + 5) = 6 (1 + 2y)

– 3y + 15 = 6 + 12y

– 3y – 12y = 6 – 15

– 15y = – 9

y = – 9 / – 15

y = 3 / 5

(ii) (p / 3) + (p / 4) = 55 – {(p + 40) / 5}

(p / 3) + (p / 4) + {(p + 40)} / 5 = 55

Here, L.C.M. of 3, 4, 5 is 60

{20p +15p + 12(p + 40)} / 60 = 55

(20p + 15p + 12p + 480) / 60 = 55

(47p + 480) / 6 = 55

47p + 480 = 55 × 60

47p + 480 = 3300

47p = 3300 – 480

47p = 2820

p = 2820 / 47

We get,

p = 60

#### Question 6.

(i) n – {(n – 1) / 2} = 1 – {(n – 2) / 3}

(ii) {(3t – 2) / 3} + {(2t + 3) / 2} = t + (7 / 6)

(i) n – {(n – 1) / 2} = 1 – {(n – 2) / 3}

(2n – n + 1) / 2 = (3 – n + 2) / 3

(n + 1) / 2 = (5 – n) / 3

3 (n + 1) = 2 (5 – n)

3n + 3 = 10 – 2n

3n + 2n = 10 – 3

5n = 7

n = 7 / 5

We get,

n = 1(2/5)

(ii) {(3t – 2) / 3} + {(2t + 3) / 2} = t + (7 / 6)

{2 (3t – 2) + 3 (2t + 3)} / 6 = (6t + 7) / 6

(6t – 4) + (6t + 9) = 6t + 7

6t + 6t + 9 – 4 = 6t + 7

12t + 5 = 6t + 7

12t – 6t = 7 – 5

6t = 2

t = 2 / 6

t = 1 / 3

#### Question 7.

(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15

(i) 4(3x + 2) – 5(6x – 1) = 2(x – 8) – 6(7x – 4)
⇒ 12x + 8 – 30x + 5 = 2x – 16 – 42x + 24
⇒ -18x + 13 = -40x + 8
⇒ -18x + 40x = 8 – 13
⇒ 22x = -5 ⇒ x = -5/22

(ii) 3(5x + 7) + 5(2x – 11) = 3(8x – 5) – 15
⇒ 15x + 21 + 10x – 55 = 24x – 15 – 15
⇒ 25x – 34 = 24x – 30
⇒ 25x – 24x = -30 + 34
⇒ x = 4

(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.1 Class 8)

#### Question 8.

(i) (3 – 2x) / (2x + 5) = – (3 / 11)

(ii) (5p + 2) / (8 – 2p) = 7 / 6

(i) (3 – 2x) / (2x + 5) = – (3 / 11)

11 (3 – 2x) = – 3 (2x + 5)

On simplification, we get,

33 – 22x = – 6x – 15

– 22x + 6x = – 15 – 33

– 16x = – 48

x = 48 / 16

x = 3

(ii) (5p + 2) / (8 – 2p) = 7 / 6

6 (5p + 2) = 7 (8 – 2p)

On further calculation, we get,

30p + 12 = 56 – 14p

30p + 14p = 56 – 12

44p = 44

p = 44 / 44

p = 1

#### Question 9.

(i) 5 / x = 7 / (x – 4)

(ii) 4 / (2x + 3) = 5 / (x + 4)

(i) 5 / x = 7 / (x – 4)

5 (x – 4) = 7x

5x – 20 = 7x

5x – 7x = 20

– 2x = 20

x = 20 / – 2

x = (- 20 / 2)

x = – 10

(ii) 4 / (2x + 3) = 5 / (x + 4)

4 (x + 4) = 5 (2x + 3)

4x + 16 = 10x + 15

4x – 10x = 15 – 16

– 6x = – 1

x = – 1 / – 6

x = 1 / 6

#### Question 10.

(i) {(2x + 5) / 2} – {5x / (x – 1)} = x

(ii) 1 / 5 {(1 / 3x) – 5} = 1 / 3 {3 – (1 / x)}

(i) {(2x + 5) / 2} – {5x / (x – 1)} = x

{(2x + 5) (x – 1) – (5x) (2)} / {2 (x – 1)} = x

{2x (x – 1) + 5 (x – 1) – 10x} / (2x – 2) = x

(2x2 – 2x + 5x – 5 – 10x) / (2x – 2) = x

(2x2 – 7x – 5) / (2x – 2) = x

2x2 – 7x – 5 = x (2x – 2)

