Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$

Question

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!

Answer 1

If you take out $x$ (clearly $x\ne 0$) we get: $$\dfrac{x(x-10+{15\over x})}{x(x-6+{15\over x})}=\dfrac{4x}{x(x-12+{15\over x})}.$$

Cancel $x$ and let $t=x+ {15\over x}$, then we have $${t-10\over t-6} = {4\over t-12}$$ or $$t^2-22t+120 = 4t-24$$ and so on...