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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequality $-12 <4 <- \large\frac{3x}{-5} \leq 2$

$\begin{array}{1 1}(A)\;[\frac{-80}{3},\frac{-10}{3}]\\(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 $'>'$ .
Given inequality is $ -12 \leq 4 - \large\frac{3x}{-5}$$ 2$
Adding -4 throughout the inequality
$=> 12 - 4 < 4 - \large\frac{-3x}{-5} $$-4 \leq 2-4$
$=> -16 < \large\frac{-3x}{-5} $$ \leq -2$
$=> -16 < \large\frac{3x}{5} $$ \leq -2$
Multiplying by 5 through out
$=> -80 <3x \leq -10$
Dividing by 3 through out
=> $ \large\frac{-80 }{3} $$ < x \leq \large\frac{-10}{3}$
Step 2:
All numbers greater than $ \large\frac{-80}{3}$ and less than or equal to $ \large\frac{-10}{3}$ are solutions to the given inequality
The solution set is $[\large\frac{-80}{3}, \frac{-10}{3} ]$
answered Aug 1, 2014 by meena.p
 
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