#### Answer

The solutions are $x=3$ and $x=5$

#### Work Step by Step

$\dfrac{2x-5}{x}=\dfrac{x-2}{3}$
Take $x$ to multiply the numerator of the right side and $3$ to multiply the numerator of the left side:
$(2x-5)(3)=(x-2)(x)$
Evaluate the indicated operations:
$6x-15=x^{2}-2x$
Take all terms to the right side and simplify:
$0=x^{2}-2x-6x+15$
$0=x^{2}-8x+15$
Rearrange:
$x^{2}-8x+15=0$
Solve by factoring:
$(x-3)(x-5)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-3=0$
$x=3$
$x-5=0$
$x=5$
The solutions are $x=3$ and $x=5$