# Linear Inequations ML Aggarwal Solutions ICSE Maths Class-10

Linear Inequations ML Aggarwal Solutions ICSE Maths Class-10 Chapter-4 . We Provide Step by Step Answer of Exercise-4 Linear Inequations , with MCQs and Chapter-Test Questions  / Problems related Exercise-4 Linear Inequations   for ICSE Class-10 APC Understanding Mathematics  .  Visit official Website CISCE  for detail information about ICSE Board Class-10.

## Linear Inequations ML Aggarwal Solutions ICSE Maths Class-10

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### How to Solve Linear Inequations Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-4 Linear Inequations of ML Aggarwal Solution. Read the Chapter Carefully then solve all example of your text book. For more practice on Linear Inequations related problems /Questions / Exercise try to solve Linear Inequations exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE)  / Concise Selina Publications Mathematics. Get the formula of Linear Inequations for ICSE Class 10 Maths  to understand the topic more clearly in effective way.

### ML Solution for ICSE Maths Class 10th Linear Inequations Chapter 4

Exercise 4

#### Question 1

Solve the inequation 3x -11 < 3 where x ∈ {1, 2, 3,……, 10}. Also represent its solution on a number line

3x – 11 < 3 => 3x < 3 + 11 => 3x < 14 x < $\\ \frac { 14 }{ 3 }$
But x ∈ 6 {1, 2, 3, ……., 10}
Solution set is (1, 2, 3, 4}. Solution set on number line

#### Question 2

Solve 2(x – 3)< 1, x ∈ {1, 2, 3, …. 10}

2(x – 3) < 1 => x – 3 < $\\ \frac { 1 }{ 2 }$ => x < $\\ \frac { 1 }{ 2 }$ + 3 => x < $3 \frac { 1 }{ 2 }$
But x ∈ {1, 2, 3 …..10}
Solution set = {1, 2, 3} Ans.

#### Question 3

Solve : 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.

5 – 4x > 2 – 3x
– 4x + 3x > 2 – 5
=> – x > – 3
=> x < 3
x ∈ w,
solution set {0, 1, 2}
Solution set on Number Line :

#### Question 4

List the solution set of 30 – 4 (2.x – 1) < 30, given that x is a positive integer.

30 – 4 (2x – 1) < 30
30 – 8x + 4 < 30
– 8x < 30 – 30 – 4
– 8x < – 4 x > $\\ \frac { -4 }{ -8 }$
=> x > $\\ \frac { 1 }{ 2 }$
x is a positive integer
x = {1, 2, 3, 4…..} Ans.

#### Question 5

Solve : 2 (x – 2) < 3x – 2, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3} .

2(x – 2) < 3x – 2
=> 2x – 4 < 3x – 2
=> 2x – 3x < – 2 + 4
=> – x < 2
=> x > – 2
Solution set = { – 1, 0, 1, 2, 3} Ans.

#### Question 6

If x is a negative integer, find the solution set of $\\ \frac { 2 }{ 3 }$+$\\ \frac { 1 }{ 3 }$ (x + 1) > 0.

$\\ \frac { 2 }{ 3 }$+$\\ \frac { 1 }{ 3 }$ x + $\\ \frac { 1 }{ 3 }$ > 0
=> $\\ \frac { 1 }{ 3 }$ x + 1 > 0
=> $\\ \frac { 1 }{ 3 }$ x > – 1

x is a negative integer
Solution set = {- 2, – 1} Ans.

#### Question 7

Solve: $\\ \frac { 2x-3 }{ 4 }$$\\ \frac { 1 }{ 2 }$, x ∈ {0, 1, 2,…,8}

$\\ \frac { 2x-3 }{ 4 }$$\\ \frac { 1 }{ 2 }$
=> 2x – 3 ≥ $\\ \frac { 4 }{ 2 }$

#### Question 8

Solve x – 3 (2 + x) > 2 (3x – 1), x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}. Also represent its solution on the number line.

x – 3 (2 + x) > 2 (3x – 1)
=> x – 6 – 3x > 6x – 2
=> x – 3x – 6x > – 2 + 6
=> – 8x > 4
=> x < $\\ \frac { -4 }{ 8 }$ => x < $- \frac { 1 }{ 2 }$
x ∈ { – 3, – 2, – 1, 0, 1, 2}
.’. Solution set = { – 3, – 2, – 1}
Solution set on Number Line :

#### Question 9

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.

x – 3 < 2x – 1
x – 2x < – 1 + 3 => – x < 2 x > – 2
But x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4, 5, 6, 7, 9} Ans.

