Math, asked by skabil, 7 months ago

solve this question correctly for class 11
question is in the picture

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Answered by Thatsomeone
3

Step-by-step explanation:

\sf \frac{x+1}{x+3} < 3 \\ \\ \sf \frac{x+ 1 }{x+3} - 3 <0 \\ \\ \sf \frac{x+1 - 3x - 9 }{x+3}<0 \\ \\ \sf \frac{-2x-8}{x+3}<0 \\ \\ \sf \frac{2(x-4)}{x+3}>0 \\ \\ \sf By\:wavy\:curve\:method \\ \\ \sf x = ( -∞ , - 3 ) U ( 4 , ∞ )

Answered by Anonymous
6

Your answer⬇️

 \frac{x + 1}{x + 3} < 3

 \frac{x + 1}{x + 3}  - 3 < 0

 \frac{x + 1 - 3x - 9}{x + 3} < 0

 \frac{ - 2x - 8}{x + 3}  > 0

Hope this will help you...

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