Question

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

Answer 1

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.