Solve {l}{2y=0}{text{Solvefor}ztext{where}}{z=-2y^2-8y^2}

$$\left. \begin{array} { l } { 2 y = 0 }\\ { \text{Solve for } z \text{ where} } \\ { z = -2 y ^ {2} - 8 y ^ {2} } \end{array} \right.$$

Solve for y, z

$$z=0$$

Solution Steps

Consider the first equation. Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$y=0$$

Consider the second equation. Insert the known values of variables into the equation.

$$z=-2\times 0^{2}-8\times 0^{2}$$

Calculate 0 to the power of 2 and get 0.

$$z=-2\times 0-8\times 0^{2}$$

Multiply -2 and 0 to get 0.

$$z=0-8\times 0^{2}$$

Calculate 0 to the power of 2 and get 0.

$$z=0-8\times 0$$

Multiply -8 and 0 to get 0.

$$z=0+0$$

Add 0 and 0 to get 0.

$$z=0$$

The system is now solved.

$$y=0 z=0$$