Question

Find all orders pairs (x, y) that satisfy the system:

Answer 1

\[x ^{3}-x ^{2}y+5xy ^{2}-y ^{3}=124\]

\[x ^{3}-5x ^{2}y+xy ^{2}-y ^{3}=4\]

Answer 2

ok finally got it :)

Answer 3

x3−x2y+5xy2−y3=124 R1
x3−5x2y+xy2−y3=4 R2
R1-R2 => -x2y+5xy2+5x2y-xy2 = 120
(-x2y - xy2) + (5xy2 + 5x2y) =120
-xy(x+y) + 5xy(x+y) =120
4xy(x+y)=120 (*)
no lets go back to R1
x3 - x2y + 5xy2 - y3 =124
x3 - 3x2y +2x2y + 3xy2 + 2xy2 - y3 = 124
(x3 - 3x2y + 3xy2 -y3) + ( 2x2y + 2xy2) =124
(x-y)3 + 2xy(x+y) =124 (**)
replace (*) into (**)
(x-y)3 + 60 = 124
(x-y)3=64
x-y = 4
x=4+y repalce this back into (*)
4(4+y)(4+y+y)=120
then solve for y

Answer 4

Doesn't R1-R2 => 4x2y+4xy2=120

Answer 5

oh yes, my bad, u're right

Answer 6

I did the whole thing out and got \[x=\sqrt{30\div(y+y ^{2})}\]

Answer 7

How does that help when I plug it back into the first equation?

Answer 8

4(4+y)(4+y+y)=120
8(y2+6y+8)=120
y2+6y+8=15
solve for y then u get y=-2 and y=-4