#### Question 10

Given A = {x : x ∈ I, – 4 ≤ x ≤ 4}, solve 2x – 3 < 3 where x has the domain A Graph the solution set on the number line.

2x – 3 < 3 => 2x < 3 + 3 => 2x < 6 => x < 3
But x has the domain A = {x : x ∈ I – 4 ≤ x ≤ 4}
Solution set = { – 4, – 3, – 2, – 1, 0, 1, 2}
Solution set on Number line :

#### Question 11

List the solution set of the inequation
$\\ \frac { 1 }{ 2 }$ + 8x > 5x $- \frac { 3 }{ 2 }$, x ∈ Z

$\\ \frac { 1 }{ 2 }$ +8x > 5x $- \frac { 3 }{ 2 }$

#### Question 12

List the solution set of $\\ \frac { 11-2x }{ 5 }$ ≥ $\\ \frac { 9-3x }{ 8 }$ + $\\ \frac { 3 }{ 4 }$,
x ∈ N

$\\ \frac { 11-2x }{ 5 }$ ≥ $\\ \frac { 9-3x }{ 8 }$ + $\\ \frac { 3 }{ 4 }$
=> 88 – 16x ≥ 45 – 15x + 30
(L.C.M. of 8, 5, 4 = 40}
=> – 16x + 15x ≥ 45 + 30 – 88
=> – x ≥ – 13
=>x ≤ 13
x ≤ N.
Solution set = {1, 2, 3, 4, 5, .. , 13} Ans.

Question 13

Find the values of x, which satisfy the inequation : $-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 }$, x ∈ N.
Graph the solution set on the number line. (2001)

$-2\le \frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le 1\frac { 5 }{ 6 }$, x ∈ N

#### Question 14

If x ∈ W, find the solution set of
$\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1$
Also graph the solution set on the number line, if possible.

$\frac { 3 }{ 5 } x-\frac { 2x-1 }{ 3 } >1$
9x – (10x – 5) > 15 (L.C.M. of 5, 3 = 15)
=> 9x – 10x + 5 > 15
=> – x > 15 – 5
=> – x > 10
=> x < – 10
But x ∈ W
Solution set = Φ
Hence it can’t be represented on number line.

#### Question 15

Solve:
(i)$\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6$ where x is a positive odd integer.
(ii)$\frac { 2x+3 }{ 3 } \ge \frac { 3x-1 }{ 4 }$ where x is positive even integer.

(i) $\frac { x }{ 2 } +5\le \frac { x }{ 3 } +6$

#### Question 16

Given that x ∈ I, solve the inequation and graph the solution on the number line :
$3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2$ (2004)

$3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 }$ and $3\ge \frac { x-4 }{ 2 } +\frac { x }{ 3 } \ge 2$

#### Question 17

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1 < x + 4.

-3 < 2x – 1 < x + 4.
=> – 3 < 2x – 1 and 2x – 1 < x + 4
=> – 2x < – 1 + 3 and 2x – x < 4 + 1
=> – 2x < 2 and x < 5
=> – x < 1
=> x > – 1
– 1 < x < 5
x ∈ {1, 2, 3, 4, 5, 6, 7, 9}
Solution set = {1, 2, 3, 4} Ans.

#### Question 18

Solve : 1 ≥ 15 – 7x > 2x – 27, x ∈ N

1 ≥ 15 – 7x > 2x – 27
1 ≥ 15 – 7x and 15 – 7x > 2x – 27

#### Question 19

If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.

2 + 4x < 2x – 5 ≤ 3x
2 + 4x < 2x – 5 and 2x – 5 ≤ 3x => 4x – 2x < – 5 – 2 ,and 2x – 3x ≤ 5

#### Question 20

Solve the inequation = 12 + $1 \frac { 5 }{ 6 } x$ ≤ 5 + 3x, x ∈ R. Represent the solution on a number line. (1999)

12 + $1 \frac { 5 }{ 6 } x$ ≤ 5 + 3x

#### Question 21

Solve : $\\ \frac { 4x-10 }{ 3 }$$\\ \frac { 5x-7 }{ 2 }$ x ∈ R and represent the solution set on the number line.

