Solve {l}{x={(3/2)}^2*{(2/3)}^-4}{y=x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x = {(\frac{3}{2})} ^ {2} \cdot {(\frac{2}{3})} ^ {-4} }\\ { y = x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{729}{64} = 11\frac{25}{64} = 11.390625$$

Solution Steps

Consider the first equation. Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$x=\frac{9}{4}\times \left(\frac{2}{3}\right)^{-4}$$

Calculate \frac{2}{3} to the power of -4 and get \frac{81}{16}.

$$x=\frac{9}{4}\times \frac{81}{16}$$

Multiply \frac{9}{4} and \frac{81}{16} to get \frac{729}{64}.

$$x=\frac{729}{64}$$

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

$$y=\frac{729}{64}$$

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

$$z=\frac{729}{64}$$

The system is now solved.

$$x=\frac{729}{64} y=\frac{729}{64} z=\frac{729}{64}$$