Question

3/(y+2) + 2/y + 3y/(2y+2)

Equação do segundo grau com fração​

Answer 1

$\frac{3}{(y + 2)} + \frac{2}{y} = \frac{3y}{(2y + 4)} \\ \\ \frac{3}{(y + 2)} + \frac{2}{y} = \frac{3}{1} . \frac{y}{(2y + 4)} \\ \\ \frac{3}{(y + 2)} + \frac{2}{y} = 3.( \frac{y}{2y + 4}) \\ \\ \frac{3}{(y + 2)} + \frac{2}{y} = \frac{3}{2} .( \frac{y}{y + 2} ) \\ \\ \frac{(3y + 2y + 4)}{y {}^{2} + 2y} = \frac{3}{2} .( \frac{y}{(y + 2)} ) \\ \\ \frac{5y + 4}{y {}^{2} + 2y} = \frac{3}{2} . (\frac{y}{y + 2} ) \\ \\ \frac{ \frac{5y + 4}{y {}^{2} + 2y } }{ \frac{3}{2} } = \frac{y}{(y + 2)} \\ \\ \frac{5y + 4}{y {}^{2} + 2y } . \frac{2}{3} = \frac{y}{(y + 2)} \\ \\ \frac{10y + 8}{3y {}^{2} + 6y } = \frac{y}{(y + 2)} \\ \\ \frac{10y + 8}{3y(y + 2)} = \frac{y}{(y + 2)} \\ \\ \frac{10y + 8}{3y} = \frac{y}{1} \\ \\ 10y + 8 = 3y {}^{2} \\ \\ 3y {}^{2} - 10y - 8 = 0 \\ \\ \Delta = b {}^{2} - 4.a.c \\ \Delta = ( - 10) {}^{2} - 4.3.( - 8) \\ \Delta = 100 + 96 \\ \Delta = 196 \\ \\ \frac{x = - b \pm \sqrt{ \Delta} }{2.a} \\ \\ x = \frac{ - ( - 10) \pm \sqrt{196} }{2.3} \\ \\ x = \frac{10 \pm14}{6} \\ \\ x {}^{1} = \frac{10 + 14}{6} \\ \\ x {}^{1} = \frac{24}{6} \\ \\ x {}^{1} = 4 \\ \\ x {}^{2} = \frac{10 - 14}{6} \\ \\ x {}^{2} = \frac{ - 4}{6} \\ \\ x {}^{2} = \frac{ - 2}{3}$