Solve {l}{x+1/3=3}{y={(2-1/3)}^2}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x + \frac{1}{3} = 3 }\\ { y = {(2 - \frac{1}{3})} ^ {2} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{25}{9} = 2\frac{7}{9} \approx 2.777777778$$

Solution Steps

Consider the first equation. Subtract \frac{1}{3} from both sides.

$$x=3-\frac{1}{3}$$

Subtract \frac{1}{3} from 3 to get \frac{8}{3}.

$$x=\frac{8}{3}$$

Consider the second equation. Subtract \frac{1}{3} from 2 to get \frac{5}{3}.

$$y=\left(\frac{5}{3}\right)^{2}$$

Calculate \frac{5}{3} to the power of 2 and get \frac{25}{9}.

$$y=\frac{25}{9}$$

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

$$z=\frac{25}{9}$$

The system is now solved.

$$x=\frac{8}{3} y=\frac{25}{9} z=\frac{25}{9}$$