Different Approach to a Set of Equations
Date: 11/13/2002 at 21:29:56 From: Kyle Hammock Subject: System of Equations Our teacher gave us a bonus question. It contains a system of equations with four variables. No one in our class has answered it yet. It is: The sum of four numbers is 22. The first number is twice the difference of the second and the fourth. The second number is five times the difference of the third and the fourth. The third number is twice the difference of the first and the fourth. What are the four numbers? I think I already have my equations, but I have no clue how to start. Would you please solve the system with steps and explanation? My equations (may be wrong) are as followed: A+B+C+D=22 A+B+C+D=22 I have no idea how to A=2(B-D) A=2B-2D start. I have tried B=5(C-D) B=5C-5D many different ways, C=2(A-D) C=2A-2D but none have worked. Thanks a lot, Kyle Hammock
Date: 11/14/2002 at 01:45:00 From: Doctor Greenie Subject: Re: System of Equations Hello, Kyle - I don't think you can solve this problem by setting up 4 equations with 4 unknowns. The reason is that the word "difference" doesn't tell you which number is larger. For example, you have, for your second equation, A = 2(B-D) whereas, in fact, it might be the case that A = 2(D-B) You could write your 4 equations as A+B+C+D = 22 A = 2|B-D| B = 5|C-D| C = 2|A-D| where |B-D| is the absolute value of (B-D). But I don't know how to solve systems of equations involving absolute values. I think you need to use a different approach to this problem. The first thing I did with this problem was to assume that, while not specifically so stated, the numbers were all positive integers. I guessed that finding a solution would be difficult without that assumption. Once I had made that assumption, I started by using the given information about the first, second, and third numbers to say A is even B is a multiple of 5 C is even Then I said: suppose B, being a multiple of 5, is in fact equal to 5. Then I know (1) D must be odd (because the sum of all 4 numbers is even) (2) A = twice the difference between B and D (3) C = twice the difference between A and D So now I try each odd possibility for D to see if one of them leads to values for A and C that satisfy the condition that the sum of the four numbers is 22: if D (odd) is then A is and C is and A+B+C+D equal to equal to equal to is equal to --------------------------------------------------- 1 2(5-1)=8 2(8-1)=14 28 3 2(5-3)=4 2(4-3)=2 14 5 2(5-5)=0 2(5-0)=10 20 7 2(7-5)=4 2(7-4)=6 22 9 2(9-5)=8 2(9-8)=2 24 The solution to the problem is (A,B,C,D) = (4,5,6,7) Checking: A+B+C+D = 4+5+6+7 = 22 check A = 2|B-D| = 2(2)=4 check B = 5|C-D| = 5(1)=5 check C = 2|A-D| = 2(3)=6 check I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.