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Algebraic expressions and fractions

Date: 04/10/97 at 18:39:36
From: D.J. Dean
Subject: Algebraic expressions and fractions

I have been working on this problem for a while now and I still 
can't figure it out.  I have asked my math teacher at school, my 
friends, and my parents.  We can't seem to find the answer.  
Could you please help?  

I am using ^ as a symbol for "to the second power". 
   X/A + Y/B + Z/C = 1 and 
   A/X + B/Y + C/Z = 0 

Then what does 
   (X^2)/(A^2) + (Y^2)/(B^2) + (Z^2)/(C^2) equal?

Thank you for your help.

Date: 04/11/97 at 19:53:55
From: Doctor Wallace
Subject: Re: Algebraic expressions and fractions

Hi D.J.!

That's quite a problem you have there.  Solving it will stretch
your algebra skills quite a bit!

I'll give you some pushes in the right direction, and then let you 
take it from there.

You are given two vital facts:

     x     y     z                     a     b     c
    --- + --- + ---  = 1    and       --- + --- + ---   = 0.    
     a     b     c                     x     y     z

We are asked to find    x^2      y^2       z^2
                       ----- +  -----  +  ----- .
                        a^2      b^2       c^2 

Now, solving this is going to require a great facility on your part to 
work with fractions, denominators, and so forth.  It's no different 
from doing similar work with fractions whose values you know, like 
1/3, 1/4, etc. The difference is that these fractions are generalized.

So, what's our first step?  Well, what I did was take a look at what 
they asked us to find.  This expression has squares in all the terms.  
How might I work with or combine the two given expressions to come up 
with squares?

I'm going to have to multiply x by x to get x^2, right?  And b by b to 
get b^2, and so on.  So multiplication has to be involved somewhere, 
and it must be term matching term, or I won't get the squares.  So how 
can I, using the first of the 2 expressions, make x/a into x^2/a^2?

I do this by multiplying it by x over a.  This gives

            x          x        x^2
           ---    *   ---   =  ------ .
            a          a        a^2

Right?  Now I have my first term.  However, in order to preserve the 
equation, I have to multiply the WHOLE side AND the other WHOLE side 
by the same amount.  I can't just multiply the one term, or my 
equation will be thrown out of balance.  So I multiply both sides by 
x/a and get:

        x^2        xy      xz       x
       -----   +  ---- +  ----  =  --- .
        a^2        ab      ac       a

Are you with me so far?  All I did was multiply the first of the two 
given equations by x/a on both sides.  Now I solve for x^2/a^2 by 
subtracting the two terms with it, and I find that:

        x^2        x       xy        xz
      -------  =  ---  -  ----   -  ----
        a^2        a       ab        ac

Now we're making progress!  We know what one of the three terms we're 
supposed to add up equals.  We can use the exact same procedure to 
find out what y^2/b^2 equals and what z^2/c^2 equals.  For the former, 
we multiply both sides of the original expression by y/b, and by z/c 
for the latter.  Then we subtract as I did above.

At this point you will have an expression for all three of the squared 
terms.  Then you add them up, since we are asked to find the sum of 
the three squared terms.

Once you have done that, the resulting expression can be simplified 

To simplify it and finish off the problem, you will use BOTH the 
original equations we are given.  Notice that we have already used the 
first three times.  You'll use it again in the final simplification.

We haven't yet used the second equation, the one that equals 0. You'll
need to play with it a bit to complete the simplification. (Hint: get 
rid of the denominators first.)

I don't want to deprive you of the satisfaction I know you'll get when
you've finally solved this problem, so I'll stop there.  Where did you 
find this problem?  It's a good one!

If you get stuck again, or just need more explanation, or both, don't 
hesitate to write again.  And, if you don't need more help, then 
please write again when you solve the problem and let us know how it 

Thanks for writing!

-Doctor Wallace,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Basic Algebra
High School Polynomials

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