Solve {l}{k=1+5}{l={(2/3)}^k}{m=l}{n=m}{o=n}{p=o}{q=p}{text{Solvefor}rtext{where}}{r=q}

$$\left. \begin{array} { l } { k = 1 + 5 }\\ { l = {(\frac{2}{3})} ^ {k} }\\ { m = l }\\ { n = m }\\ { o = n }\\ { p = o }\\ { q = p }\\ { \text{Solve for } r \text{ where} } \\ { r = q } \end{array} \right.$$

Solve for k, l, m, n, o, p, q, r

$$r=\frac{64}{729}\approx 0.087791495$$

Solution Steps

Consider the first equation. Add 1 and 5 to get 6.

$$k=6$$

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

$$l=\left(\frac{2}{3}\right)^{6}$$

Calculate \frac{2}{3} to the power of 6 and get \frac{64}{729}.

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

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

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

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

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

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

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

Consider the equation (6). Insert the known values of variables into the equation.

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

Consider the equation (7). Insert the known values of variables into the equation.

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

Consider the equation (8). Insert the known values of variables into the equation.

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

The system is now solved.

$$k=6 l=\frac{64}{729} m=\frac{64}{729} n=\frac{64}{729} o=\frac{64}{729} p=\frac{64}{729} q=\frac{64}{729} r=\frac{64}{729}$$