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

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

Solve for x, y

$$y=-24$$

Solution Steps

Consider the first equation. Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x=-2$$

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

$$y=4\left(-2\right)^{3}+2\left(-2\right)^{2}$$

Calculate -2 to the power of 3 and get -8.

$$y=4\left(-8\right)+2\left(-2\right)^{2}$$

Multiply 4 and -8 to get -32.

$$y=-32+2\left(-2\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$y=-32+2\times 4$$

Multiply 2 and 4 to get 8.

$$y=-32+8$$

Add -32 and 8 to get -24.

$$y=-24$$

The system is now solved.

$$x=-2 y=-24$$