Question

solve
for class 11
basic algebra​

Answer 1

Step-by-step explanation:

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

Answer 2

Your answer ⬇️

] \: \: \: \: \: \: \: \: \frac{x + 1}{x + 3} < 3[/tex]

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

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

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

$\: \: \: \: \: \: \: \: \: \: \frac{2(x - 4)}{x + 3} > 0$