2x2 – 7x – 5 = 2x2 – 2x

– 7x – 5 = – 2x

– 7x + 2x = 5

– 5x = 5

x = 5 / – 5

x = – 1

(ii) 1 / 5 {(1 / 3x) – 5} = 1 / 3 {3 – (1 / x)}

1 / 5 [{1 – 5 (3x)} / 3x] = 1 / 3 [{(3x – 1)} / x]

1 / 5 {(1 – 15x) / 3x} = 1 / 3 {(3x – 1) / x}

(1 – 15x) / 15x = (3x – 1) / 3x

3x (1 – 15x) = 15x (3x – 1)

3 (1 – 15x) = 15 (3x – 1)

3 – 45x = 45x – 15

– 45x – 45x = – 15 – 3

– 90x = – 18

x = 18 / 90

x = 1 / 5

#### Question 11.

(i) {(2x – 3) / (2x – 1)} = {(3x – 1) / (3x + 1)}

(ii) {(2y + 3) / (3y + 2)} = {(4y + 5) / (6y + 7)}

(i) {(2x – 3) / (2x – 1)} = {(3x – 1) / (3x + 1)}

(2x – 3) (3x + 1) = (3x – 1) (2x – 1)

6x2 + 2x – 9x – 3 = 6x2 – 3x – 2x + 1

6x2 – 7x – 3 = 6x2 – 5x + 1

6x2 – 7x – 6x2 + 5x = 1 + 3

– 7x + 5x = 4

– 2x = 4

x = 4 / – 2

x = – 2

(ii) {(2y + 3) / (3y + 2)} = {(4y + 5) / (6y + 7)}

(2y + 3) (6y + 7) = (4y + 5) (3y + 2)

12y2 + 14y + 18y + 21 = 12y2 + 8y + 15y + 10

32y + 21 = 23y + 10

32y – 23y = 10 – 21

9y = – 11

y = – 11 / 9

(ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.1 Class 8)

#### Question 12. If x = p + 1, find the value of p from the equation (1 / 2) (5x – 30) – (1/ 3) (1 + 7p) = 1 / 4

x = p + 1 ….. (1)

(1 / 2) (5x – 30) – (1 / 3) (1 + 7p) = 1 / 4 ………. (2)

Substituting the value of x from (1) in (2), we get,

(1 / 2) {5 (p + 1) – 30} – (1 / 3) (1 + 7p) = 1 / 4

1 / 2 (5p + 5 – 30) – 1 / 3 (1 + 7p) = 1 / 4

On further calculation, we get,

1 / 2 (5p – 25) – 1 / 3 (1 + 7p) = 1 / 4

(5p – 25) / 2 – (1 + 7p) / 3 = 1 / 4

{3 (5p – 25) – 2 (1 + 7p)} / 6 = 1 / 4

(15p – 75 – 2 – 14p) / 6 = 1 / 4

(p – 77) / 6 = 1 / 4

On simplification, we get,

4 (p – 77) = 6 (1)

4p – 308 = 6

4p = 6 + 308

4p = 314

p = 314 / 4

We get,

p = 157 / 2

p = 78(1/2)

#### Question 13. Solve {(x + 3) / 3} – {(x – 2) / 2} = 1, Hence find p if (1 / x) + P = 1

{(x + 3) / 3} – {(x – 2) / 2} = 1

{2 (x + 3) – 3(x – 2)} / 6 = 1

(2x + 6 – 3x + 6) / 6 = 1

(- x + 12) / 6 = 1

– x + 12 = 1 × 6

– x + 12 = 6

– x = 6 – 12

– x = – 6

x = 6

(1 / x) + P = 1

x = 6,

(1 / 6) + P = 1

(1 + 6P) / 6 = 1

(1 + 6P) = 6

6P = 6 – 1

6P = 5

P = 5 / 6

— End of Linear Equations and Inequalities in One Variable Exe-12.1 Class 8 ICSE Maths Solutions :–