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Questions  >>  CBSE XI  >>  Math  >>  Linear Inequalities

Solve the inequality for real $x: \large\frac{(2x-1)}{3} \geq \frac{(3x-2)}{4} -\frac{(2-x)}{5}$

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

1 Answer

  • 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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