System of quadratic equations $x^2 + y = 4$ and $x + y^2 = 10$

Question

How would you solve the following system of equations: $$ x^2 + y = 4 \\ x + y^2 = 10 $$ Thanks very much!

I tried defining y in terms of x and then inserting in to the second equation:

$$ y = 4 - x^2 \\ x + (4 - x^2)^2 = 10 $$

Expand the second equation:

$$ x + 16 - 8x^2 + x^4 = 10 $$

Rearrange the terms:

$$ x^4 - 8x^2 +x + 6 = 0 $$

I tried factoring out this polynomial to simplify it for solving, but didn't succeed :(

Answer 1

$$y=4-x^2$$ put it in 2nd equation

$$x+(4-x^2)^2=10$$ $$x+16+x^4-8x^2=10$$ $$x^4-8x^2+x+6=0$$ $$x^4-x^3+x^3-7x^2-x^2+7x-6x+6=0$$ $$x^4-x^3+x^3-x^2-7x^2+7x-6x+6=0$$

$$x^3(x-1)+x^2(x-1)-7x(x-1)-6(x-1)=0$$ $$(x-1)(x^3+x^2-7x-6)=0\implies x-1=0\implies x=1$$

solve the above equation one of the value of x=1 and y=3

Answer 2

To solve general equations of the form $ax^4+bx^3+cx^2+dx+e=0$ requires quartic formula.