Solve and find the numbers of positive integral solutions of the following inequation.

$2 x - 3 < x + 2 \leq 3 x + 5, x ϵ R$

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Solution

The correct option is C
4

we have

$\Rightarrow 2 x - 3 < x + 2 \leq 3 x + 5 \Rightarrow 2 x - 3 < x + 2 a n d x + 2 \leq 3 x + 5 \Rightarrow 2 x - x < 3 + 2 a n d 2 - 5 \leq 3 x - x \Rightarrow x < 5 a n d (- 3 / 2) \leq x \Rightarrow (- 3 / 2) \leq x a n d x < 5 \Rightarrow (- 3 / 2) \leq x < 5$

But, $x ϵ R,$ So, the solution set is ${x : (- 3 / 2) \leq x < 5, x ϵ R}$

So, positive integers which are belongs to the solution set are 1, 2 , 3, 4.

Hence, the total number of positive integers are 4.

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Solve the following inequation and graph the solution set on the number line :

$2 x - 3 < x + 2 \leq 3 x + 5; x \in R .$

2 x - 3 < x + 2 \leq 3 x + 5

k

k = | x + | 2 x - 1 | | - | x - | 2 x - 1 | |

2 x - 3 < x + 2 \leq 3 x + 5, x ϵ R

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