Solve -6/7x+4/9leq1/3+5/3x

$$- \frac { 6 } { 7 } x + \frac { 4 } { 9 } \leq \frac { 1 } { 3 } + \frac { 5 } { 3 } x$$

Solve for x

$$x\geq \frac{7}{159}$$

Solution Steps

Subtract \frac{5}{3}x from both sides.

$$-\frac{6}{7}x+\frac{4}{9}-\frac{5}{3}x\leq \frac{1}{3}$$

Combine -\frac{6}{7}x and -\frac{5}{3}x to get -\frac{53}{21}x.

$$-\frac{53}{21}x+\frac{4}{9}\leq \frac{1}{3}$$

Subtract \frac{4}{9} from both sides.

$$-\frac{53}{21}x\leq \frac{1}{3}-\frac{4}{9}$$

Least common multiple of 3 and 9 is 9. Convert \frac{1}{3} and \frac{4}{9} to fractions with denominator 9.

$$-\frac{53}{21}x\leq \frac{3}{9}-\frac{4}{9}$$

Since \frac{3}{9} and \frac{4}{9} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{53}{21}x\leq \frac{3-4}{9}$$

Subtract 4 from 3 to get -1.

$$-\frac{53}{21}x\leq -\frac{1}{9}$$

Multiply both sides by -\frac{21}{53}, the reciprocal of -\frac{53}{21}. Since -\frac{53}{21} is negative, the inequality direction is changed.

$$x\geq -\frac{1}{9}\left(-\frac{21}{53}\right)$$

Multiply -\frac{1}{9} times -\frac{21}{53} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{-\left(-21\right)}{9\times 53}$$

Do the multiplications in the fraction \frac{-\left(-21\right)}{9\times 53}.

$$x\geq \frac{21}{477}$$

Reduce the fraction \frac{21}{477} to lowest terms by extracting and canceling out 3.

$$x\geq \frac{7}{159}$$