Answer
The equation has one solution, $\frac{16}{3}$.
Work Step by Step
$\frac{4}{3(c+4)}+1=\frac{2c}{c+4}$
$\frac{4}{3(c+4)}+\frac{3(c+4)}{3(c+4)}-\frac{3\times2c}{3(c+4)}=0$
$\frac{4}{3(c+4)}+\frac{3c+12}{3(c+4)}-\frac{6c}{3(c+4)}=0$
$\frac{4+3c+12-6c}{3(c+4)}=0$
$16-3c=0\times3(c+4)$
$16-3c=0$
$3c=16$
$c=\frac{16}{3}$
Check:
$4\div[3(16/3+4)]+1=4/28+1=8/7;(2\times16/3)\div(16/3+4)=(32/3)\div(28/3)=32/28=8/7$
The equation has one solution, $\frac{16}{3}$.
