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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the given inequality for real $x: 4x+3 < 5x +7$

$\begin{array}{1 1} (A)\;x \in (-6,\infty)\\(B)\;x \in (-2,\infty)\\(C)\;x \in (-4,\infty)\\(D)\;x \in (-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 $'>'$ .
Step 1:
$4x+3 < 5x+ 7$
adding -7 to both sides
=> $4x+3-7 < 5x+7-7$
=> $ 4x-4 < 5x$
adding $-4x $ to both sides
=> $ -4 < 5-4x$
=> $ -4 < x $
Step 2:
All real numbers which are greater than $-4$ are the solution for the inequality .
The solution set is $x \in (-4,\infty)$
Hence C is the correct answer.
answered Jul 25, 2014 by meena.p
 
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