Question

The number of ordered pairs $(x,y)$ of real numbers that satisfy the simultaneous equations
$x+y^2=x^2+y=12$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is D $4$

$x+y^2=x^2+y=12x+y^2=x^2+y\Rightarrow x-y=x^2-y^2\Rightarrow x=y,x+y=1$
when $y=x\Rightarrow x^2+x-12=0$
$x=-4,3\therefore (-4,-4),(3,3)$
$\text{When}y=1-x\Rightarrow x^2-x-11=0x=\frac{1\pm \sqrt{45}}{2}$
$\therefore$ For two irrational value of $x$ corresponds to two value of $y.$
So, there are four ordered pair that satisfy the given equations