# Class 11 NCERT Solutions- Chapter 6 Linear Inequalities – Exercise 6.1 | Set 1

**Question 1. Solve 24x < 100, when**

**(i) x is a natural number. (ii) x is an integer.**

**Solution:**

(i)when x is a natural number.Clearly x>0 because from definition (N =1,2,3,4,5,6…..)

Now we have to divide the inequation by 24 we get x<25/6

But x is a natural number that is the solution will be {1,2,3,4}

Which is less than 25/6 and greater than 0.

Hence, {…, -2,-1, 0, 1, 2, 3, 4} is the solution set.x={1,2,3,4}

(ii)Given 24x<100

Now we will divide the equation by 24 we get x<25/6

but according to question x is an integer then the solution

less than 25/6 are…-2,-1, 0, 1, 2, 3, 4

Hence, {…, -2,-1, 0, 1, 2, 3, 4} is the solution set.

**Question 2. Solve – 12x > 30, when**

**(i) x is a natural number.****(ii) x is an integer.**

**Solution:**

(i) Given, – 12x > 30

Now by dividing the equation by -12 on both sides we get, x < -5/2(as per the rule if we divide by a negative integer then the inequality sign changes)

As per question when x is a natural integer then

It is clear that there is no natural number less than -2/5 since, the result of -2/5 will be negative and x is smaller than the result but the natural number contains only positive number

Therefore, there would be no any solution of the given equation when x is a natural number(x>0).

(ii) Given that, – 12x > 30

Now by dividing the equation by -12 on both sides we get, x < -5/2 (As explained earlier why sign changes)

As per question now x is an integer then

It is clear that the integer number less than -5/2 are…, -6.-5, -4, – 3

Thus, the solution of – 12x > 30 is …,-6,-5, -4, -3, when x is an integer.

Therefore, the solution set is {…,-6, -5, -4, -3}

**Question 3. Solve 5x – 3 < 7, when**

**(i) x is an integer****(ii) x is a real number**

**Solution:**

(i) Given 5x – 3 < 7

Now let us add 3 both side we get,

5x – 3 + 3 < 7 + 3

Above equation becomes

5x < 10

Again let us divide both sides by 5 we get,

5x/5 < 10/5

x < 2

As per question x is an integer then

It is clear that the integer number less than 2 are…, -3, -2, -1, 0, 1.

Thus, the solution of 5x – 3 < 7 is …, -3,-2, -1, 0, 1, when x is an integer.

Therefore, the solution set is {…, -3.-2, -1, 0, 1}

(ii) Given that, 5x – 3 < 7

Now let us add 3 both side we get,

5x – 3 + 3 < 7 + 3

Above equation becomes

5x < 10

Again let us divide both the sides by 5 we get,

5x/5 < 10/5

x < 2

As per question x is a real (x ∈ R) number then

It is clear that the solutions of 5x – 3 < 7 will be given by x < 2 i.e it states that all the real numbers that are less than 2.

Hence, the solution set is x ∈ (-∞, 2)

**Question 4. Solve 3x + 8 > 2, when**

**(i) x is an integer.****(ii) x is a real number.**

**Solution:**

(i) Given, 3x + 8 > 2

Now let us subtract 8 from both the sides we get,

3x + 8 – 8 > 2 – 8

The above equation becomes,

3x > – 6

Again let us divide both the sides by 3 we get,

3x/3 > -6/3

Hence, x > -2

As per question x is an integer then

It is clear that the integer number greater than -2 are -1, 0, 1, 2, 3,…

Thus, the solution of 3x + 8 > 2is -1, 0, 1, 2, 3,… when x is an integer.

Hence, the solution set is {-1, 0, 1, 2, 3,…}

(ii) Given3x + 8 > 2

Now let us subtract 8 from both sides we get,

3x + 8 – 8 > 2 – 8

The above equation becomes,

3x > – 6

Again let us divide both the sides by 3 we get,

3x/3 > -6/3

Hence, x > -2

As per question x is a real number.

It is clear that the solutions of 3x + 8 >2 will be given by x > -2 which states that all the real numbers that are greater than -2.

Therefore, the solution set is x ∈ (-2, ∞)

**Question 5. 4x + 3 < 5x + 7**

**Solution:**

Given, 4x + 3 < 5x + 7

Now let us subtract 7 from both the sides, we get

4x + 3 – 7 < 5x + 7 – 7

The above equation becomes,

4x – 4 < 5

Again let us subtract 4x from both the sides, we get

4x – 4 – 4x < 5x – 4x

x > – 4Thus, solution of the given equation is defined by all the real numbers greater than -4.

