Question

The number of distinct pairs $(x,y)$ of the real numbers satisfying $x=x^3+y^4$ and $y=2xy$ is

• A. $5$
• B. $12$
• C. $3$
• D. $7$

Answer 1

Answer: (A)

$5$

$x=x^3+y^4$

Now, putting $y=0$in this equatio we get

$x=x^3\to x=0,1,-1$

Thus the Solutions are $(0,0),(1,0),(-1,0)$

these also satisfy the equation $y=2xy$

now $y=2xy$

Thus putting $x=\frac{1}{2}[$as we get $1=2x$ cancelling $y]$

The solutions are $\frac{1}{2}={\left(\frac{1}{2}\right)}^3+y^4$

$y^4=\frac{1}{2}-\frac{1}{8}$

$y^4=\frac{3}{8}$

thus $y=\pm {\left(\frac{3}{8}\right)}^{\frac{-1}{4}}$

Solution set is $(\frac{1}{2},{\left(\frac{3}{8}\right)}^{\frac{-1}{4}}),(\frac{1}{2},-{\left(\frac{3}{8}\right)}^{\frac{-1}{4}})$

Sets of solutions are 5