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Four positive integers satisfy the following equations:
pq+2p+q = 348

qr+4q+3r = 208

rs+8r+6s = 722

 

What are p, q, r, and s?

 Nov 15, 2020
 #1
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p q + 2 p + q = 348,
q r + 4 q + 3 r = 208,
r s + 8 r + 6 s = 722, solve for p, q, r, s

 

r = 16 and s = 27 and p = 34 and q = 8

 Nov 15, 2020
 #2
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I/m using a technique that I "borrowed" from a really good  mathematician on  here, Alan  !!!!

 

pq  + 2p + q  =  348

qr  + 4q  + 3r  = 208

rs  + 8r  + 6s =  722

 

Manipulating the first equation

p (q + 2)  =  (348 - 1q)

p = (348 - 1q) / (q + 2)

p = [ (350 - 1( q + 2) ] / (q + 2)

p = (350) / ( q + 2) -  1 

 

Manipulating the  second equation

qr + 4q + 3r = 208

q ( r + 4) =  208 - 3r

q = ( 208 - 3r) ( r + 4)

q =  [ (220 -3(r + 4) ] / (r + 4)

q = (220) / (r + 4) - 3

 

Manipulating the  third  equation

rs + 8r + 6s  = 722

r (s + 8)  =  722 - 6s

r = (722 - 6s) / ( s + 8)

r = [722 - 6(s + 8)] / (s + 8)

r = [ 770 - 6(s +8) ] / (s + 8)

r = [ 770 ] / (s + 8) -  6

 

Note  that  770  factors  as   2 * 5 * 7  * 11 =  2 * (35) * 11 = 2 * (27 + 8) * 11

 

If we let s = 27     then we have that

 

r = 770 / 35 - 6  =   22 - 6  =  16

q=  (220) / (16 + 4)  - 3  =  11 - 3 = 8

p = (350) / ( 8 + 2) - 1  =  35 - 1  =  34

 

 

cool cool cool

 Nov 15, 2020

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