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

Solve the given inequality for real $x: \large\frac{3(x-2)}{5} \leq \large\frac{5(2-x)} {3}$

$\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 $\large\frac{3(x-2)}{5}$$ \leq 5 \bigg( \large\frac{2-x}{3}\bigg)$
Multiplying both sides of the inequality by 15.
$=> \large\frac{3(x-2)}{5} $$ \times 15 \leq \large\frac{5(2-x)}{3}$$ \times 15$
=> $ 9(x-2) \leq 25 (2-x)$
=> $ 9x -18 \leq 50 -25 x$
Adding 25 x to both sides of the inequality => $9x-18+25 x \leq 50$
Adding 18 to both sides of inequality
$=> 9x +25 x \leq 50+18$
$=> 34 x \leq 68$
dividing x by 34 on both sides,
$=> \large\frac{34 x}{34} \leq \frac{68}{34}$
$ x \leq 2$
Step 2:
All real number less than or equal to $2$ satisfy the given inequality
The solution set is $(-\infty ,2]$
Hence B is the correct answer.
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