Solve 81/64geq15Q-6/3

$$\frac { 81 } { 64 } \geq \frac { 15 Q - 6 } { 3 }$$

Solve for Q

$$Q\leq \frac{209}{320}$$

Solution Steps

Multiply both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$\frac{81}{64}\times 3\geq 15Q-6$$

Express \frac{81}{64}\times 3 as a single fraction.

$$\frac{81\times 3}{64}\geq 15Q-6$$

Multiply 81 and 3 to get 243.

$$\frac{243}{64}\geq 15Q-6$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$15Q-6\leq \frac{243}{64}$$

Add 6 to both sides.

$$15Q\leq \frac{243}{64}+6$$

Convert 6 to fraction \frac{384}{64}.

$$15Q\leq \frac{243}{64}+\frac{384}{64}$$

Since \frac{243}{64} and \frac{384}{64} have the same denominator, add them by adding their numerators.

$$15Q\leq \frac{243+384}{64}$$

Add 243 and 384 to get 627.

$$15Q\leq \frac{627}{64}$$

Divide both sides by 15. Since 15 is positive, the inequality direction remains the same.

$$Q\leq \frac{\frac{627}{64}}{15}$$

Express \frac{\frac{627}{64}}{15} as a single fraction.

$$Q\leq \frac{627}{64\times 15}$$

Multiply 64 and 15 to get 960.

$$Q\leq \frac{627}{960}$$

Reduce the fraction \frac{627}{960} to lowest terms by extracting and canceling out 3.

$$Q\leq \frac{209}{320}$$