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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequality $-15 < \large\frac{3(x-2)}{5} \leq 0$

$\begin{array}{1 1}(A)\;[2,8]\\(B)\;[-4,2]\\(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 inequality is $ -15 < \large\frac{3(x-2)}{5}$$ \leq 0$
Multiplying by 5 throughout the inequality $-75 <3 (x-2) \leq 0$
Dividing by 3 throughout the inequality $-25 < x -2 \leq 0$
Adding 2 thoughout the inequality
$-25 +2 < x -2 +2 \leq 0+2$
=> $ -23 < x \leq 2$
Step 2:
All numbers greater than -23 and less than or equal to 2 are solutions to the given inequality .
The solution set is $(-23,2 ]$
Hence C is the correct answer.
answered Aug 1, 2014 by meena.p
 
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