Answer
The solutions are $-1-\sqrt{5}$ and $-1+\sqrt{5}$.
Work Step by Step
Factor the denominator of the expression on the right side of the equation to obtain:
$$\dfrac{5}{x-3} + \dfrac{x}{x+3} = \dfrac{19}{(x-3)(x+3)}$$
Multiply the LCD $(x-3)(x+3)$ to both sides of the equation to get rid of the denominators:
\begin{array}{ccc}
&(x-3)(x+3)\left[\dfrac{5}{x-3}+\dfrac{x}{x+3}\right] &= &\dfrac{19}{(x-3)(x+3)} \cdot (x-3)(x+3)
\\&(x-3)(x+3) \cdot \dfrac{5}{x-3}+(x-3)(x+3) \cdot \dfrac{x}{x+3} &= &\dfrac{19}{(x-3)(x+3)} \cdot (x-3)(x+3)
\end{array}
Simplify by cancelling common factors to obtain:
\begin{array}{ccc}
\require{cancel}
\\&\cancel{(x-3)}(x+3) \cdot \dfrac{5}{\cancel{x-3}}+(x-3)\cancel{(x+3)} \cdot \dfrac{x}{\cancel{x+3}} &= &-\dfrac{19}{\cancel{(x-3)(x+3)}} \cdot \cancel{(x-3)(x+3)}
\\&(x+3)(5)+(x-3)(x) &= &19
\\&5(x) + 5(3) + x(x)-x(3) &= &19
\\&5x+15+x^2-3x &= &19
\\&x^2+2x+15 &= &19
\end{array}
Put all terms on the left side. Note that when a term is transferred to the other side of an equation, it changes it sign to its opposite.
\begin{array}{ccc}
\\&x^2+2x+15 &= &19
\\&x^2+2x+15-19 &= &0
\\&x^2+2x-4 &= &0
\end{array}
Solve the equation using the quadratic formula with $a=1, b=2, c=-4$ to obtain:
$x=\dfrac{-2\pm \sqrt{2^2-4(1)(-4)}}{2(1)}
\\x=\dfrac{-2\pm\sqrt{4+16}}{2}
\\x=\dfrac{-2\pm \sqrt{20}}{2}
\\x=\dfrac{-2\pm \sqrt{4(5)}}{2}
\\x=\dfrac{-2\pm 2\sqrt{5}}{2}
\\x=\dfrac{-2}{2} \pm \dfrac{2\sqrt{5}}{2}
\\x=-1\pm \sqrt5$
Therefore, the solutions are $-1-\sqrt{5}$ and $-1+\sqrt{5}$.
