Linear Inequalities

9. x + $\frac{x}{2}+\frac{x}{3}$ < 11

$\Rightarrow \frac{6x+3x+2x}{6}<11$

$\Rightarrow \frac{(6x+3x+2x)*6}{6}<11*6$

11x < 66

$\frac{11x}{11}<\frac{66}{11}$

x < 6.

So, x ∈ (–∞, 6)

8. 3 (2 – x) ≥ 2 (1 – x)

6 – 3x ≥ 2 – 2x

6– 2 ≥ 3x– 2x

4 ≥ x

So, x ∈ (–∞, 4]

7. 3 (x –1) ≤ 2 (x –3)

3x– 3 ≤ 2x –6

=> 3x– 2x ≤ –6 + 3

x ≤ –3

So, x ∈ (–∞, – 3]

6. Given, 3x – 7 > 5x – 1.

7+1 > 5x – 3x

6 > 2x

$\Rightarrow \frac{-6}{2}>\frac{2x}{2}$

–3 > x

So, x ∈ (–∞, – 3)

5. Given,

4 x + 3 < 5 x+ 7.

3 – 7 < 5 x 4 x.

–4 < x.

So, x ( –4, ∞)

4. Given, 3x + 8 > 2

3x + 8 – 8 > 2 – 8

3x > – 6

$\Rightarrow \frac{3x}{3}>-\frac{6}{3}$

x > – 2

(i) when x is an integer, x ∈ z all integer greater than – 2 are the soln. so, x = { – 1, 0, 1, 2, 3, …}.

(ii) when x is a real number, all real number greater than – 2 are the soln. so, x ( –2, ∞  )

3. Given, 5x – 3 < 7.

5x – 3 + 3 < 7 + 3

5x< 10

$\frac{5x}{5}<\frac{10}{5}$

x< 2.

(i) when x ∈ z, i. e, x is an integer all integers less than 2 are the solⁿ so, x = {…., – 3, – 2, – 1, 0, 1}

(ii) when x ∈ R i, e x is a real number all t real number less than 2. are the solⁿ so, x∈ (– ∞, 2)

2. – 12 x> 30.

Dividing both sides by 12 we get,

$-x>\frac{30}{12}$

$-x>\frac{5}{2}$

Multiplying both side by ( – 1) the inequality will change.

i e, x< $\frac{5}{2}$ = – 2.5

(i) As x is a natural number, the soln of the given inequality does not exist in natural numbers.

(ii) As x is an integer, the soln of the given inequality will be all the integer less than – 5/2

i. e, – 3, – 4, – 5, ….

1. 24x < 100.

Dividing both sides by 24 we get,

$\frac{24x}{24}<\frac{150}{24.}$

x< $\frac{25}{6}$ = 4.166

(i) As x is a natural no the solⁿ of the given inequality are 1, 2, 3, 4.

(ii) As x is an integer the solⁿ of the given inequality are 4, 3, 2, 1, 0, 1, 2, 3, ….