Suppose x and y are real numbers both greater than 1 that satisfy xy + x + y = 19 and x2y + y2x = 84. Find x2 + y2.
xy + x + y = 19
xy + (x+y) = 19
x^2y + y^2x = 84
(xy)(x+y) = 84
p = xy
s = x+y
p + s = 19
p * s = 84
(19-s)*s = 84
s^2 - 19s + 84 = 0
s = [19 ±√(361 - 336)]/2
s = [19 ± 5]/2
{s, p} = {7, 12}
case 1:
x+y = 7
xy = 12
(7-x)x = 12
x^2 - 7x + 12 = 0
{x,y} = {3,4}
x^2 + y^2 = 25
case 2:
x+y = 12
xy = 7
(12-x)x = 7
x^2 - 12x + 7 = 0
x = 6 +/- √29
y = 6 -/+ √29
36 + 29
x^2 + y^2 = 2*36 + 2*29=130
However case 2 is rejected because 6 - √29 < 1.
The solution set for x and y is {x,y} = {3,4} and
the only solution for x^2 + y^2 is 25.
https://www.desmos.com/calculator/kdxppc3eql
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Posted by Larry
on 2025-01-13 11:26:11 |
Solution
