<s>[INST] 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.
 [/INST] Taking from where you left off, and rewrite the left hand side of your last line:
$\begin{align}(x^2+15)^2 - 18x(x^2+15)  + 81x^2 - x^2 -3x(x^2+15)+18x^2&=0\\ \implies (x^2+15)^2-21x(x^2+15)+98x^2&=0\\ \implies (x^2+15)^2 - 2\cdot 10x\cdot (x^2+15)+(10x)^2 - x(x^2+15)-2x^2&=0\\ \implies (x^2+15-10x)^2-x(x^2+15-10x)-12x^2&=0\\ \implies (x^2-10x+15 -4x)(x^2-10x+15 + 3x) &= 0 \\ \implies (x^2-14x+15)(x^2-7x+15)&=0 \end{align}$
From this you can use quadratic formula to calculate the roots.
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