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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequality for real $x: 2(2x+3) -10 < 6(x-2)$

$\begin{array}{1 1} (A)\;(4,\infty)\\(B)\;(\infty,-6)\\(C)\;(-4,\infty)\\(D)\; (-9,\infty)\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 $'>'$ .
The given inequlaity is $2(2x+3) -10 < 6 (x-2)$
$=> 4x+6 -10 < 6 (x-2)$
$=> 4x-4 < 6x -12$
Adding 12 on both sides,
$=> 4x-4+12 < 6x-12+12$
Adding $-4x$ on both sides,
=> $4x+8-4x < 6x -4x $
=> $ 8 < 2x$
dividing by a positive number 2 on both sides, => $4 < x$
Step 2:
All numbers greater than $4$ satisfy the given inequality
The solution set is $(4 , \infty)$
Hence A is the correct answer.
answered Jul 25, 2014 by meena.p
 
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