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Q4.    Solve 3x + 8 >2, when

                (ii) x is a real number.

Answers (1)

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Given :   3x + 8 >2

\Rightarrow      3x + 8 >2

\Rightarrow \, \, \, 3x> -6

Divide by 3 from both side 

\Rightarrow \, \, \, \frac{3}{3}x> \frac{-6}{3}

 \Rightarrow \, \, \, x> - 2

 

x are  real numbers  greater  than -2

Hence , values of x can be as  x\in (-2,\infty )

Posted by

seema garhwal

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