How do you solve #1=1/(1-a)+a/(a-1)#?

2 Answers
Mar 12, 2018

Answer:

The solution is #x inRR;x!=1#.

(This means #x# can be any number but #1#.)

Explanation:

Get a common denominator, then simplify the expression:

# 1=1/(1-a)+a/(a-1) #

# 1color(blue)(\*(a-1))=1/(1-a)color(blue)(\*(a-1))+a/color(red)cancelcolor(black)(a-1)color(red)cancelcolor(blue)(*(a-1)) #

# a-1=(a-1)/(1-a)+a #

# a-1=(a-1)/(-a+1)+a #

# a-1=(a-1)/(-1(a-1))+a #

# a-1=color(red)cancelcolor(black)(a-1)/(-1color(red)cancelcolor(black)((a-1)))+a #

# a-1=1/(-1)+a #

# a-1=-1+a #

# a-1=a-1 #

# a=a #

Since this is true for all values of #a#, the solution should be #x inRR# (#x# is all real numbers), but it isn't because of the original problem. Take a look at the denominator of those fractions:

# 1=1/(1-a)+a/(a-1) #

If #a# is #1#, then there is a division by zero, and that can't happen. Therefore, #x# can be any number but #1#.

The final solution set is #x inRR;x!=1#.

Mar 12, 2018

Answer:

You can't solve this, because this is an identity, valid for all values of #a# (other than #a=1#) . You can, of course, prove this - see below

Explanation:

#1 = (1-a)/(1-a) = 1/(1-a) -a/(1-a) =1/(1-a)+a/(a-1)#