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Q4.    Solve 3x + 8 >2, when
                (i) x is an integer.

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Given :   3x + 8 >2

\Rightarrow      3x + 8 >2

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

Divide by 3 from both sides 

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

 \Rightarrow \, \, \, x> - 2

 

x are  integers greater  than -2

Hence, the values of x can be \left \{-1,0,1,2,3,4...............\right \} .

Posted by

seema garhwal

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