Required solution set is (-4, ∞)

**Question 6. 3x – 7 > 5x – 1**

**Solution:**

Given, 3x – 7 > 5x – 1

Now let us add 7 to both the sides, we get

3x – 7 +7 > 5x – 1 + 7

3x > 5x + 6

Again let us subtract 5x from both the sides,

3x – 5x > 5x + 6 – 5x

-2x > 6

Now let us divide both the sides by -2 to simplify we get

-2x/-2 < 6/-2

x < -3

Solutions of the given inequality are defined by all the real numbers less than -3.

Hence, the required solution set is (-∞, -3)

**Question 7. 3(x – 1) ≤ 2 (x – 3)**

**Solution:**

Given, 3(x – 1) ≤ 2 (x – 3)

After multiplying, the above inequation can be written as

3x – 3 ≤ 2x – 6

Now let us add 3 to both the sides, we get

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Again let us subtract 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

x ≤ -3

Therefore, the solutions of the given equation is defined by all the real numbers less than or equal to -3.

Hence, the required solution set is (-∞, -3]

**Question 8. 3 (2 – x) **≥ **2 (1 – x)**

**Solution:**

Given, 3 (2 – x) ≥ 2 (1 – x)

After multiplying, the above equation can be written as

6 – 3x ≥ 2 – 2x

Now let us add 2x to both the sides,

6 – 3x + 2x ≥ 2 – 2x + 2x

6 – x ≥ 2

Again let us subtract 6 from both the sides, we get

6 – x – 6 ≥ 2 – 6

– x ≥ – 4

Now multiplying the equation by negative sign we get

x ≤ 4

Thus, solutions of the given equation is defined by all the real numbers greater than or equal to 4.

Hence, the required solution set is (- ∞, 4]

**Question 9. x + x/2 + x/3 < 11**

**Solution:**

Given, x + x/2 + x/3 < 11

Now taking 6 as the lcm we will simplify the equation,(6x+3x+2x)/6 <11

11x/6<11

Now let us multiply 6 at both the sides

11x<66

Now let us divide the equation by 11 at both the sides we get,

x<6Thus, the solution of the given equation is defined by all the real numbers less than 6.

Hence, the solution set is (-∞, 6)

**Question 10. x/3 > x/2 + 1**

**Solution:**

Given x/3 > x/2 + 1

Firstly, we will move all the terms containing x to the left-hand side we get,

x/3-x/2>1

Now taking 6 as the lcm we get

(2x-3x)/6>1

-x/6>1

Now multiplying 6 at both the sides we get,

-x>6

Now multiplying by -1 at both the ends

x<6

Thus, the solution of the given equation is defined by all the real numbers less than – 6.

Hence, the required solution set is (-∞, -6)

**Question 11. 3(x – 2)/5 **≤** 5 (2 – x)/3**

**Solution:**

Given, 3(x – 2)/5 ≤ 5 (2 – x)/3

Now by cross – multiplying the denominators, we get

9(x- 2) ≤ 25 (2 – x)

9x – 18 ≤ 50 – 25x

Now let us add 25x both the sides,

9x – 18 + 25x ≤ 50 – 25x + 25x

34x – 18 ≤ 50

Let us add 25x both the sides,

34x – 18 + 18 ≤ 50 + 18

34x ≤ 68

Dividing both the sides by 34,

34x/34 ≤ 68/34

x ≤ 2

Thus, the solution of the given equation is defined by all the real numbers less than or equal to 2.

Hence, solution set is (-∞, 2]

**Question 12. 1/2(3x/5+4)>=1/3(x-6)**

**Solution:**

Given, 1/2(3x/5+4)>=1/3(x-6)

Now let us cross multiply,

3(3x/5+4)>=2(x-6)

Now multiply the respective terms at both the sides

9x/5+12>=2x-12

Now subtracting 12 at both the sides

9x/5+12-12 >= 2x-12-12

9x/5>=2x-24

Now multiplying by 5 both the sides,

9x>=10x-120

Now subtracting 10x both the sides.

-x>=-120

Now multiplying with -1 both the sides

x <= 120

Thus, the solutions of the given equation is defined by all the real numbers less than or equal to 120.

Thus, (-∞, 120] is the required solution set.