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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the inequality for real $x: \large\frac{x}{4} < \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\begin{array}{1 1} (A)\;(-3,\infty)\\(B)\;(-\infty,2]\\(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 $\large\frac{x}{4} < \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$
=> $ \large\frac{x}{4} < \frac{5(5x-2)-3(7x-3)}{15}$
=> $ \large\frac{x}{4} < \frac{(25x-10-2)x+9}{15}$
=> $ \large\frac{x}{4} < \frac{4x-1}{15}$
Multiplying by a positive number 60 on both sides
=> $15 x < 4(4x-1)$
=> $15 x < 16 x -4$
=> $4 < x $
Step 2:
Thus all real numbers which are greater than 4 are solutions of the given inequality.
The solution set is $(4, \infty)$
Hence C is the correct answer.
answered Jul 26, 2014 by meena.p
 
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