First clear denominators. The only denominator is x - 3, so that is the smallest common denominator. Multiply both sides by x - 3.

On the left, this means multiplying both terms by x - 3

After cancelling we are left with

Remove parentheses.

Combine like terms.

At this point we see that we have a quadratic, so we transpose all terms to one side leaving a 0 on the other. In this case if we transpose all terms to the right, the square term will be positive.

This will factor

Set the factors = 0

Check: First lets check x = 3. Copy down the original equation

except where there is an x, copy down a 3 in parentheses.

Do the arithmetic

and we get bad news. There are zeroes in denominators. If a solution gives us a zero denomnator, we consider that the solution does not check. So 3 is not a solution.

There is a practical application for this situation. 6/0 is how long it will take you to travel 6 miles if you don't move. 6/0 - 9 is how much longer it will take you to travel 6 miles if you don't move if you started 9 hours ago. In either case, it will take forever, so in this situation, one could argue that the solution checks. But to do so would involve questions of infinite quantities, and most authors will simply tell us that if you get a zero denominator, the solution does not check.

But just because one solution does not check does not mean that we can abandon all hope at this point. Let's check the other solution. Copy down the original equation,

except that wherever there is an x, copy down a 2/3 in parentheses.

It would be advantageous to find common denominators for the subtractions on the bottoms. At the same time we can simplify the other term on the left.

Finish the arithmetic on the tops and bottoms.

We are now ready to invert and multiply on both sides. First we consder the signs. A positive divided by a negative is a negative.

This simplifies to

The left side has been simplified down to a subtraction of fractions. Find common denominators,

and it will check

So this is an example of a rational equation where there are two answers, one of which will check and the other will not. This illustrates that the student will need to check all solutions to a rational equation to see which ones will check and which ones will not.