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Rath111

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    \[\frac{a-x}{a-b} - 2 = \frac{c-x}{b-c}\] \[\frac{(b-c)(a-x)}{a-b} - 2(b-c) = c-x\] \[\frac{(b-c)(a-x)}{a-b} + (x-c) = 2(b-c)\] \[ab-bx-ac+cx + (x-c)(a-b) = 2(b-c)(a-b)\] \[x(c-b+a-b) + ab -ac -ac+bc = 2(ab -b^2-ac+bc)\] \[x(a-2b+c) = 2(ab -b^2-ac+bc) - ab +2ac -bc\] \[x(a-2b+c) = ab - 2b^2 +bc\] \[x=\frac{b(a-2b+c)}{(a-2b+c)} = b\]

  2. anonymous
    • 5 years ago
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    sorry tat last bit is this: \[x=\frac{b(a-2b^2+c)}{(a-2b+c)}\]

  3. anonymous
    • 5 years ago
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    it's not =b

  4. Rath111
    • 5 years ago
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    The book says the answer is x=b.. So i'm confused as hell.

  5. anonymous
    • 5 years ago
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    Try \[\frac{a-x}{a-b} - 2 = \frac{c-x}{b-c}\] \[\frac{a-x}{a-b} - \frac{c-x}{b-c} = 2\] \[(a-x)(b-c) - (a-b)(c-x) = 2(a-b)(b-c)\] \[ab -ac -bx + cx -ac+ax+bc-bx = 2ab-2ac-2b^2+2bc\] \[x(c-b+a-b) +ab -2ac + bc= 2ab-2ac-2b^2+2bc\] \[x(a-2b+c) = ab -2b^2+bc\] \[x(a-2b+c) = b(a-2b+c)\]

  6. anonymous
    • 5 years ago
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    then just divide by (a-2b+c)

  7. anonymous
    • 5 years ago
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    to get x=b

  8. anonymous
    • 5 years ago
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    LOL I factored it wrong at the very end - lapse o concentration obviously

  9. anonymous
    • 5 years ago
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    my first answer was right. sorry about doing it twice lol

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