#### Answer

the solutions are -1 and 16

#### Work Step by Step

$\frac{15}{x+4}$=$\frac{x-4}{x}$
x(x+4)$\frac{15}{x+4}$=x(x+4)$\frac{x-4}{x}$
$\frac{15}{x+4}$=$\frac{x-4}{x}$
15x=$x^{2}$-16
0=$x^{2}$-15x-16
0=$x^{2}$+x-16x-16
0=x(x+1)-16(x+1)
0=(x-16)(x+1)
x-16=0 or x+1=0
x=16 or x=-1
the solutions are -1 and 16
Check
Let x=-1
$\frac{15}{-1+4}$=$\frac{-1-4}{-1}$
$\frac{15}{-3}$=$\frac{-5}{-1}$
-5=-5
Let x=16
$\frac{15}{16+4}$=$\frac{16-4}{16}$
$\frac{15}{20}$=$\frac{12}{16}$
$\frac{3}{4}$=$\frac{3}{4}$