Solve {l}{x=3}{y=1}{z=x^3/3-y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x = 3 }\\ { y = 1 }\\ { z = x ^ {3} / 3 - y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=8$$

Solution Steps

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

$$z=\frac{3^{3}}{3}-1$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 1 from 3 to get 2.

$$z=3^{2}-1$$

Calculate 3 to the power of 2 and get 9.

$$z=9-1$$

Subtract 1 from 9 to get 8.

$$z=8$$

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

$$a=8$$

The system is now solved.

$$x=3 y=1 z=8 a=8$$