This example is intended to show you what not to do when solving rational equations. Not because it is wrong – everything done here is correct – but because it makes the problem much harder than it needs to be.
Your work will always be easier if you clear the denominators in the equation by multiplying by the LCM of all the denominators. So practice finding the LCM if necessary. Here is a Khan Academy video example.
This is one of the versions of problem 4 on Test 3. To see a solution which avoids the complexities of this, look at the Test 3 solutions
Solve $3 + \frac{9}{x-3} = \frac{27}{x^{2} – 3x}$
Suppose my first thought is to combine the two terms on the left-hand side. Nothing wrong with that, provided I change to a common denominator first.
$3 + \frac{9}{x-3} = \frac{27}{x^{2} – 3x}$
$\frac{3(x-3)}{x-3} + \frac{9}{x-3} = \frac{27}{x^{2} – 3x}$
$\frac{3x – 9}{x-3} + \frac{9}{x-3} = \frac{27}{x^{2} – 3x}$
$\frac{3x}{x-3} = \frac{27}{x^{2} – 3x}$
And now, instead of clearing the denominators using the LCM, I will just cross-multiply. Nothing wrong with that either. Cross-multiplying is the same as clearing the denominators, except that instead of the LCM it (covertly) uses the product of the denominators.
$3x\right(x^{2} – 3x\left) = 27(x-3)$
$3x^{3} – 9x^{2} = 27x – 81$
$3x^{3} – 9x^{2} – 27x + 81 = 0$
This is now a thrid-degree (cubic) equation! They can be very hard to solve in general. If I am lucky, this can be solved by factoring and using the Zero Product Property. Let’s see.
I notice that I can divide the whole equation by 3, and get
$x^{3} – 3x^{2} – 9x + 27 = 0$
The only technique of factoring that we have learned in this course which applies to a polynomial with four terms is factoring by grouping. It does not always work, but let’s try it.
Group:
$\left(x^{3} – 3x^{2}\right) – (9x – 27) = 0$
Factor out the common factor in each group:
$x^{2}\left(x – 3\right) – 9(x – 3) = 0$
And how do I know that my plan will work?* Because the factors in parentheses are the same! (If they were not the same, I could not factor this way.)
$\left(x^{2} – 9\right)\left(x – 3\right) = 0$
factor some more:
$\left(x – 3\right)(x + 3)\left(x – 3\right) = 0$
Two of those factors are the same, so the Zero Product Property says that
either $x-3 = 0$ or $x+3 = 0$
$\implies x = 3$ or $x = -3$.
But I have to check that these solve the original equation. $x=3$ gives a zero denominator,
so the only solution is $x = -3$
The complication could have been avoided by using the LCM of the denominators.
* At this point, I said, of course this reminds me of something, everything reminds me of something, and everything reminds me of my dog even though I don’t have a dog. Can you guess what I was reminded of? Hint: Hamilton, as usual.
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