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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequality for real $x, \large\frac{x}{3} > \frac{x}{2} $$+1$

$\begin{array}{1 1} (A)\;(-3,\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 inequality is $\large\frac{x}{3} > \frac{x}{2} $$+1$
Subtracting $\large\frac{x}{2}$ on both sides,
$=> \large\frac{x}{3} -\frac{x}{2} > \frac{ x}{2} $$+1 -\large\frac{x}{2}$
$=> \large\frac{x}{3}-\frac{x}{2} $$ > 1$
$=> \large\frac{2x-3x }{6} $$ > 1$
$ => \large\frac{-x}{6} $$ > 1$
Multiplying both sides of the inequality by a negative number $-6$.
$=> \large\frac{-x}{6} $$ \times -6 < 1 \times -6$
$=> x < -6$
Step 2:
Hence all real number less than $-6$ satisfy the given inequality
The solution set is $(-\infty, -6)$
Hence B is the correct answer.
answered Jul 25, 2014 by meena.p
edited Jul 25, 2014 by meena.p
 
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