Solving Rational Equations by Using the LCD

Date: 09/27/2007 at 19:32:27
From: Nikki
Subject: Literal Equations/Quadratic Equations

How do you solve the equation 3/(n-2) + 6/(n^2-5n+6) = 4/(n-3)?

Whenever I try to solve the equation and try to put it in quadratic 
form, the solution I get from the quadratic formula doesn't work 
when I substitute it back into the equation.  Please, I also need the 
steps to solving the equation right from the beginning. 

Solving these types of equations are entirely new to me and it is a 
school homework assignment and I have no clue where to begin.  The 
only thing I know is that I have to put it in quadratic form.

Date: 09/27/2007 at 22:49:39
From: Doctor Peterson
Subject: Re: Literal Equations/Quadratic Equations

Hi, Nikki.

You say you got a solution, but it didn't work.  What was that
solution?  I'd want to be sure you didn't just make a mistake in the
check before saying too much else.

I would approach this by first factoring the quadratic denominator,
and then finding the LCD, which will be a lot simpler than it would
have looked without having factored.  Then I'd multiply each term by
that LCD and cancel before doing any actual multiplying.  That would
leave me with an equation, possibly quadratic, to solve.

But it is entirely possible to get an extraneous root when you do
this, because you have multiplied by a variable expression, which
could be worth zero.  So you do have to check, and if each solution 
you found would result in division by zero, then they are not really
solutions of the original equation.  That's perfectly normal.

In this case, though, you must have made a mistake, because when I
solve it I don't get a quadratic equation at all.  It will help if you
can show me what you did, so I can identify what went wrong.  My guess
is that you didn't cancel correctly, or you used the wrong LCD.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/28/2007 at 16:44:15
From: Nikki
Subject: Literal Equations/Quadratic Equations

Hi,

Thanks for your answer.  I didn't solve it your way.  I solved it by
first multiplying all the denominators together.  Then I multiplied
them to the numerator.

I got 3n^3 - 15n^2 + 18n - 9n^2 + 45n - 54 + 6n^2 - 18n - 12n + 36 =
4n^3 - 20n^2 + 24n - 8n^2 + 40n - 48, and that is all over a common
denominator that I got when I multiplied them together.

Then I combined the like terms:

-n^3 + 10n^2 - 31n + 30 = 0

I'm not sure my solution is right because I then divided 0 by -n^3 
to get a quadratic equation. 

Are my steps to solving correct?  If not, could you please direct me 
in the right steps?  And I'm not sure what you mean by finding the 
LCD of the denominators.  Do I have to find the LCD of each of the 
denominators, or just the one big denominator?

Thanks, 

Nikki

Date: 09/28/2007 at 21:20:36
From: Doctor Peterson
Subject: Re: Literal Equations/Quadratic Equations

Hi, Nikki.
 
>I didn't solve it your way.  I solved it by first multiplying all the 
>denominators together.  Then I multiplied them to the numerator.

You can do that, but it makes the problem a LOT harder.  You're 
finding a common denominator by just multiplying everything together, 
where I'm finding the LEAST common denominator, which means a lot less 
multiplication and a lot less factoring to get things to simplify 
again.  You really need to learn my method, rather than practicing the
hard way.  Math is not about making life hard; it's about learning how
to simplify things and save work.

>I got 3n^3 - 15n^2 + 18n - 9n^2 + 45n - 54 + 6n^2 - 18n - 12n + 36 =
>4n^3 - 20n^2 + 24n - 8n^2 + 40n - 48, and that is all over a common
>denominator that I got when I multiplied them together.

It looks like you got this from

  3(n-3)(n^2-5n+6) + 6(n-2)(n-3) = 4(n-2)(n^2-5n+6)

This is the correct numerator if you use your denominator.

>Then I combined the like terms:
>
>-n^3 + 10n^2 - 31n + 30 = 0

That is what I get.

>I'm not sure my solution is right because I then divided 0 by -n^3 
>to get a quadratic equation. 

No, you can't do that.  The -n^3 is ADDED to the rest; if you divide
the left side by -n^3, it won't just remove that term.  Try it!  Make
sure you always use the inverse operation to undo something, and check
that it really does what you want to do.  When something is added, you
can only remove it by subtracting, not by dividing; when something is
multiplied, you can only remove it by dividing, not by subtracting.

>Are my steps to solving correct?  If not, could you please direct me 
>in the right steps?  And I'm not sure what you mean by finding the 
>LCD of the denominators.  Do I have to find the LCD of each of the 
>denominators, or just the one big denominator?

Now you have to solve a cubic equation.  That's not easy; you pretty
much have to guess at a factor, and it can take a long time.  It's not
incorrect, really, but it wastes a lot of time.

What you've done is like adding two ordinary fractions without using
the LCD:

   7     5    7*84 + 5*120    1188
  --- + --- = ------------ = -----
  120    84     120*84       10080

Now you've got to simplify that monstrosity!  You can do it, but it's 
a lot of unnecessary work, considering that you just MADE these huge
numbers by multiplying, and now you have to undo them by factoring.
That's hard with numbers; with polynomials, it can be impossible.  The
smart thing to do is to use the LCD, which in this case is 840:

   7     5    7*7 + 5*10    99
  --- + --- = ---------- = ---
  120    84      840       840

Now you have a lot less simplifying; I get 33/280.

Presumably before you would be given a problem like yours, you were
shown how to find the LCD, because that is THE method both for solving
rational equations like this, and for adding rational expressions,
just as it is for adding fractions.  There is no such thing as the LCD
of ONE denominator; the term LCD means Least COMMON Denominator, that
is, the smallest number that can be used as a denominator by ALL of
the fractions.  It is the least common multiple of the denominators.
The way to find it is to factor each denominator, and then choose only
those factors that you need in order to get a multiple of each.  In
your case, the denominators are

  n-2   (n-2)(n-3)   n-3

To make a common multiple, you need at least one (n-2), and at least
one (n-3); the LCM is therefore (n-2)(n-3).  Use this, and the whole
thing will be incredibly easier.  You can either do the addition and
then multiply by the (new) denominator, or just multiply by the LCD
immediately and cancel, so you never have to write the denominator
again.  Your text probably shows how to do this.

It's probably a good experience for you to have done all that work,
and then to see how much easier it can be.  Hopefully you'll learn 
well that this is a useful method!

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/28/2007 at 22:55:48
From: Nikki
Subject: Thank you (Literal Equations/Quadratic Equations)

Thanks!  The solution was 5!  This was a really big help to me!

Nikki

Date: 09/28/2007 at 23:15:21
From: Doctor Peterson
Subject: Re: Thank you (Literal Equations/Quadratic Equations)

Hi, Nikki.

Let's do one final step, which you always have to do: check your
answer. Plugging in 5, we have

    3          6           4
  ----- + ------------ = -----
  n - 2   n^2 - 5n + 6   n - 3

    3           6           4
  ----- + ------------- = -----
  5 - 2   5^2 - 5*5 + 6   5 - 3

   3     6     4
  --- + --- = ---
   3     6     2

   1  +  1  =  2

Yes, it does work.  And by doing the check, we made sure that we 
didn't end up dividing by 0, which would have made it an extraneous 
solution, which means not really a solution at all.

Good work.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/