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Subtract From Sum And Multiply, Get Number(s) (Posted on 2007-07-19) |
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Determine all possible real pairs (p, q) satisfying the following system of equations:
4p(p+q-5) = 3q
q(p+q-4) = 16p
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Submitted by K Sengupta
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Rating: 3.0000 (1 votes)
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Solution:
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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.
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Comments: (
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Subject |
Author |
Date |
| Quick | Federico Kereki | 2007-07-20 08:17:25 |
| re: Solution | xdog | 2007-07-19 22:13:14 |
| Solution | Praneeth Yalavarthi | 2007-07-19 10:00:09 |
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