Question

Me ajudem por favor eu imploro

Answer 1

$\frac{4}{x+3} -\frac{2}{x+1} =\frac{5}{2x+6} -\frac{2}{x+3}$

$\frac{4(x+1)}{(x+3)(x+1)} -\frac{2(x+3)}{(x+3)(x+1)} =\frac{5(x+3)}{(2x+6)(x+3)} -\frac{2(2x+6)}{(2x+6)(x+3)}$

$\frac{4(x+1)-2(x+3)}{(x+3)(x+1)} =\frac{5(x+3)-2(2x+6)}{(2x+6)(x+3)}$

$\frac{4(x+1)-2(x+3)}{(x+1)} =\frac{5(x+3)-2(2x+6)}{(2x+6)}$

$\frac{4(x+1)}{(x+1)}-\frac{2(x+3)}{(x+1)} =\frac{5(x+3)}{(2x+6)}-\frac{2(2x+6)}{(2x+6)}$

$4-\frac{2(x+3)}{(x+1)} =\frac{5(x+3)}{(2x+6)}-2$

$4-\frac{2(x+3)}{(x+1)} =\frac{5(x+3)}{2(x+3)}-2$

$-\frac{2(x+3)}{(x+1)} =-\frac{7}{2}$

$\frac{2(x+3)}{(x+1)} =\frac{7}{2}$

$4(x+3)=7(x+1)\\4x+12=7x+7\\4x-7x=7-12\\-3x=-5\\3x=5\\\\\boxed{x=\frac{5}{3}}$