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Solve the given inequality for real $x. x +\large\frac{x}{2} +\frac{x}{3} $$<11 \begin{array}{1 1} (A)\;(-3,\infty)\\(B)\;(-\infty,6)\\(C)\;(-4,\infty)\\(D)\; (-9,\infty)\end{array} Can you answer this question? 1 Answer 0 votes Toolbox: • Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality. • Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality. • If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e '>' sign changes to '<' and '<' changes '>' . The given inequality is x + \large\frac{x}{2} +\frac{x}{3}$$< 11$
$=> x [1+ \large\frac{1}{2} +\frac{1}{3} \bigg] $$<11 => x \bigg[ \large\frac{6+ 3+2}{6} \bigg]$$ <11$
$=> \large \frac{11 x}{6} $$<11 Dividing both sides of the inequality by 11. => \large\frac{11 x}{6 \times 11} < \frac{11}{11} => \large\frac{x}{6}$$<1$
=> $x < 6$
Step 2:
All real numbers which are less than 6 satisfy the given in equality.
The solution set is $(-\infty, 6)$
Hence B is the correct answer.