Math, asked by shradhayadav301159, 5 months ago

solve these inequailities ​

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Answered by DevyaniKhushi
2

 \rm{i) \:  \: x + 3 > 3 } \\ \rm{}x  + 3 - 3 > 3 - 3 \\ \rm{}x > 0 \\  \\  \therefore{  \:  \:  \: \sf{ \green{x :  \{ 1,2,3 \: ... \:  \:  \infin\}}}}

 \rm{ii) \:  \: x - 5 <  - 5 } \\   \rm{}x - 5  + 5<  - 5 + 5 \\  \rm{}x < 0 \\  \\  \therefore \:  \:  \green{ \sf{ \:  \: x :  \{   - 1, - 2, - 3 \: ... \:  \infin\}}}

 \rm{iii) \:  \: x - 2 <  - 3}  \\  \rm{}x - 2 + 2 <  - 3 + 2 \\  \rm{} \:  \: x <  - 1 \\  \\  \therefore \:  \:  \green{ \sf{x :  \{ - 2, - 3, - 4 \: ... \infin \}}}

 \rm{iv) \:  \: 1 - x > 1} \\\rm 1 - x - 1 > 1 - 1 \\ \rm - x > 0 \\   \\ \rm \frac{ - x}{ - 1}   <  \frac{0}{ - 1}    \:  \:  \:  \:  \:  \:  \:   \\ \tiny\{ \because  \text{on dividing both side by \:  --ve number, the sign gets reversed}\} \\ \rm  x < 0 \\  \\   \therefore  \{ \sf  x :   - 1, - 2, - 3, \: ... \:  \infin\}

 \rm{v)} \:  \:  - 3  <   - 2 + x \\   \rm{} - 3 + 2 <  - 2 + 2 + x \\  \rm{} - 1 < x

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