ML Aggarwal Linear Equations and Inequalities in One Variable Exe-12.1 Class 8 ICSE Ch-12 Maths Solutions
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
Page-206
Solve the following equations (1 to 12):
Question 1.
(i) 5x – 3 = 3x – 5
(ii) 3x – 7 = 3(5 – x)
Answer:
(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)
Answer:
(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)
Answer:
(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}
Answer:
(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}
Answer:
(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)
Answer:
(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
Answer:
(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
Answer:
(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)
Answer:
(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)}
Answer:
(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)}
Answer:
(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
Answer:
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
Answer:
{(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 :–
Return to : – ML Aggarwal Maths Solutions for ICSE Class -8
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