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

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

Solve for x, y

$$y=32$$

Solution Steps

Consider the first equation. Divide both sides by 2.

$$\frac{4}{2}=x$$

Divide 4 by 2 to get 2.

$$2=x$$

Swap sides so that all variable terms are on the left hand side.

$$x=2$$

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

Calculate 2 to the power of 3 and get 8.

$$y=8+3\times 2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$y=8+3\times 8$$

Multiply 3 and 8 to get 24.

$$y=8+24$$

Add 8 and 24 to get 32.

$$y=32$$

The system is now solved.

$$x=2 y=32$$