Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$ Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$
First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^2-6x+15)\\x^4-12x^3+15x^2-10x^3+120x-150x+15x^2-180x+225=\\=4x^3-24x^2+60x$$ which is an equation I can't solve. I tried to simplify the LHS by $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{(x^2-6x+15)-4x}{x^2-6x+15}=1-\dfrac{4x}{x^2-6x+15}$$ but this isn't helpful at all. Any help would be appreciated! :) Thank you in advance!

let $p=x^2+15,q=x$ then we have to solve $$\frac{p-10q}{p-6q}=\frac{4q}{p-12q}$$ $$\iff (p-10q)(p-12q)-4q(p-6q)=0$$ $$\iff (p - 8 q) (p - 18 q)=0$$ provided that $p\neq 6q,p\neq 12q$
Can you end it now?
