# The question says:"Find all real solutions of the equation":(3/x)-(2/(x-3))=(-12/(x^2-9)) In the text, it says to find the LCD and then expand, eventually moving all terms to the left/setting the equation to zero. From there, it says to factor. Each time I try this out, I get a slightly different solution, and I think I'm skipping up on the very first step, to find the LCD, and possiby still one or two places after that. In the text, the possible solutions are x=3, -9, the only real solution being -9. I understad how they got those, but I'm not sure how they get from the beginning to step 2, with finding the LCD.

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The question says:

"Find all real solutions of the equation":

(3/x)-(2/(x-3))=(-12/(x^2-9))

In the text, it says to find the LCD and then expand, eventually moving all terms to the left/setting the equation to zero. From there, it says to factor. Each time I try this out, I get a slightly different solution, and I think I'm skipping up on the very first step, to find the LCD, and possiby still one or two places after that. In the text, the possible solutions are x=3, -9, the only real solution being -9. I understad how they got those, but I'm not sure how they get from the beginning to step 2, with finding the LCD.

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Step 1

Let us simplify the equation : help_outlineImage Transcriptionclose3 2 12 (x2-32) (x-3) 3 or 2 12 (x232) (x-3) 3 2 12 x0, x3 or (*+3)(х- 3) (x 3) 3(x-3) 2x) or 12 x)(x-3) (х +3)(х- 3) х — 9 12 = (x)x-3) (x 3)(x-3) х — 9 12 [Cancelling (x-3) from both side] (x) (x3) (x -9)x3) -12(x) 2 12x) or, x* + 3x -9х — 27 x6x-27 0 or = II N fullscreen
Step 2

Now, let us factorize the q... help_outlineImage Transcriptionclosex26x 27 x(9)x3x + (9)(-3) or xx+9)(3)x +9) 0 (x+9)x+(-3)) 0 or (x+9)0 x (-3)) 0 Thus x -9 x= 3 or fullscreen

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### Calculus 