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Solve the inequality for real x.

    Q10.    \frac{x}{3} > \frac{x}{2} + 1

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Given :   \frac{x}{3} > \frac{x}{2} + 1

\Rightarrow      \frac{x}{3} > \frac{x}{2} + 1

\Rightarrow \, \, \, \frac{x}{3}-\frac{x}{2}> 1

\Rightarrow \, \, \,x (\frac{1}{3}-\frac{1}{2})> 1

 \Rightarrow \, \, \,x (-\frac{1}{6})> 1

\Rightarrow \, \, \, -x > 6

\Rightarrow \, \, \, x< -6

x are  real numbers less than -6

Hence, values of x can be  as  x\in (-\infty ,-6)

Posted by

seema garhwal

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