Determine if @POINT is a Solution x^2+3y=0 , y=5 , (3,4)

$x^{2} + 3 y = 0$ , $y = 5$ , $(3, 4)$

Simplify $3^{2} + 3 (4)$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$9 + 3 (4) = 0$

$4 = 5$

Multiply $3$ by $4$ .

$9 + 12 = 0$

$4 = 5$

$9 + 12 = 0$

$4 = 5$

Add $9$ and $12$ .

$21 = 0$

$4 = 5$

$21 = 0$

$4 = 5$

Since $21 \neq 0$ , there are no solutions.

No solution

Since $4 \neq 5$ , there are no solutions.

No solution

The ordered pair is not a solution to the system of equations.

$(3, 4)$ is not a solution