Redundant Sets of EquationsDate: 02/07/97 at 01:00:13 From: Harry Crowder Subject: Multi-Variable Equations Hello, I'm hoping you can solve this problem: x + y + z = 50 2x + 3y + 4z = 158 4x + 3y + 2z = 142 Find the values of x, y and z. People here have been tring to solve it for 5 years now, all unsuccessfully. Thanks, Jesse Date: 02/07/97 at 06:09:24 From: Doctor Mitteldorf Subject: Re: Multi-Variable Equations Dear Jesse, Sometimes two equations contain contradictory information, and it's impossible to solve both at the same time. For example: x + y = 7 x + y = 8 Obviously, if x and y have the same values in each equation, x+y can't be both 7 and 8. So you can't solve both equations at the same time. Similarly, it's possible that two equations can really be "saying the same thing," so that if you have both equations, you don't have any more information than you have from each of them separately. For example: x + y = 7 2x + 2y = 14 Whenever the first one is true, the second is also true. Whenever the second is true, the first is also true. There are lots of pairs of numbers x and y that solve this pair of equations because the equations are multiples of each other. With 3 equations, the fact that there is contradictory information or redundant information in the equations can be a little more hidden. It might not be obvious that the three equations are really saying only as much as two equations could say. In your problem, it turns out that if you add the last two equations up (just by adding all the terms on the left and all the terms on the right) and then divide the result by 6, you get exactly the first equation. This means that there is no extra information in the first equation compared with what you'd have if you just had the last two. Therefore, there are lots of triplets of numbers that satisfy all 3 equations. There is no one solution. For example, you could arbitrarily say x = 1, and then you'd have: y + z = 49 (first equation) 3y + 4z = 156 (second equation) If you solve these two, then you'd automatically have a solution to the third equation. Of course, you could also say arbitrarily that x = 2, and do the same thing... By the way, there's a famous test for whether 3 equations (or n equations) have just one solution or whether there's a hidden redundancy. You take the coefficients out of the equations and make them into an array called a matrix. In this case: 1 1 1 2 3 4 4 3 2 Then you find the determinant of the matrix. If the determinant is 0, then the equations are either redundant or contradictory. If the determinant is non-zero, the equations have one solution (they are "independent"). What's a matrix? What's a determinant? These are big questions, and they have interesting answers. If you're interested in exploring further, you can find out more in a second year high school algebra text or right here in the Dr. Math Archives. Just do a search under "matrix" or "determinant" (just the words, not the quotes) at http://mathforum.org/mathgrepform.html -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/