Equations with Rational Expressions in Two VariablesDate: 12/07/2002 at 10:36:27 From: Tanis Glass Subject: Solving equations with rational expressions in two variables Determine all positive integers a and b that satisfy the equation: 1/a + a/b + 1/ab = 1. I have determined (by trial and error) two possible (a,b) solutions: (a=2, b=5) and (a =3, b=5) I also realize that certain restrictions must be placed on the variables, such as: a, b cannot equal 0; a cannot equal 1; a cannot equal b; and b must be greater than a. However, I have no clue as to how to coordinate this information to come up with a solution that would allow me to figure out more positive integers except by trial and error. This is a grade 11 math course in functions so my background knowledge is limited. Thanks for any help you can give me. Date: 12/07/2002 at 12:10:36 From: Doctor Ian Subject: Re: Solving equations with rational expressions in two variables Hi Tanis, Note that the restriction that a and b be positive integers rules out a=0 and b=0. The first thing that occurs to me would be to use something like Mathematica to make a three-dimensional plot of the surface described by the equation. That is, you'd have values of a along one axis, values of b along the other, and the z axis would contain the values of the function f(a,b) = 1/a + a/b + 1/(ab) - 1 The points where the surface intersects the xy plane would be solutions of the equation. If you plot it only for integer values of a and b, you won't have to worry about non-integer solutions. This also suggests that you could write a small computer program to start checking values of a and b. But what else can we do? One thing we can do is solve for one variable in terms of the other. 1/a + a/b + 1/ab = 1 b/(ab) + a^2/(ab) + 1/(ab) = 1 b + a^2 + 1 = ab a^2 + 1 = ab - b a^2 + 1 = b(a - 1) (a^2 + 1)/(a - 1) = b Now, you still have to use trial and error. But this form makes it a little easier to do. You can start choosing values of a, and see what values of b are forced: a (a^2 + 1)/(a - 1) b --- ----------------- ----- 1 undefined 2 5/1 5 3 10/2 5 4 17/3 5 26/4 6 37/5 On the other hand, unless there is a bug in the computer program I just wrote a few minutes ago, trial and error isn't going to find you any more solutions for values of a up to a million. Which doesn't prove, but strongly suggests, that there are no other solutions. So why would that be the case? Let's try doing the division, and see what happens: a + 1 _____________ a - 1 ) a^2 + 0a + 1 a^2 - a -------- a + 1 a - 1 ----- 2 So (a^2 + 1)/(a - 1) = (a + 1) + 2/(a - 1) and now our table looks like this: a (a + 1) + 2/(a - 1) b --- ------------------- ----- 1 undefined 2 3 + 2/1 5 3 4 + 2/2 5 4 5 + 2/3 5 6 + 2/4 Does this make it easier to see how many other solutions there will be? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 12/07/2002 at 12:33:26 From: Tanis Glass Subject: Thank you (Solving equations with rational expressions in two variables) Thanks very much, Doctor Ian. The use of a graph in 3-D was a bit too complex (for me, anyway) because we have not studied that in school; however, the rest of your help was just great and I understand your explanation. You were a big help. I have consulted the Dr. Math archives in the past for help with other topics and have had success in locating similar problems; but this one had me stumped. Thanks again! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/