Answer
{$\frac{1}{2},4$}
Work Step by Step
First, we add the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $x(x-3)$:
$\frac{5}{x-3}+\frac{4}{x}=6$
$\frac{5(x)+4(x-3)}{x(x-3)}=6$
$\frac{5x+4x-12}{x(x-3)}=6$
$\frac{9x-12}{(x^{2}-3x)}=6$
Now, we cross multiply the two fractions in order to create a quadratic equation:
$\frac{9x-12}{(x^{2}-3x)}=\frac{6}{1}$
$1(9x-12)=6(x^{2}-3x)$
$9x-12=6x^{2}-18x$
$6x^{2}-18x=9x-12$
$6x^{2}-18x-9x+12=0$
$6x^{2}-27x+12=0$
$2x^{2}-9x+4=0$
Now, we use rules of factoring trinomials to solve the equation:
$2x^{2}-9x+4=0$
$2x^{2}-8x-1x+4=0$
$2x(x-4)-1(x-4)=0$
$(x-4)(2x-1)=0$
$(x-4)=0$ or $(2x-1)=0$
$x=4$ or $x=\frac{1}{2}$
Therefore, the solution is {$\frac{1}{2},4$}.
