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Redundant Sets of Equations

Date: 02/07/97 at 01:00:13
From: Harry Crowder
Subject: Multi-Variable Equations


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 
         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 

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   

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

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