Question

2≤3(x-2)+5 <8 , x belongs to w

Answer 1

Question :

Solve for x ;

If 2 ≤ 3(x - 2) + 5 < 8 , x € W.

Answer :

x = 1 , 2.

Solution :

We have;

2 ≤ 3(x - 2) + 5 < 8 , x € W

Case(1);

When ,2 ≤ 3(x - 2) + 5

=> 2 ≤ 3(x - 2) + 5

=> 2 ≤ 3x - 6 + 5

=> 2 ≤ 3x - 1

=> 3x - 1 ≥ 2

=> 3x ≥ 2 + 1

=> 3x ≥ 3

=> x ≥ 3/3

=> x ≥ 1 -------------(1)

{ x can be 1,2,3.... }

Case(2)

When ,3(x - 2) + 5 < 8

=> 3(x - 2) + 5 < 8

=> 3x - 6 + 5 < 8

=> 3x - 1 < 8

=> 3x < 8 + 1

=> 3x < 9

=> x < 9/3

=> x < 3 ------------(2)

{ x can be 0,1,2 }

Note:

The intersection of case(1) and case (2) will give the solution of the given inequations.

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

The whole numbers which are 'greater than or equal to 1' and 'smaller than 3' simultaneously are 1 and 2.

Hence,

The required solution of the given inequations is ,

x = 1, 2 .

Answer 2

Answer:-

→ x= 1,2

→ x € [ 1 , 3)

Here, according to the given condition 3 will not be included.

Step - by - step explanation:-

Conditions applied :-

Here we solved for two solutions .

for less then equal to ,and for greater then.

Solution :-

• In case (1)

if ,

→ 2 ≤ 3 (x - 2)+5

→ 2 ≤ 3x - 6 + 5

→ 3 ≤ 3x

→ x ≥ 1

Means here no values of x is less then 1.

• In case (2)

if,

→ 3(x - 2) +5 < 8

→ 3x - 6 + 5 < 8

→ 3x - 1 < 8

→ 3x < 9

→ x< 3

Means, here no value x is greater then 3 .

Hence ,

When we add both cases (1) and (2).

Then, we get

→ x € [ 1 , 3)

Here, 3 will not include.

when we take integer of x < 3 then ,

3 will counted in integer of [2≤ x < 3]

= [2]