+0

# find the solution set

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(2x+1)/(x^(2)-1)+(x)/(x^(2)-3x+2)=4

Guest Jul 16, 2017
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#1
+90192
+2

(2x+1)/(x^(2)-1)+(x)/(x^(2)-3x+2)=4

I'll try and make sense of your question first

$$\frac{(2x+1)}{(x^{2}-1)}+\frac{(x)}{(x^{2}-3x+2)}=4\qquad x\ne\pm1\\ \frac{(2x+1)}{(x-1)(x+1)}+\frac{(x)}{(x-2)(x-1)}=4 \qquad x\ne\pm1,\quad x\ne2\\ \frac{(2x+1)(x-2)}{(x-1)(x+1)(x-2)}+\frac{(x)(x+1)}{(x-2)(x-1)(x+1)}=4\\ (2x+1)(x-2)+(x)(x+1)=4(x-1)(x+1)(x-2)\\ 2x^2-4x+x-2+x^2+x=4(x-1)(x+1)(x-2)\\ 3x^2-3x+x-2=(x-2)(4x^2-4)\\ 3x^2-2x-2=4x^3-4x-8x^2+8\\ 0=4x^3-4x+2x-8x^2-3x^2+8+2\\ 0=4x^3-11x^2-2x+10\\$$

https://www.wolframalpha.com/input/?i=4x%5E3-11x%5E2-2x%2B10%3D0

Melody  Jul 16, 2017
#2
+1

Thank you. I'm having doubts about my answer but now I see where I am wrong

Guest Jul 16, 2017
#3
+26102
+2

.

Alan  Jul 16, 2017

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