Question

2.

Find all real solutions to the system of equations. * (5 Points)
√ x ³² y
= -
16
2
x² + 4y + 16 = 0
O The solution set is {(√8, 2)}.
O The solution set is {(√8, 2), (-√8, 2)}
O {The solution set is {(√8, 2), (-√8, -4√2)}, (√8,-2), (-√8,-2)}
O The system has no solutions.
O I don't know.

Answer 1

Answer: the system has no solution.

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\\eq 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.`