if xy=6 and (x^2)y +(y^2)x + x + y =63 find x^2 + y^2=??????/?

Answer 1

Feeling my way here.
x^2 + y^2
= (x+y)^2 - 2xy
=(x+y)^2 - 12

Second equation can be written as
xy(x+y) + x+y = 63.
Since xy = 6 this becomes
7(x+y) = 63
so
x+y = 9

Substitute this in the top equation:
x^2 + y^2 = 81-12
=69

Answer 2

xy = 6
(x^2)y + (y^2)x + x + y = 63

xy = 6
y = 6/x

(x^2)(6/x) + ((6/x)^2)x + x + (6/x) = 63
6x + (36/x) + x + (6/x) = 63
7x + (42/x) = 63
7x^2 + 42 = 63x
7x^2 - 63x + 42 = 0
x^2 - 9x + 6 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (9 ± sqrt(81 - 24))/2
x = (9 ± sqrt(57))/2

y = 6/((9 ± sqrt(57))/2) = 12/(9 ± sqrt(57))

Now just plug those values in

(((9 ± sqrt(57))/2)^2 + (12/(9 ± sqrt(57))^2

((9 ± sqrt(57))(9 ± sqrt(57)))/4
(81 ± 9sqrt(57) ± 9sqrt(57) + 57)/4
(138 ± 18sqrt(57))/4
(69 ± 9sqrt(57))/2

(12/(9 ± sqrt(57)))^2 =
144/(138 ± 18sqrt(57)) =
72/(69 ± 9sqrt(57))

((69 ± 9sqrt(57))/2) + (72/(69 ± 9sqrt(57)))
((69 ± 9sqrt(57))^2 + 144)/(2(69 ± 9sqrt(57)))

(69 ± 9sqrt(57))(69 ± 9sqrt(57)) + 144 =
4761 ± 621sqrt(57) ± 621sqrt(57) + (81 * 57) + 144 =
4905 ± 1242sqrt(57) + 4617 =
9522 ± 1242sqrt(57)

(9522 ± 1242sqrt(57))/(2(69 ± 9sqrt(57))) =
(3174 ± 414sqrt(57))/(2(23 ± 3sqrt(57))) =
(1587 ± 207sqrt(57))/(23 ± 3sqrt(57)) =
69

x^2 + y^2 = 69

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you can also do it like this

(x^2)y + (y^2)x + x + y = 63
(xy)x + (xy)y + x + y = 63
(xy)(x + y) + (x + y) = 63
(xy + 1)(x + y) = 63
(6 + 1)(x + y) = 63
7(x + y) = 63
x + y = 9

xy = 6
x + y = 9

x + y = 9
y = -x + 9

x(-x + 9) = 6
-x^2 + 9x = 6
x^2 - 9x + 6 = 0

And the rest follows like above.