Solve 30geq6(2/3z+1/3)

$$30 \geq 6( \frac{ 2 }{ 3 } z+ \frac{ 1 }{ 3 } )$$

Solve for z

$$z\leq 7$$

Solution Steps

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

$$\frac{30}{6}\geq \frac{2}{3}z+\frac{1}{3}$$

Divide 30 by 6 to get 5.

$$5\geq \frac{2}{3}z+\frac{1}{3}$$

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

$$\frac{2}{3}z+\frac{1}{3}\leq 5$$

Subtract \frac{1}{3} from both sides.

$$\frac{2}{3}z\leq 5-\frac{1}{3}$$

Convert 5 to fraction \frac{15}{3}.

$$\frac{2}{3}z\leq \frac{15}{3}-\frac{1}{3}$$

Since \frac{15}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2}{3}z\leq \frac{15-1}{3}$$

Subtract 1 from 15 to get 14.

$$\frac{2}{3}z\leq \frac{14}{3}$$

Multiply both sides by \frac{3}{2}, the reciprocal of \frac{2}{3}. Since \frac{2}{3} is positive, the inequality direction remains the same.

$$z\leq \frac{14}{3}\times \frac{3}{2}$$

Multiply \frac{14}{3} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$z\leq \frac{14\times 3}{3\times 2}$$

Cancel out 3 in both numerator and denominator.

$$z\leq \frac{14}{2}$$

Divide 14 by 2 to get 7.

$$z\leq 7$$