How do you solve the rational equation #(3a-2)/(2a+2)=3/(a-1)#?

1 Answer
Jan 7, 2016

Answer:

Multiply both sides by #(2a+2)(a-1)#, simplify and solve to find:

#a=-1/3# or #a=4#

Explanation:

Multiply both sides by #(2a+2)(a-1)# to get:

#(3a-2)(a-1) = 3(2a+2)#

That is:

#3a^2-5a+2 = 6a+6#

Subtract #6a+6# from both sides to get:

#3a^2-11a-4 = 0#

Factor this by noticing that #AC=3*4=12# has factors #12# and #1# which differ by #B=11#.

So:

#0 = 3a^2-11a-4#

#= (3a^2-12a) + (a-4)#

#= 3a(a-4)+1(a-4)#

#= (3a+1)(a-4)#

Hence #a=-1/3# or #a=4#

It remains to check that when we multiplied both sides by #(2a+2)(a-1)# we were not multiplying both sides by #0#.

That's OK since #(2a+2) = 0# when #a=-1# and #(a-1) = 0# when #a=1#, so neither of the solutions of the multiplied equation were introduced by the multiplication step.