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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 2$

$\begin{array}{1 1}(A)\;[\frac{-80}{3},\frac{-10}{3}]\\(B)\;[1, \frac{11}{3}]\\(C)\;(-23,2]\\(D)\;[2,3]\end{array} $

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1 Answer

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Toolbox:
  • 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 :
Given $7 \;\leq \large\frac{3x+11}{2}$$ \leq 2$
Multiplying by 2 throughout the inequality $14 \leq 3x+11 \leq 22$
$=> 14 \leq 3x+11 \leq 22$
Subtracting 11 throughout the inequality
$ => 7-11 \leq 3x+11 -11 \leq 22-11$
$=>3 \leq 3x \leq 11$
Dividing by 3 throughout $=> 1 \leq x \leq \large\frac{11}{3}$
Step 2 :
All number greater than r equal to 1 and less than or equal to $ \large\frac{11}{3}$ are solutions of the given inequality .
The solution set is $ [1, \large\frac{11}{3} ]$
Hence B is the correct answer.
answered Aug 4, 2014 by meena.p
 
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