Expert: Josh - 2/9/2005

Question
3/x-2=1/x-1 - 7/(x-1)(x-2)
trying to help my son with a college course, it has been too long since i dealt with this type of math

Answer
Hi Wendy,

The key to this problem is writing everything over a common denominator.

We have 3/(x-2)=1/(x-1)-7/(x-1)(x-2).
You can see that the (x-1) term is no where to be found on the left hand side (LHS). To introduce this factor (i.e., x-1) into 3/(x-2) without changing this quantity, we multiply 3/(x-2) by (x-1) both in the numerator (top) and denominator (bottom) of the fraction.

The idea behind this is that (x-1)/(x-1)=1 and multiplying anything by 1 changes nothing. It still retains the same ratio. [Wendy, please make sure that you get this point across. Use this to lead him to think about what needs to be done to the first term appearing on the right hand side of the equation. Repeat this idea until it sinks in.]

Similarly, with the 1/(x-1) term, the (x-2) term is no where to be found. So, we multiply it by the identity (x-2)/(x-2).

Now, we have
3(x-1)/[(x-2)(x-1)] = (x-2)/[(x-1)(x-2)]-7/[(x-1)(x-2)].

At this point, we can merge the two fractions on the RHS, because the parts (x-2) and 7 are both expressed relative to the same quantity (x-1)(x-2) in the denominator.

We can rewrite the original equation as,
3(x-1)/[(x-2)(x-1)] = [(x-2)-7]/[(x-1)(x-2)]

Furthermore, now, both sides of the quations are expressed relative to the same quantity (x-1)(x-2), so, we can multiply both sides by (x-1)(x-2) to CANCEL OUT the factors in the denominator, leaving us with
3(x-1) = (x-2)-7
[emphasis] without changing the nature of the equation,...although surely, this is a lot easier to manipulate.

Expanding the LHS,
3x-3 = (x-2)-7
3x-3 = x-9 ...next, subtracting x from both sides
2x-3 = -9  ...adding 3 to both sides
2x = -6 ...dividing by 2
answer: x=-3

Cheers