Systems with More Variables than Equations
Date: 12/11/2002 at 11:51:14 From: Sridar Subject: Solving systems Dr. Math. How do you solve this system?: 187y + 98x + 45z = 48 2y + 9x + 3z = 198 I tried solving it by trying to multiply equation 2 by 15 to eliminate Z, but do you multiply again by another number to eliminate another variable? I am totally stuck here. P.S. I am in 4th grade but I am learning high school math! Isn't that cool?
Date: 12/11/2002 at 13:11:27 From: Doctor Greenie Subject: Re: Solving systems Hello, Sridar - With three variables and only two equations, you can't do anything more than eliminate one variable. You will end up with a single equation with two variables; the graph of such an equation is a straight line, indicating that there are infinitely many solutions. Usually, with problems like this, it is either stated or implied that the solutions should be integers; in that case, you can perform an analysis to determine the family of distinct solutions. In your particular example, eliminating z gives you the equation 37x - 157y = 2922 The standard approach to finding the integer solutions of an equation like this looks something like this: 37x - 157y = 2922 37x = 157y + 2922 x = (157/37)y + (2922/37) x = 4y + (9/37)y + 78 + 36/37 x = (4y + 78) + (9y + 36)/37 Now since x must be an integer, and since 4y + 78 is an integer, (9y + 36)/37 must also be an integer. 9y + 36 ------- = n for some integer n 37 9(y + 4) -------- = n 37 Since 9 and 37 are relatively prime, y + 4 must be a multiple of 37 for this expression to have an integer value. The smallest positive value for which this is the case is y=33. So we have as our first solution in which both x and y are integers: y = 33 x = (4y + 78) + (9y + 36)/37 = (132 + 78) + (333/37) = 210 + 9 = 219 2y + 9x + 3z = 198 66 + 1971 + 3z = 198 3z = -1839 z = -613 These integer values of x and y yield an integer value for z, so we have our first solution to the original pair of equations in integers. For the family of solutions, we have 9(y + 4) -------- = n 37 where n is an integer. The general solution, for this expression to be an integer, is y = 33 + 37k where k is any integer. To find the general solution, we then have y = 33 + 37k x = (4y + 78) + (9y + 36)/37 x = (132 + 148k + 78) + (297 + 333k + 36)/37 x = (148k + 210) + (9k + 9) x = 157k + 219 2y + 9x + 3z = 198 (66 + 74k) + (1413k + 1971) + 3z = 198 3z = -1487k - 1839 z = -(1487/3)k - 613 The coefficient 1487 is not divisible by 3; this tells us that only values of k which are multiples of 3 will yield integer values for z. So now we can form the expressions for the family of integer solutions to your problem. y = 33 + 37(3k) where k is any integer (not the same k as above) y = 111k + 33 x = (4y + 78) + (9y + 36)/37 x = (444k + 132 + 78) + (999k + 297 + 36)/37 x = (444k + 210) + (27k + 9) x = 471k + 219 2y + 9x + 3z = 198 (222k + 66) + (4239k + 1971) + 3z = 198 4461k + 2037 + 3z = 198 3z = -4461k -1839 z = -1487k - 613 The family of integer solutions to your pair of equations is (x, y, z) = (471k + 219, 111k + 33, -1487k - 613) where k is any integer. Equations like these, where there are more variables than equations, and where solutions are to be integers, are called Diophantine equations. You can find many pages in the Dr. Math archives where similar problems are discussed by performing a search of the archives using the keyword diophantine I hope all this helps. If you have any questions about any of this, try looking at some more examples in the archives. Then write back if you still have questions. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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