Solve {l}{B=frac{3^{8}*11^{2}}{3^{9}*11^3}=}{C=9^2*17^3/17*3^5*7=}

$$\left. \begin{array} { l } { B = \frac { 3 ^ { 8 } \times 11 ^ { 2 } } { 3 ^ { 9 } \times 11 ^ { 3 } } = } \\ { C = \frac { 9 ^ { 2 } \times 17 ^ { 3 } } { 17 \times 3 ^ { 5 } \times 7 } = } \end{array} \right.$$

Solve for B, C

$$B=\frac{1}{33}\approx 0.03030303$$

$$C = \frac{289}{21} = 13\frac{16}{21} \approx 13.761904762$$

Solution Steps

Consider the first equation. Cancel out 3^{8}\times 11^{2} in both numerator and denominator.

$$B=\frac{1}{3\times 11}$$

Multiply 3 and 11 to get 33.

$$B=\frac{1}{33}$$

Consider the second equation. Cancel out 17 in both numerator and denominator.

$$C=\frac{9^{2}\times 17^{2}}{7\times 3^{5}}$$

Calculate 9 to the power of 2 and get 81.

$$C=\frac{81\times 17^{2}}{7\times 3^{5}}$$

Calculate 17 to the power of 2 and get 289.

$$C=\frac{81\times 289}{7\times 3^{5}}$$

Multiply 81 and 289 to get 23409.

$$C=\frac{23409}{7\times 3^{5}}$$

Calculate 3 to the power of 5 and get 243.

$$C=\frac{23409}{7\times 243}$$

Multiply 7 and 243 to get 1701.

$$C=\frac{23409}{1701}$$

Reduce the fraction \frac{23409}{1701} to lowest terms by extracting and canceling out 81.

$$C=\frac{289}{21}$$

The system is now solved.

$$B=\frac{1}{33} C=\frac{289}{21}$$