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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