Solve {l}{x={(5/3)}^{2}*{(3/5)}^-3}{y={(7^-1*3^-1)}/7^-1}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x = {(\frac{5}{3})} ^ {2} \cdot {(\frac{3}{5})} ^ {-3} }\\ { y = {(7 ^ {-1} \cdot 3 ^ {-1})} / 7 ^ {-1} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=\frac{1}{3}\approx 0.333333333$$

Solution Steps

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

$$x=\frac{25}{9}\times \left(\frac{3}{5}\right)^{-3}$$

Calculate \frac{3}{5} to the power of -3 and get \frac{125}{27}.

$$x=\frac{25}{9}\times \frac{125}{27}$$

Multiply \frac{25}{9} and \frac{125}{27} to get \frac{3125}{243}.

$$x=\frac{3125}{243}$$

Consider the second equation. Cancel out \frac{1}{7} in both numerator and denominator.

$$y=\frac{1}{3}$$

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

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

The system is now solved.

$$x=\frac{3125}{243} y=\frac{1}{3} z=\frac{1}{3}$$