1/(x+a) = 2/(x-1) - 1/(x-a). Solve for x? (where a,b are constants)

Question

1/(x+a) = 2/(x-1)  -  1/(x-a). Solve for x? (where a,b are constants)

myininaya · Answer

i seen this problem the other day 
let me see if i can find that question

zarkon · Answer

"where a,b are constants"...there are no b's in your equation

anonymous · Answer

yes b is missing

anonymous · Answer

i think it says that just if there happen to be a b. but yes there is no b so just ignore that part. *given that a is constant*

myininaya · Answer

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e5960d70b8b1f45b47a4e7b

anonymous · Answer

can u plz rectify ur question

anonymous · Answer

x=a

anonymous · Answer

myininaya are you sure it couldnt be simpler?

anonymous · Answer

yeah i remember this one.  was happy for you to do it!

anonymous · Answer

x=a^2

anonymous · Answer

x=a^2

anonymous · Answer

start with 
$$\frac{1}{x+a}=\frac{2}{x-1}-\frac{1}{x-a}$$ multiply all this nonsense on  both sides by 
$$(x+a)(x-1)(x-a)$$ do a raft of algebra and then get 
$$x=a^2$$

anonymous · Answer

yeah right u r.i missed it.

anonymous · Answer

if you need all the annoying algebra steps, click on "show steps" http://www.wolframalpha.com/input/?i=solve+for+x+1%2F%28x%2Ba%29+%3D+2%2F%28x-1%29+-+1%2F%28x-a%29.

anonymous · Answer

by simplifying 
lcd=(x-1)(x+a)(x-a)
(x-1)(x-a)-2[(x+a)(x-a)]+(x-1)(x-a)=0
2x^2+a-x-a-2x^2+a^2-ax-x+a=0
x=a^2