MENU

Chapter 6 Linear Inequalities Ex 6.1 Solutions

Question - 11 : - 3(x – 2)/5 ≤ 5 (2 – x)/3

Answer - 11 : - Given that,

Now by cross – multiplying the denominators, we get
9(x- 2) ≤ 25 (2 – x)
9x – 18 ≤ 50 – 25x
Now adding 25x both the sides,
9x – 18 + 25x ≤ 50 – 25x + 25x
34x – 18 ≤ 50
Adding 25x both the sides,
34x – 18 + 18 ≤ 50 + 18
34x ≤ 68
Dividing both sides by 34,
34x/34 ≤ 68/34
x ≤ 2
The solutions of the given inequality are defined by all the real numbers less than or equal to 2.
Required solution set is (-∞, 2]

Question - 12 : -

Answer - 12 : -


120 ≥ x
∴ The solutions of the given inequality are defined by all the real numbers less than or equal to 120.
Thus, (-∞, 120] is the required solution set.

Question - 13 : - 2 (2x + 3) – 10 < 6 (x – 2)

Answer - 13 : -

Given that,
2 (2x + 3) – 10 < 6 (x – 2)
By multiplying we get
4x + 6 – 10 < 6x – 12
On simplifying we get
4x – 4 < 6x – 12
– 4 + 12 < 6x – 4x
8 < 2x
4 < x
∴ The solutions of the given inequality are defined by all the real numbers greater than or equal to 4.
Hence the required solution set is (4, -∞)

Question - 14 : - 37 – (3x + 5) ≥ 9x – 8 (x – 3)

Answer - 14 : -

Given that, 37 – (3x + 5) ≥ 9x – 8 (x – 3)
On simplifying we get
= 37 – 3x – 5 ≥ 9x – 8x + 24
= 32 – 3x ≥ x + 24
On rearranging
= 32 – 24 ≥ x + 3x
= 8 ≥ 4x
= 2 ≥ x
All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality
Hence, (-∞, 2] will be the solution for the given inequality

Question - 15 : -

Answer - 15 : -


= 15x < 4 (4x – 1)
= 15x < 16x – 4
= 4 < x
All the real numbers of x which are greater than 4 are the solutions of the given inequality
Hence, (4, ∞) will be the solution for the given inequality

Question - 16 : -

Answer - 16 : -


= 20 (2x – 1) ≥ 3 (19x – 18)
= 40x – 20 ≥ 57x – 54
= – 20 + 54 ≥ 57x – 40x
= 34 ≥ 17x
= 2 ≥ x
∴ All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality
Hence, (-∞, 2] will be the solution for the given inequality
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.

Question - 17 : - 3x – 2 < 2x + 1

Answer - 17 : -

Given,
3x – 2 < 2x + 1
Solving the given inequality, we get
3x – 2 < 2x + 1
= 3x – 2x < 1 + 2
= x < 3
Now, the graphical representation of the solution is as follows:

Question - 18 : - 5x – 3 ≥ 3x – 5

Answer - 18 : -

We have,
5x – 3 ≥ 3x – 5
Solving the given inequality, we get
5x – 3 ≥ 3x – 5
On rearranging we get
= 5x – 3x ≥ -5 + 3
On simplifying
= 2x ≥ -2
Now divide by 2 on both sides we get
= x ≥ -1
The graphical representation of the solution is as follows:

Question - 19 : - 3 (1 – x) < 2 (x + 4)

Answer - 19 : -

Given,
3 (1 – x) < 2 (x + 4)
Solving the given inequality, we get
3 (1 – x) < 2 (x + 4)
Multiplying we get
= 3 – 3x < 2x + 8
On rearranging we get
= 3 – 8 < 2x + 3x
= – 5 < 5x
Now by dividing 5 on both sides we get
-5/5 < 5x/5
= – 1 < x
Now, the graphical representation of the solution is as follows:

Question - 20 : -

Answer - 20 : -


On computing we get

= 15x ≥ 2 (4x –1)

= 15x ≥ 8x -2

= 15x -8x ≥ 8x -2-8x

= 7x ≥ -2

= x ≥ -2/7

Now, the graphicalrepresentation of the solution is as follows:

Free - Previous Years Question Papers
Any questions? Ask us!
×