Problem:
If ( x , y ) (x, y)( x , y ) is a solution to the system
x y = 6 and x 2 y + x y 2 + x + y = 63 x y=6 \quad \text { and } \quad x^{2} y+x y^{2}+x+y=63
x y = 6 and x 2 y + x y 2 + x + y = 6 3
find x 2 + y 2 x^{2}+y^{2}x 2 + y 2 .
Answer Choices:
A. 13 131 3
B. 1173 32 \dfrac{1173}{32}3 2 1 1 7 3 ​
C. 55 555 5
D. 69 696 9
E. 81 818 1
Solution:
One can solve for x xx and y yy individually, and then substitute, but it's a mess: x = 9 ± 57 2 x=\dfrac{9 \pm \sqrt{57}}{2}x = 2 9 ± 5 7 ​ ​ and y = 9 ∓ 57 2 y=\dfrac{9 \mp \sqrt{57}}{2}y = 2 9 ∓ 5 7 ​ ​ . It's sufficient, and easier, to solve for x + y x+yx + y :
63 = x 2 y + x y 2 + x + y = x y ( x + y ) + ( x + y ) = ( 6 + 1 ) ( x + y ) 63=x^{2} y+x y^{2}+x+y=x y(x+y)+(x+y)=(6+1)(x+y)
6 3 = x 2 y + x y 2 + x + y = x y ( x + y ) + ( x + y ) = ( 6 + 1 ) ( x + y )
so x + y = 9 x+y=9x + y = 9 and
x 2 + y 2 = ( x + y ) 2 − 2 x y = 81 − 12 = 69 x^{2}+y^{2}=(x+y)^{2}-2 x y=81-12=69
x 2 + y 2 = ( x + y ) 2 − 2 x y = 8 1 − 1 2 = 6 9