Solve 2/5x-11/3+7/2leq3/2-frac{4}{5}

$$\frac { 2 } { 5 } x - \frac { 11 } { 3 } + \frac { 7 } { 2 } \leq \frac { 3 } { 2 } - \frac { 4 } { 5 }$$

Solve for x

$$x\leq \frac{13}{6}$$

Solution Steps

Least common multiple of 3 and 2 is 6. Convert -\frac{11}{3} and \frac{7}{2} to fractions with denominator 6.

$$\frac{2}{5}x-\frac{22}{6}+\frac{21}{6}\leq \frac{3}{2}-\frac{4}{5}$$

Since -\frac{22}{6} and \frac{21}{6} have the same denominator, add them by adding their numerators.

$$\frac{2}{5}x+\frac{-22+21}{6}\leq \frac{3}{2}-\frac{4}{5}$$

Add -22 and 21 to get -1.

$$\frac{2}{5}x-\frac{1}{6}\leq \frac{3}{2}-\frac{4}{5}$$

Least common multiple of 2 and 5 is 10. Convert \frac{3}{2} and \frac{4}{5} to fractions with denominator 10.

$$\frac{2}{5}x-\frac{1}{6}\leq \frac{15}{10}-\frac{8}{10}$$

Since \frac{15}{10} and \frac{8}{10} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2}{5}x-\frac{1}{6}\leq \frac{15-8}{10}$$

Subtract 8 from 15 to get 7.

$$\frac{2}{5}x-\frac{1}{6}\leq \frac{7}{10}$$

Add \frac{1}{6} to both sides.

$$\frac{2}{5}x\leq \frac{7}{10}+\frac{1}{6}$$

Least common multiple of 10 and 6 is 30. Convert \frac{7}{10} and \frac{1}{6} to fractions with denominator 30.

$$\frac{2}{5}x\leq \frac{21}{30}+\frac{5}{30}$$

Since \frac{21}{30} and \frac{5}{30} have the same denominator, add them by adding their numerators.

$$\frac{2}{5}x\leq \frac{21+5}{30}$$

Add 21 and 5 to get 26.

$$\frac{2}{5}x\leq \frac{26}{30}$$

Reduce the fraction \frac{26}{30} to lowest terms by extracting and canceling out 2.

$$\frac{2}{5}x\leq \frac{13}{15}$$

Multiply both sides by \frac{5}{2}, the reciprocal of \frac{2}{5}. Since \frac{2}{5} is positive, the inequality direction remains the same.

$$x\leq \frac{13}{15}\times \frac{5}{2}$$

Multiply \frac{13}{15} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{13\times 5}{15\times 2}$$

Do the multiplications in the fraction \frac{13\times 5}{15\times 2}.

$$x\leq \frac{65}{30}$$

Reduce the fraction \frac{65}{30} to lowest terms by extracting and canceling out 5.

$$x\leq \frac{13}{6}$$