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

Solve the inequality for real $x: \large\frac{1}{2} \bigg(\large\frac{3x}{5} +4 \bigg) $$ \geq \large\frac{1}{3} $$(x-6)$

$\begin{array}{1 1} (A)\;(-3,\infty)\\(B)\;(-\infty,120]\\(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 $'>'$ .
Step 1:
The given inequality is $\large\frac{1}{2} \bigg(\large\frac{3x}{5} +4 \bigg) $$ \geq \large\frac{1}{3} $$(x-6)$
Multiplying by 6 on both sides,
$=> 3 \bigg( \large\frac{3x}{5} +4 \bigg) \geq 2(x-6)$
$=> \large\frac{9x}{5} $$+12 \geq 2(x-6)$
Adding 12 on both sides ,
$\large\frac{9x}{5}$$+12 \geq 2x-12$
Adding $ -\large\frac{9x}{5}$ on both sides
=> $ 24 \geq 2x -\large\frac{9x}{5}$
$=> 24 \geq \large\frac{10x -9x}{5}$
$=> 24 \geq \large\frac{x}{5}$
Multiplying by 5 on both sides,
$120 \geq x$
Step 2:
All real numbers Which are less than or equal to $120$satisfy the given inequality
The solution set is $(-\infty,120]$
Hence B is the correct answer.
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