Solve the given inequality for real x: x+frac | vedclass

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Question

(A) Given inequality: $x+\frac{x}{2}+\frac{x}{3} < 11$
Taking $x$ as a common factor:
$x(1+\frac{1}{2}+\frac{1}{3}) < 11$
Finding the common

Vedclass · Accepted Answer

$(-\infty, 6)$

Vedclass · Answer

(A) Given inequality: $x+\frac{x}{2}+\frac{x}{3} < 11$
Taking $x$ as a common factor:
$x(1+\frac{1}{2}+\frac{1}{3}) < 11$
Finding the common denominator for the terms inside the bracket:
$x(\frac{6+3+2}{6}) < 11$
$x(\frac{11}{6}) < 11$
Multiply both sides by $\frac{6}{11}$:
$x < 11 	imes \frac{6}{11}$
$x < 6$
Thus,all real numbers $x$ which are less than $6$ are the solutions of the given inequality.
Hence,the solution set is $(-\infty, 6)$.

Solve the given inequality for real $x$: $x+\frac{x}{2}+\frac{x}{3} < 11$

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