Anyone with the knowledge on how to do [(a-x)/(a-b)] -2 = (c-x)/(b-c)? Stuck on this problem in my HW and I need to solve to find x...

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Anyone with the knowledge on how to do [(a-x)/(a-b)] -2 = (c-x)/(b-c)? Stuck on this problem in my HW and I need to solve to find x...

Mathematics
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\[\frac{a-x}{a-b} - 2 = \frac{c-x}{b-c}\] \[\frac{(b-c)(a-x)}{a-b} - 2(b-c) = c-x\] \[\frac{(b-c)(a-x)}{a-b} + (x-c) = 2(b-c)\] \[ab-bx-ac+cx + (x-c)(a-b) = 2(b-c)(a-b)\] \[x(c-b+a-b) + ab -ac -ac+bc = 2(ab -b^2-ac+bc)\] \[x(a-2b+c) = 2(ab -b^2-ac+bc) - ab +2ac -bc\] \[x(a-2b+c) = ab - 2b^2 +bc\] \[x=\frac{b(a-2b+c)}{(a-2b+c)} = b\]
sorry tat last bit is this: \[x=\frac{b(a-2b^2+c)}{(a-2b+c)}\]
it's not =b

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The book says the answer is x=b.. So i'm confused as hell.
Try \[\frac{a-x}{a-b} - 2 = \frac{c-x}{b-c}\] \[\frac{a-x}{a-b} - \frac{c-x}{b-c} = 2\] \[(a-x)(b-c) - (a-b)(c-x) = 2(a-b)(b-c)\] \[ab -ac -bx + cx -ac+ax+bc-bx = 2ab-2ac-2b^2+2bc\] \[x(c-b+a-b) +ab -2ac + bc= 2ab-2ac-2b^2+2bc\] \[x(a-2b+c) = ab -2b^2+bc\] \[x(a-2b+c) = b(a-2b+c)\]
then just divide by (a-2b+c)
to get x=b
LOL I factored it wrong at the very end - lapse o concentration obviously
my first answer was right. sorry about doing it twice lol

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