$\\ \frac { 4x-10 }{ 3 }$$\\ \frac { 5x-7 }{ 2 }$
=> 8x – 20 ≤ 15x – 21
(L.C.M. of 3, 2 = 6)

#### Question 22

Solve $\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 }$ > 1, x ∈ R and represent the solution set on the number line.

$\frac { 3x }{ 5 } -\frac { 2x-1 }{ 3 }$ > 1
=> 9x – (10x – 5) > 15
=> 9x – 10x + 5 > 15
=> – x > 15 – 5
=> – x > 10
=> x < – 10
x ∈ R.
.’. Solution set = {x : x ∈R, x < – 10}
Solution set on the number line

#### Question 23

Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line. (2000)

– 3 ≤ 3 – 2x < 9
– 3 ≤ 3 – 2x and 3 – 2x < 9
2x ≤ 3 + 3 and – 2x < 9 – 3
2x ≤ 6 and – 2x < 6 => x ≤ 3 and – x < 3 => x ≤ – 3 and – 3 < x
– 3 < x ≤ 3.
Solution set= {x : x ∈ R, – 3 < x ≤ 3)
Solution on number line

#### Question 24

Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line. (2003)

2 ≤ 2x – 3 ≤ 5 .or 2 ≤ 2x – 3 and 2x – 3 ≤ 5 or 2 + 3 ≤ 2x and 2x ≤ 5 + 3
5 ≤ 2x and 2x ≤ 8.

#### Question 25

Given that x ∈ R, solve the following inequation and graph the solution on the number line: – 1 ≤ 3 + 4x < 23. (2006)

We have
– 1 ≤ 3 + 4x < 23 => – 1 – 3 ≤ 4x < 23 – 3 => – 4 ≤ 4x < 20 => – 1 ≤ x < 5, x ∈ R
Solution Set = { – 1 ≤ x < 5; x ∈ R}

#### Question 26

Solve tlie following inequation and graph the solution on the number line. (2007)
$-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 }$ x∈R

Given $-2\frac { 2 }{ 3 } \le x+\frac { 1 }{ 3 } <3+\frac { 1 }{ 3 }$ x∈R
$-\frac { 8 }{ 3 } \le x+\frac { 1 }{ 3 } <\frac { 10 }{ 3 }$
Multiplying by 3, L.C.M. of fractions, we get

#### Question 27

Solve the following inequation and represent the solution set on the number line :
$-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R$

$-3<-\frac { 1 }{ 2 } -\frac { 2x }{ 3 } \le \frac { 5 }{ 6 } ,x\in R$

#### Question 28

Solve $\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R$. Also graph the solution set on the number line

$\frac { 2x+1 }{ 2 } +2(3-x)\ge 7,x\in R$

#### Question 29

Solving the following inequation, write the solution set and represent it on the number line. – 3(x – 7)≥15 – 7x > $\\ \frac { x+1 }{ 3 }$, n ∈R

– 3(x – 7)≥15 – 7x > $\\ \frac { x+1 }{ 3 }$, n ∈R

#### Question 30

Solving the following inequation , write the solution set and represent it on the real number line. -2+10x ≤13x +10<24+10x, x ∈ Z.

#### Question 31

Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line. (2011)

2x – 5 ≤ 5x + 4 < 11 2x – 5 ≤ 5x + 4
=> 2x – 5 – 4 ≤ 5x and 5x + 4 < 11
=> 2x – 9 ≤ 5x and 5x < 11 – 4
and 5x < 7
=> 2x – 5x ≤ 9 and x < $\\ \frac { 7 }{ 5 }$
=> 3x > – 9 and x< 1.4
=> x > – 3

#### Question 32

If x ∈ I, A is the solution set of 2 (x – 1) < 3 x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩B.

2 (x – 1) < 3 x – 1
2x – 2 < 3x – 1
2x – 3x < – 1 + 2 => – x < 1 x > – 1
Solution set A = {0, 1, 2, 3, ..,.}
4x – 3 ≤ 8 + x
4x – x ≤ 8 + 3
=> 3x ≤ 11
=> x ≤ $\\ \frac { 11 }{ 3 }$
Solution set B = {3, 2, 1, 0, – 1…}
A ∩ B = {0, 1, 2, 3} Ans.

