$\begin{array}{1 1} (A)\;(-3,\infty)\\(B)\;(-\infty,2]\\(C)\;(4,\infty)\\(D)\; (-9,\infty]\end{array} $

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- Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
- Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
- If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $ '>'$ sign changes to $'<' $ and $'<'$ changes $'>'$ .

The given inequality is

$\large\frac{(2x-1)}{3} \geq \frac{(3x-2)} {4} - \frac{(2-x)}{5}$

=> $ \large\frac{(2x-1)}{3} \geq \frac{5(3x-2) -4(2-x)}{20}$

=> $ \large\frac{2x-1)}{3} \geq \frac{5(3x-2)-4(2-x)}{20}$

=> $ \large\frac{(2x-1)}{3} \geq \frac{19 x -18}{20}$

Multiplying both sides if the inequality by a positive number 60.

=> $20 (2x-1) \geq 3( 19 x -18)$

=> $40 x -20 \geq 57 x -54$

Adding 54 and -40 on both sides of the inequality

$-20 +54 \geq 57 x -40 x$

$34 \geq 17 x$

$2 \geq x$

Step 2:

All numbers which are less than or equal to 2 satisfy the inequality .

The solution set is $(-\infty,2]$

Hence B is the correct answer.

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