[tex]\frac{1}{x} +\frac{1}{x+4}[/tex] (by taking the common factor for the two elements of the equation
[tex]\frac{1]{x} *\frac{x+4}{x+4} +\frac{1}{x+4}*\frac{x}{x}=\frac{x+4}{x*(x+4)}+\frac{x}{x*(x+4)}[/tex] now we can add both of the equation as the denominators have the same value.
[tex]\frac{x+4+x}{x*(x+4)}=\frac{2x+4}{x^{2}+4x} = \frac{1}{5}[/tex] by multiplying both sides by [tex]5*(x^{2}+4x)[/tex]
[tex] 5*(2x+4)=x^{2} +4x[/tex]
[tex] 10x+20=x^{2}+4x[/tex] by subtracting both sides by (10x+20)
[tex] x^{2} -6x -20 = 0[/tex] (by using the quadratic formula)
we will find two answers for this equation
x={[tex]\frac{-6- \sqrt{116}}{-2}[/tex] and [tex] \frac{-6+\sqrt{116}}{-2}[/tex]