#### Question 33

If P is the solution set of – 3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find
(i) P ∩ Q
(ii) Q – P.

(i) – 3 x + 4 < 2 x – 3
– 3x – 2x < – 3 – 4 => – 5x < – 7

#### Question 34

A = {x : 11x – 5 > 7x + 3, x ∈R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}
Find the range of set A ∩ B and represent it on a number line

A = {x : 11x – 5 > 7x + 3, x ∈R}
B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}
Now, A = 11x – 5 > 7x + 3
=> 11x – 7x > 3 + 5
=> 4x > 8
=>x > 2, x ∈ R

#### Question 35

Given: P {x : 5 < 2x – 1 ≤ 11, x∈R)
Q{x : – 1 ≤ 3 + 4x < 23, x∈I) where
R = (real numbers), I = (integers)
Represent P and Q on number line. Write down the elements of P ∩ Q. (1996)

P= {x : 5 < 2x – 1 ≤ 11}
5 < 2x – 1 ≤ 11

#### Question 36

If x ∈ I, find the smallest value of x which satisfies the inequation $2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2$

$2x+\frac { 5 }{ 2 } >\frac { 5x }{ 3 } +2$
=>$2x-\frac { 5x }{ 3 } >2-\frac { 5 }{ 2 }$
=>12x – 10x > 12 – 15
=> 2x > – 3
=>$x>-\frac { 3 }{ 2 }$
Smallest value of x = – 1 Ans.

#### Question 37

Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when
(i) x ∈ I
(ii) x ∈ W
(iii) x ∈ N.

20 – 5 x < 5 (x + 8)
⇒ 20 – 5x < 5x + 40
⇒ – 5x – 5x < 40 – 20
⇒ – 10x < 20
⇒ – x < 2
⇒ x > – 2
(i) When x ∈ I, then smallest value = – 1.
(ii) When x ∈ W, then smallest value = 0.
(iii) When x ∈ N, then smallest value = 1. Ans.

#### Question 38

Solve the following inequation and represent the solution set on the number line :
$4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R$

We have
$4x-19<\frac { 3x }{ 5 } -2\le -\frac { 2 }{ 5 } +x,x\in R$
Hence, solution set is {x : -4 < x < 5, x ∈ R}
The solution set is represented on the number line as below.

#### Question 39

Solve the given inequation and graph the solution on the number line :
2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.

2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.
(a) 2y – 3 < y + 1
⇒ 2y – y < 1 + 3
⇒ y < 4
⇒ 4 > y ….(i)
(b) y + 1 ≤ 4y + 7

#### Question 40

Solve the inequation and represent the solution set on the number line.
$-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I$

Given : $-3+x\le \frac { 8x }{ 3 } +2\le \frac { 14 }{ 3 } +2x,Where\quad x\in I$

#### Question 41

Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.

Let the greatest integer = x
According to the condition,
2x + 7 > 3x
⇒ 2x – 3x > – 7
⇒ – x > – 7
⇒ x < 7
Value of x which is greatest = 6 Ans.

#### Question 42

One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.

Let the length of the shortest pole = x metre
Length of pole which is burried in mud = $\\ \frac { x }{ 3 }$
Length of pole which is in the water = $\\ \frac { x }{ 6 }$

### MULTIPLE  CHOICE QUESTIONS

Choose the correct answer from the given four options (1 to 5) :

#### Question 1

If x ∈ { – 3, – 1, 0, 1, 3, 5}, then the solution set of the inequation 3x – 2 ≤ 8 is
(a) { – 3, – 1, 1, 3}
(b) { – 3, – 1, 0, 1, 3}
(c) { – 3, – 2, – 1, 0, 1, 2, 3}
(d) { – 3, – 2, – 1, 0, 1, 2}

x ∈ { -3, -1, 0, 1, 3, 5}
3x – 2 ≤ 8
….⇒ 3x ≤ 8 + 2
and ⇒ 3x ≤ 10
so ⇒ x ≤ $\\ \frac { 10 }{ 3 }$
therefore ⇒ x < $3 \frac { 1 }{ 3 }$
Solution set = { -3, -1, 0, 1, 3} (b)

