<s>[INST] Is there another way to solve this quadratic equation? $$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$
Steps I took:
$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } $$
$$4(x^{ 2 }-2x+4)+7(x^{ 2 }-2x+1)=2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1)$$
$$11x^{ 2 }-22x+23=2x^{ 4 }-8x^{ 3 }+18x^{ 2 }-20x+8$$
I can keep going with all the steps I took, but is there a more elegant way to arrive at the solution for this equation? It seems as if I keep going the way I am, I will hit a dead end. No actual solution, please. Hints are much better appreciated.  
 [/INST] To simplify computations a little bit, I would put $y=(x-1)^2$, so that the equation becomes $$\frac{4}{y}+\frac{7}{y+3}=2$$ which you can easily solve to find the values of $y$ and finally the corresponding values of $x$.
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