Solve {l}{2x=0}{text{Solvefor}ytext{where}}{y=x^4-3x^2+2x^2}

$$\left. \begin{array} { l } { 2 x = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = x ^ {4} - 3 x ^ {2} + 2 x ^ {2} } \end{array} \right.$$

Solve for x, y

$$y=0$$

Solution Steps

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

$$x=0$$

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

$$y=0^{4}-3\times 0^{2}+2\times 0^{2}$$

Calculate 0 to the power of 4 and get 0.

$$y=0-3\times 0^{2}+2\times 0^{2}$$

Calculate 0 to the power of 2 and get 0.

$$y=0-3\times 0+2\times 0^{2}$$

Multiply -3 and 0 to get 0.

$$y=0+0+2\times 0^{2}$$

Add 0 and 0 to get 0.

$$y=2\times 0^{2}$$

Calculate 0 to the power of 2 and get 0.

$$y=2\times 0$$

Multiply 2 and 0 to get 0.

$$y=0$$

The system is now solved.

$$x=0 y=0$$