## 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...

1. anonymous

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

2. anonymous

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

3. anonymous

it's not =b

4. Rath111

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

5. anonymous

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

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

7. anonymous

to get x=b

8. anonymous

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

9. anonymous