Question

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...

Answer 1

\[\frac{a-x}{a-b} - 2 = \frac{c-x}{b-c}\]
\[\frac{(b-c)(a-x)}{a-b} - 2(b-c) = c-x\]
\[\frac{(b-c)(a-x)}{a-b} + (x-c) = 2(b-c)\]
\[ab-bx-ac+cx + (x-c)(a-b) = 2(b-c)(a-b)\]
\[x(c-b+a-b) + ab -ac -ac+bc = 2(ab -b^2-ac+bc)\]
\[x(a-2b+c) = 2(ab -b^2-ac+bc) - ab +2ac -bc\]
\[x(a-2b+c) = ab - 2b^2 +bc\]
\[x=\frac{b(a-2b+c)}{(a-2b+c)} = b\]

Answer 2

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

Answer 3

it's not =b

Answer 4

The book says the answer is x=b.. So i'm confused as hell.

Answer 5

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

Answer 6

then just divide by (a-2b+c)

Answer 7

to get x=b

Answer 8

LOL I factored it wrong at the very end - lapse o concentration obviously

Answer 9

my first answer was right. sorry about doing it twice lol