Question

2.

Answer 1

Answer: the system has no solution.

Step-by-step explanation:

\displaystyle\\

\left \{ {{x^2y=16\ \ \ \ \ (1)} \atop {x^2+4y+16=0\ \ \ \ \ (2)}} \right. .\\

Multiply\  both\  sides\  of\  the\  equation\ (2)\   by\  y\  (y\neq 0):\\

x^2y+4y^2+16y=0\\

We\  substitute\  equation\  (1)\  into\  equation\  (2):\\

16+4y^2+16y=0\\

4y^2+16y+16=0\\

4*(y^2+4y+4)=0\\

4*(y^2+2*y*2+2^2)=0\\

4*(y+2)^2=0\\

Divide\  both\  sides\  of\  the \ equation\  by\  4:\\

(y+2)^2=0\\

(y+2)*(y+2)=0\\

So,\ y+2=0\\

y=-2.\\

$\displaystyle\\We\ substitute\ the\ value\ of\ y\ into\ equation\ (1):\\x^2*(-2)=16\\Divide\ both\ sides\ of\ the \ equation\ by\ -2:\\x^2=-8\\x^2\geq 0\\ Hence,\ the\ system\ has\ no\ solution.$