Solve for $x$: $\frac{x^2-10x+15}{x^2-6x+15}=\frac{3x}{x^2-8x+15}$

Solve the equation: $\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{3x}{x^2-8x+15}$.

When $x\ne3$ and $x\ne5$ we get $$(x^2-10x+15)(x^2-8x+15)=3x(x^2-6x+15)\\(x^2-9x+15-x)(x^2-9x+15+x)-3x(x^2-6x+15)=0\\(x^2-9x+15)^2-x^2-3x(x^2-6x+15)=0.$$ I am stuck here. The Rational Root Theorem won't be useful as the equation does not have such roots.

I got $-9x$ by averaging $-10x$ and $-8x$.

I don't know if it makes sense.

Let $y:=x^2-6x+15$ so$$\frac{y-4x}{y}=\frac{3x}{y-2x}\implies y^2-6xy+8x^2=3xy\implies (y-x)(y-8x)=0.$$That gives you two quadratics to solve.

Hint: divide the numerators and denominators by $x \ne 0$ and let $z = x + \dfrac{15}{x}\,$, then:

$$ \frac{z-10}{z-6} = \frac{3}{z-8} $$