Solve 8q>9^3/3000+200/9

$$8 q > \frac { \frac { 9 ^ { 3 } } { 3000 } + 200 } { 9 }$$

Solve for q

$$q>\frac{200243}{72000}$$

Solution Steps

Multiply both sides of the equation by 9. Since 9 is positive, the inequality direction remains the same.

$$72q>\frac{9^{3}}{3000}+200$$

Calculate 9 to the power of 3 and get 729.

$$72q>\frac{729}{3000}+200$$

Reduce the fraction \frac{729}{3000} to lowest terms by extracting and canceling out 3.

$$72q>\frac{243}{1000}+200$$

Convert 200 to fraction \frac{200000}{1000}.

$$72q>\frac{243}{1000}+\frac{200000}{1000}$$

Since \frac{243}{1000} and \frac{200000}{1000} have the same denominator, add them by adding their numerators.

$$72q>\frac{243+200000}{1000}$$

Add 243 and 200000 to get 200243.

$$72q>\frac{200243}{1000}$$

Divide both sides by 72. Since 72 is positive, the inequality direction remains the same.

$$q>\frac{\frac{200243}{1000}}{72}$$

Express \frac{\frac{200243}{1000}}{72} as a single fraction.

$$q>\frac{200243}{1000\times 72}$$

Multiply 1000 and 72 to get 72000.

$$q>\frac{200243}{72000}$$