Solve 2/5+1/3x>4/5x-7/3

$$\frac{ 2 }{ 5 } + \frac{ 1 }{ 3 } x > \frac{ 4 }{ 5 } x- \frac{ 7 }{ 3 }$$

Solve for x

$$x<\frac{41}{7}$$

Solution Steps

Subtract \frac{4}{5}x from both sides.

$$\frac{2}{5}+\frac{1}{3}x-\frac{4}{5}x>-\frac{7}{3}$$

Combine \frac{1}{3}x and -\frac{4}{5}x to get -\frac{7}{15}x.

$$\frac{2}{5}-\frac{7}{15}x>-\frac{7}{3}$$

Subtract \frac{2}{5} from both sides.

$$-\frac{7}{15}x>-\frac{7}{3}-\frac{2}{5}$$

Least common multiple of 3 and 5 is 15. Convert -\frac{7}{3} and \frac{2}{5} to fractions with denominator 15.

$$-\frac{7}{15}x>-\frac{35}{15}-\frac{6}{15}$$

Since -\frac{35}{15} and \frac{6}{15} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{7}{15}x>\frac{-35-6}{15}$$

Subtract 6 from -35 to get -41.

$$-\frac{7}{15}x>-\frac{41}{15}$$

Multiply both sides by -\frac{15}{7}, the reciprocal of -\frac{7}{15}. Since -\frac{7}{15} is negative, the inequality direction is changed.

$$x<-\frac{41}{15}\left(-\frac{15}{7}\right)$$

Multiply -\frac{41}{15} times -\frac{15}{7} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{-41\left(-15\right)}{15\times 7}$$

Do the multiplications in the fraction \frac{-41\left(-15\right)}{15\times 7}.

$$x<\frac{615}{105}$$

Reduce the fraction \frac{615}{105} to lowest terms by extracting and canceling out 15.

$$x<\frac{41}{7}$$