#### Question 2

If x ∈ W, then the solution set of the inequation 3x + 11 ≥ x + 8 is
(a) { – 2, – 1, 0, 1, 2, …}
(b) { – 1, 0, 1, 2, …}
(c) {0, 1, 2, 3, …}
(d) {x : x∈R,x≥$- \frac { 3 }{ 2 }$}

x ∈ W
3x + 11 ≥ x + 8
⇒ 3x – x ≥ 8 – 11

#### Question 3

If x ∈ W, then the solution set of the inequation 5 – 4x ≤ 2 – 3x is
(a) {…, – 2, – 1, 0, 1, 2, 3}
(b) {1, 2, 3}
(c) {0, 1, 2, 3}
(d) {x : x ∈ R, x ≤ 3}

x ∈ W
5 – 4x < 2 – 3x
⇒ 5 – 2 ≤ 3x + 4x
⇒ 3 ≤ x
Solution set = {0, 1, 2, 3,} (c)

#### Question 4

If x ∈ I, then the solution set of the inequation 1 < 3x + 5 ≤ 11 is
(a) { – 1, 0, 1, 2}
(b) { – 2, – 1, 0, 1}
(c) { – 1, 0, 1}
(d) {x : x ∈ R, $- \frac { 4 }{ 3 }$ < x ≤ 2}

x ∈ I
1 < 3x + 5 ≤ 11
⇒ 1 < 3x + 5
⇒ 1 – 5 < 3x

#### Question 5

If x ∈ R, the solution set of 6 ≤ – 3 (2x – 4) < 12 is
(a) {x : x ∈ R, 0 < x ≤ 1}
(b) {x : x ∈ R, 0 ≤ x < 1}
(c) {0, 1}
(d) none of these

x ∈ R
6 ≤ – 3(2x – 4) < 12
⇒ 6 ≤ – 3(2x – 4)
⇒ 6 ≤ – 6x + 12

### Solution of ML Aggarwal  Linear Inequations Chapter 4 for ICSE Maths Class 10

CHAPTER TEST

#### Question 1

Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.

5x – 2 < 3(3 – x)
⇒ 5x – 2 ≤ 9 – 3x
⇒ 5x + 3x ≤ 9 + 2

#### Question 2

Solve the inequations :
6x – 5 < 3x + 4, x ∈ I.

6x – 5 < 3x + 4
6x – 3x < 4 + 5
⇒ 3x <9
⇒ x < 3
x ∈ I
Solution Set = { -1, -2, 2, 1, 0….. }

#### Question 3

Find the solution set of the inequation
x + 5 < 2 x + 3 ; x ∈ R
Graph the solution set on the number line.

x + 5 ≤ 2x + 3
x – 2x ≤ 3 – 5
⇒ -x ≤ -2
⇒ x ≥ 2

#### Question 4

If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.

-1 < 3 – 2x ≤ 7
-1 < 3 – 2x and 3 – 2x ≤ 7
⇒ 2x < 3 + 1 and – 2x ≤ 7 – 3
⇒ 2x < 4 and -2x ≤ 4
⇒ x < 2 and -x ≤ 2
and x ≥ -2 or -2 ≤ x
x ∈ R
Solution set -2 ≤ x < 2
Solution set on number line

#### Question 5

Solve the inequation :

$\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } ,x\in R$

$\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 }$
$\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le \frac { 8 }{ 5 } +\frac { 3x-1 }{ 7 }$

#### Question 6

Find the range of values of a, which satisfy 7 ≤ – 4x + 2 < 12, x ∈ R. Graph these values of a on the real number line.

7 < – 4x + 2 < 12
7 < – 4x + 2 and – 4x + 2 < 12

#### Question 7

If x∈R, solve $2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x$

$2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x$
$2x-3\ge x+\frac { 1-x }{ 3 }$ and $x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x$

#### Question 8

Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.

Let the positive integer = x
According to the problem,
5a – 6 < 4x
⇒ 5a – 4x < 6
⇒ x < 6
Solution set = {x : x < 6}
= { 1, 2, 3, 4, 5, 6}

#### Question 9

Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.

Let first least natural number = x
then second number = x + 1
and third number = x + 2

— : End of Linear Inequations Solutions :–

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