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

$$y=4-x^2$$
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
