Solve {l}{-x-3=0}{text{Solvefor}ytext{where}}{y=4x^2+12x}

$$\left. \begin{array} { l } { -x - 3 = 0 }\\ { \text{Solve for } y \text{ where} } \\ { y = 4 x ^ {2} + 12 x } \end{array} \right.$$

Solve for x, y

$$y=0$$

Solution Steps

Consider the first equation. Add 3 to both sides. Anything plus zero gives itself.

$$-x=3$$

Divide both sides by -1.

$$x=-3$$

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

$$y=4\left(-3\right)^{2}+12\left(-3\right)$$

Calculate -3 to the power of 2 and get 9.

$$y=4\times 9+12\left(-3\right)$$

Multiply 4 and 9 to get 36.

$$y=36+12\left(-3\right)$$

Multiply 12 and -3 to get -36.

$$y=36-36$$

Subtract 36 from 36 to get 0.

$$y=0$$

The system is now solved.

$$x=-3 y=0$$