Solve 3p+8/21leq71

$$3 p + \frac { 8 } { 21 } \leq 71$$

Solve for p

$$p\leq \frac{1483}{63}$$

Solution Steps

Subtract \frac{8}{21} from both sides.

$$3p\leq 71-\frac{8}{21}$$

Convert 71 to fraction \frac{1491}{21}.

$$3p\leq \frac{1491}{21}-\frac{8}{21}$$

Since \frac{1491}{21} and \frac{8}{21} have the same denominator, subtract them by subtracting their numerators.

$$3p\leq \frac{1491-8}{21}$$

Subtract 8 from 1491 to get 1483.

$$3p\leq \frac{1483}{21}$$

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

$$p\leq \frac{\frac{1483}{21}}{3}$$

Express \frac{\frac{1483}{21}}{3} as a single fraction.

$$p\leq \frac{1483}{21\times 3}$$

Multiply 21 and 3 to get 63.

$$p\leq \frac{1483}{63}$$