# 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, {â€¦, -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 areâ€¦-2,-1, 0, 1, 2, 3, 4
Hence, {â€¦, -2,-1, 0, 1, 2, 3, 4} is the solution set.

### Question 2. Solve â€“ 12x > 30, when

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

Solution:

(i) Given, â€“ 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, â€“ 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 areâ€¦, -6.-5, -4, â€“ 3
Thus, the solution of â€“ 12x > 30 is â€¦,-6,-5, -4, -3, when x is an integer.
Therefore, the solution set is {â€¦,-6, -5, -4, -3}

### Question 3. Solve 5x â€“ 3 < 7, when

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

Solution:

(i) Given 5x â€“ 3 < 7
Now let us add 3 both side we get,
5x â€“ 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 areâ€¦, -3, -2, -1, 0, 1.
Thus, the  solution of 5x â€“ 3 < 7 is â€¦, -3,-2, -1, 0, 1, when x is an integer.
Therefore, the solution set is {â€¦, -3.-2, -1, 0, 1}

(ii) Given that, 5x â€“ 3 < 7
Now let us add 3 both side we get,
5x â€“ 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 âˆˆ  R) number then
It is clear that the solutions of 5x â€“ 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 âˆˆ (-âˆž, 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 â€“ 8 > 2 â€“ 8
The above equation becomes,
3x > â€“ 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,â€¦
Thus, the solution of 3x + 8 > 2is -1, 0, 1, 2, 3,â€¦ when x is an integer.
Hence, the solution set is {-1, 0, 1, 2, 3,â€¦}

(ii) Given3x + 8 > 2
Now let us subtract 8 from both sides we get,
3x + 8 â€“ 8 > 2 â€“ 8
The above equation becomes,
3x > â€“ 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 âˆˆ (-2, âˆž)

### 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 â€“ 7 < 5x + 7 â€“ 7
The above equation becomes,
4x â€“ 4 < 5
Again let us subtract 4x from both the sides, we get
4x â€“ 4 â€“ 4x < 5x â€“ 4x
x > â€“ 4

Thus, solution of the given equation is defined by all the real numbers greater than -4.
Required solution set is (-4, âˆž)

### Question 6. 3x â€“ 7 > 5x â€“ 1

Solution:

Given, 3x â€“ 7 > 5x â€“ 1
Now let us add 7 to both the sides, we get
3x â€“ 7 +7 > 5x â€“ 1 + 7
3x > 5x + 6
Again let us subtract 5x from both the sides,
3x â€“ 5x > 5x + 6 â€“ 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 (-âˆž, -3)

### Question 7. 3(x â€“ 1) â‰¤ 2 (x â€“ 3)

Solution:

Given, 3(x â€“ 1) â‰¤ 2 (x â€“ 3)
After multiplying, the above inequation can be written as
3x â€“ 3 â‰¤ 2x â€“ 6
Now let us add 3 to both the sides, we get
3x â€“ 3+ 3 â‰¤ 2x â€“ 6+ 3
3x â‰¤ 2x â€“ 3
Again let us subtract 2x from both the sides,
3x â€“ 2x â‰¤ 2x â€“ 3 â€“ 2x
x â‰¤ -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 (-âˆž, -3]

### Question 8. 3 (2 â€“ x) â‰¥ 2 (1 â€“ x)

Solution:

Given, 3 (2 â€“ x) â‰¥ 2 (1 â€“ x)
After multiplying, the above equation can be written as
6 â€“ 3x â‰¥ 2 â€“ 2x
Now let us add 2x to both the sides,
6 â€“ 3x + 2x â‰¥ 2 â€“ 2x + 2x
6 â€“ x â‰¥ 2
Again let us subtract 6 from both the sides, we get
6 â€“ x â€“ 6 â‰¥ 2 â€“ 6
â€“ x â‰¥ â€“ 4
Now multiplying the equation by negative sign we get
x â‰¤ 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 (- âˆž, 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<6

Thus, the solution of the given equation is defined by all the real numbers less than 6.
Hence, the solution set is (-âˆž, 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 â€“ 6.
Hence, the required solution set is (-âˆž, -6)

### Question 11. 3(x â€“ 2)/5 â‰¤ 5 (2 â€“ x)/3

Solution:

Given, 3(x â€“ 2)/5 â‰¤ 5 (2 â€“ x)/3
Now by cross â€“ multiplying the denominators, we get
9(x- 2) â‰¤ 25 (2 â€“ x)
9x â€“ 18 â‰¤ 50 â€“ 25x
Now let us add 25x both the sides,
9x â€“ 18 + 25x â‰¤ 50 â€“ 25x + 25x
34x â€“ 18 â‰¤ 50
Let us add 25x both the sides,
34x â€“ 18 + 18 â‰¤ 50 + 18
34x â‰¤ 68
Dividing both the sides by 34,
34x/34 â‰¤ 68/34
x â‰¤ 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 (-âˆž, 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, (-âˆž, 120] is the required solution set.

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