Answer to Question #125928 in Algebra for Ojugbele Daniel

Question

x⁴+x²y²+y⁴ = 133
x²+xy+y² = 19

Solve for x and y

Answer 1

Let $x^2+y^2=u, xy=v.$ Then $x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=u^2-v^2$

$u^2-v^2=133$ $u+v=19$

$(u-v)(u+v)=133$ $u+v=19$

$u-v=7$ $u+v=19$

$u=13$ $v=6$

Hence

$x^2+y^2=13$ $xy=6$

$(x+y)^2 -2xy=13$ $xy=6$

$(x+y)^2=25$ $xy=6$

If $x+y=5$

$(2, 3), (3, 2)$

If $x+y=-5$

$(-3, -2), (-2, -3)$

Therefore

$(-3, -2), (-2, -3), (2, 3), (3, 2)$