All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Subtract From Sum And Multiply, Get Number(s) (Posted on 2007-07-19) Difficulty: 2 of 5
Determine all possible real pairs (p, q) satisfying the following system of equations:
                     4p(p+q-5) = 3q
                      q(p+q-4)  = 16p

  Submitted by K Sengupta    
Rating: 3.0000 (1 votes)
Solution: (Hide)
Multiplying both sides of the two equations and simplifying, we obtain:

pq(b^2 -9b+ 8) =0, whenever b=p+q
Or, pq(b-8)(b-1) = 0
Thus, p=0, or q=0, or p+q= 1, 8

We observe that p=0 gives q=0 from the two given equations. Likewise, q=0 yields p=0

Now, substituting p+q=8 in the first equation, we obtain: 4p=q, so that:
(p,q) = (8/5, 32/5)

Substituting p+q =1 in the first equation, we obtain:
16p+3q =0, so that: (p, q) = (-3/13, 16/13)

Consequently, (p, q) = (0,0); (8/5, 32/5); (-3/13, 16/13) constitutes all possible solutions to the given problem.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionQuickFederico Kereki2007-07-20 08:17:25
re: Solutionxdog2007-07-19 22:13:14
SolutionSolutionPraneeth Yalavarthi2007-07-19 10:00:09
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information