Solve the following inequation and write down the solution set. 11x-4 < 15x+4 <= 13x+14 ; xεW. Represent the solution on a number line. [icse 2019]

Question: Solve the following inequation and write down the solution set.

11x - 4 < 15x + 4 ≤ 13x + 14 ; x εW

Represent the solution on the number line.

Given that:

11x - 4 < 15x + 4 ≤13x + 14 - - - - - - - (1)

=> 11x - 4 < 15x + 4 and 15x + 4 ≤13x + 14

Case I.

11x - 4 < 15x + 4

Subtracting 11x on both sides, we get,

-4 < 4x +4

Subtracting '4' on both sides, we get,

-8 < 4x

Dividing both sides by 4, we get,

-2 < x 

or x > -2. - - - - - - (2)

Case II.

15x + 4 ≤ 13x + 14

Subtracting both sides by 13x, we get,

2x + 4 ≤ 14

Subtracting 4 on both sides, we get,

2x  ≤10

Dividing both sides by 2, we get

x ≤ 5 - - - - - - - (3)

From,(1), (2) and (3), the solution set is

- 2 < x ≤ 5 :xεW

=(-2,5], here W is a whole number.

W = 0, 1, 2, 3 ,- - - - - - - - .

So the solution set will be {0,1,2,3,4,5}

Solve the following inequation and represent the solution set on a number line.(-8*1/2)< -1/2 -4x<=7*1/2 ; x€I [icse 2017]

Solve the following inequation and represent the solution set on a number line:

- 8*1/2 < - 1/2 - 4x ≤ 7 *1/2 , , x εI

Solutions:

Given that:- 

-8 *1/2 < - 1/2 - 4x≤7 *1/2

⇒ Convert mixed fractions to improper fractions:

- 17/2 < - 1/2 - 4x ≤15/2

Adding 1/2 to each term, we get:

- 17/2 + 1/2 < - 4x ≤ 15/2 + 1/2

Simplify the fractions:

- 16/2 < - 4x ≤16/2

=> - 8 < - 4x ≤ 8

Dividing each term by -4, we get:

2 > x >= - 2

[Sign reverses when each term is divided by same negative number]

We can write it as

- 2 ≤ x < 2

The solution set is {x:  - 2 ≤ x < 2; x Ɛ I}

= {-2,-1,0,1}

Use the remainder theorem to factorise the following expression: 2x^3+x^2-13x+6. [icse 2010]

 

Rules for solving Inequations

Rules for Solving a Linear Inequality

<, >, ≤ and ≥ are known as signs or an inequality.

Rule 1

If a positive number is added to both sides of an inequality, the sign of inequality remains the same.

(a) x+3<5 => x + 3 + 2 < 5 + 2

(b) x+1>3 => x + 1 + 1 > 3 + 1

(c) x+3≥7 =>  x + 3 + 1≥7 +1

(d) x+5≤9 => x + 5 + 2 <= 9 + 2

Rule 2:-

If a negative number is subtracted on both sides of an inequality, the sign of that  ineq -uality remains the same.

(a) x+1<5 => x+1-2< 5-2

(b) x+3>7 => x+3-2>7-2

(c) x + 2≼ 11 =>x+2-1≼ 11-1

(d) x + 5≥9 => x+5-2 ≥ 9-2

Rule 3:- 

If we multiply the both sides of an inequality with the same positive number, then the sign of the inequality remains the same.

(a) x < 2 => x×5<2×5 

(b) x > 5 => x×2>5×2

(c)x ≤ 3 => x× 2 ≤ 2×3

(d) x ≤7 => x ×3≤ 7×3

Rule 4 :-

If we multiply both sides of an inequality with the same negative number, then the inequality will reverse i.e. the sign of inequality reversed.

(a) x < 5 => (-2)× x > (-2)×5

(b) x > 4 =>  (-3) × x < (-3) × 4 

(c) x ≤ 9 => (-1) × x ≥ (-1) × 9 

(d) x≽3 => (-2)×x ≼ (-2)×3

Rule 5:- 

If both sides of an inequality are divided by the same positive number, then the sign of inequality remains the same.

(a) x<5  => x/2 < 5/2

(b)x>3 => x/3 > 3/3

(c) x ≤2  =>   x/2 ≤  2/2

(d) x≥6 => x/3 ≥6/3

Rule 6 :-

If both the sided of an inequality are divided by same negative number, then the sigh of inequality will reverse. 

(a) x<4 => x/(-2) > 4/(-2)

(b) x>5 => x/(-3) < 5/(-3)

(c) x≤6 => x/(-2) ≥ 6/(-2)

(d) x≥8 => x/(-4)≤ 8/(-4)

Find the value of x, which satisfy the inequation - 2*5/6 less than1/2 -2x/3 less than aur equal to 2 and x€W and represent the solution on the numbe line.[icse 10th 2014]

Find the value of x, which satisfy the inequation and x ∈ W and represent the solution on the number line. 

Given that - 2 *5/6 < 1/2 - (2x)/3 ≤ 2 

=>  - 17/6 < 1/2 - (2x)/3 ≤ 2 

=> - 17/6 < 1/2 - (2x)/3 and 1/2 - (2x)/3≤ 2 

Case I  - 17/6 < 1/2 - (2x)/3

Multiplying both sides by 6, we get, 

- 17 < 3 - 4x

Subtracting both sides by 3, we get,

- 17 - 3 < 3 - 4x - 3

- 20 < - 4x

Dividing both sides by -4, we get,

5>x

or x < 5  -  - - - - - (1) 

[It both sides of  an inequality is divided by the same negative number,then sign of inequality reverses.]

Case 2:- 1/2 -2x/3 ≤ 2 

Subtracting both sides  1/2 we get,

=> 1/2 - (2x)/3 - 1/2 ≤ 2 - 1/2

=>  - (2x)/3 ≤ 3/2

Multiplying both sides by 3, we get

- 2x ≤ 9/2

Dividing both sides by -2, we get,

x ≥ - 9 /4   =>  - 9/4 ≤ x   - - - - - - - - (2)

From (1) and (2), we get,

- 9/4 ≤ x < 5

=> x=[ -9/4, 5)

But x is a whole Numbers, i.e  {W=0, 1, 2, 3, 4, 5, ...}

So the solution set ={x ; x = 0, 1, 2, 3, 4}