How do you solve for x in #[ (a-cx)/(b+dx) ]+ a = 0#?

1 Answer
Jun 17, 2018

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(a)# from each side of the equation:

#[(a - cx)/(b + dx)] + a - color(red)(a) = 0 - color(red)(a)#

#[(a - cx)/(b + dx)] + 0 = -a#

#(a - cx)/(b + dx) = -a#

Next, multiply each side of the equation by #color(red)((b + dx))# to eliminate the fraction while keeping the equation balanced:

#color(red)((b + dx)) xx (a - cx)/(b + dx) = -acolor(red)((b + dx))#

#cancel(color(red)((b + dx))) xx (a - cx)/color(red)(cancel(color(black)(b + dx))) = (-a xx color(red)(b)) - (a xx color(red)(dx))#

#a - cx = -ab - adx#

Then, add #color(red)(cx)# and add #color(blue)(ab)# to each side of the equation to isolate the #x# terms while keeping the equation balanced:

#a + color(blue)(ab) - cx + color(red)(cx) = -ab + color(blue)(ab) + color(red)(cx) - adx#

#a + ab - 0 = 0 + cx - adx#

#a + ab = cx - adx#

#a + ab = (c - ad)x#

Now, divide each side of the equation by #color(red)(c - ad)# to solve for #x# while keeping the equation balanced:

#(a + ab)/color(red)(c - ad) = ((c - ad)x)/color(red)(c - ad)#

#(a + ab)/(c - ad) = (color(red)(cancel(color(black)((c - ad))))x)/cancel(color(red)(c - ad))#

#(a + ab)/(c - ad) = x#

#x = (a + ab)/(c - ad)#

Or

#x = (a(1 + b))/(c - ad)#