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

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