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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequality $7 \leq \large\frac{(3x+11)}{2} \leq 11$

$\begin{array}{1 1}(A)\;[1,\frac{11}{3}]\\(B)\;[-4,2]\\(C)\;(-23,2]\\(D)\;[2,3]\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 $'>'$ .
Given inequality is $ 7 \leq \large\frac{(3x+11)}{2}$$ \leq 11$
Multiplying by 2 throughout the inequality
$=> 14 \leq 3x+11 \leq 22$
Subtracting 11 throughout the inequality
$=> 7-11 \leq 3x+11 -11 \leq 22-11$
$=> 3 \leq 3x \leq 11$
Dividing by 3 through out
$=> 1 \leq x \leq \large\frac{11}{3}$
All numbers greater than equal to 1 and less than or equal to 11/3 are solutions of the given inequality .
The solution set is $[1, 11/3]$
Hence A is the correct answer.
answered Aug 1, 2014 by meena.p
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