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

$\begin{array}{1 1} (A)\;(-6,\infty)\\(B)\; (-2,\infty)\\(C)\;(-\infty,-3)\\(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 $'>'$ .
Step 1:
$3x -7+7 > 5x + 6-5x$
Adding 7 to both sides of the inequality
=> $ 3x-7 +7 > 5x -1+7$
=> $ 3x > 5x +6 $
subtracting 5x from both sides of the inequality .
$=> 3x -5x > 5x +6-5x$
=> $ -2x >6$
Dividing both sides by a negetive number -2.
=> $ \large\frac{-2x}{-2} < \frac{6}{-2}$
=> $ x < -3$
Step 2:
All real numbers which are less than $-3$ satisfy the given inequality .
The solution set is $(-\infty, -3)$
Hence C is the correct answer.
answered Jul 25, 2014 by meena.p
 
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