The number of distinct pairs (x,y) of real numbers satisfying x = x^3 + y^4 and y = 2xy is:
A)5
B)12
C)3
D)7
$\displaystyle y = 2xy$
$\displaystyle y - 2xy = 0$
$\displaystyle y(1 - 2x) = 0$
$\displaystyle y = 0$ or $\displaystyle x = \frac{1}{2}$
for $\displaystyle y = 0$ ...
$\displaystyle x = x^3 + y^4$
$\displaystyle x = x^3$
$\displaystyle x - x^3 = 0$
$\displaystyle x(1 - x^2) = x(1 - x)(1 + x) = 0$
$\displaystyle (0,0) $, $\displaystyle (1,0)$ and $\displaystyle (-1,0)$
for $\displaystyle x = \frac{1}{2}$ ...
$\displaystyle \frac{1}{2} = \frac{1}{8} + y^4$
$\displaystyle y^4 - \frac{3}{8} = 0$
$\displaystyle \left(y^2 + \sqrt{\frac{3}{8}}\right)\left(y^2 - \sqrt{\frac{3}{8}}\right) = 0$
$\displaystyle \left(y^2 + \sqrt{\frac{3}{8}}\right)\left(y - \sqrt[4]{\frac{3}{8}}\right)\left(y + \sqrt[4]{\frac{3}{8}}\right) = 0$
$\displaystyle \left(\frac{1}{2}, \sqrt[4]{\frac{3}{8}}\right)$ and $\displaystyle \left(\frac{1}{2}, -\sqrt[4]{\frac{3}{8}}\right)$
looks like 5 distinct solutions to me ... hope I didn't miss any.