Solve 2-5/3ugeq7/3-8/9u

$$2 - \frac { 5 } { 3 } u \geq \frac { 7 } { 3 } - \frac { 8 } { 9 } u$$

Solve for u

$$u\leq -\frac{3}{7}$$

Solution Steps

Add \frac{8}{9}u to both sides.

$$2-\frac{5}{3}u+\frac{8}{9}u\geq \frac{7}{3}$$

Combine -\frac{5}{3}u and \frac{8}{9}u to get -\frac{7}{9}u.

$$2-\frac{7}{9}u\geq \frac{7}{3}$$

Subtract 2 from both sides.

$$-\frac{7}{9}u\geq \frac{7}{3}-2$$

Convert 2 to fraction \frac{6}{3}.

$$-\frac{7}{9}u\geq \frac{7}{3}-\frac{6}{3}$$

Since \frac{7}{3} and \frac{6}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{7}{9}u\geq \frac{7-6}{3}$$

Subtract 6 from 7 to get 1.

$$-\frac{7}{9}u\geq \frac{1}{3}$$

Multiply both sides by -\frac{9}{7}, the reciprocal of -\frac{7}{9}. Since -\frac{7}{9} is negative, the inequality direction is changed.

$$u\leq \frac{1}{3}\left(-\frac{9}{7}\right)$$

Multiply \frac{1}{3} times -\frac{9}{7} by multiplying numerator times numerator and denominator times denominator.

$$u\leq \frac{1\left(-9\right)}{3\times 7}$$

Do the multiplications in the fraction \frac{1\left(-9\right)}{3\times 7}.

$$u\leq \frac{-9}{21}$$

Reduce the fraction \frac{-9}{21} to lowest terms by extracting and canceling out 3.

$$u\leq -\frac{3}{7}$$