Posted by ayan on Tuesday, April 16, 2013 at 3:41am.
The point (x,y) lies on both conics x2+xy+x=81 and y2+xy+y=51. Given that x+y is positive, determine the value of x+y.

math  Reiny, Tuesday, April 16, 2013 at 8:59am
mmmh, been messing around this for a while
from 1st:
x(x + y + 1) = 81
from 2nd:
y(y + x + 1) = 51
divide them:
x/y = 81/51
81y = 51x > y = 51x/81
sub that into x^2 + xy + x = 81
81x^2 + x(51x/81) + x = 81
81x^2 + 51x^2 + 81x = 6561
132x^2 + 81x  6561 = 0
x = (81 ± √3470769)/ 264
= (81 ± 1863)/264 , which reduces to
= 27/4 or 81/11
ahhh, I guess we could have factored it, lol
132x^2 + 81x  6561 = 0
44x^2 + 27x  2187 = 0
(4x  27)(11x + 81) = 0
anyway, too late,
if x = 27/4 , from y = 51x/81 > y = 17/4
if x = 81/11, > y = 51/11
for first point:
x + y = 27/4 + 17/4 = 44
for 2nd point:
x+y = negative
so for your condition:
x+y = 44
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