Solve x+4/5geq1/3

$$x + \frac { 4 } { 5 } \geq \frac { 1 } { 3 }$$

Solve for x

$$x\geq -\frac{7}{15}$$

Solution Steps

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

$$x\geq \frac{1}{3}-\frac{4}{5}$$

Least common multiple of 3 and 5 is 15. Convert \frac{1}{3} and \frac{4}{5} to fractions with denominator 15.

$$x\geq \frac{5}{15}-\frac{12}{15}$$

Since \frac{5}{15} and \frac{12}{15} have the same denominator, subtract them by subtracting their numerators.

$$x\geq \frac{5-12}{15}$$

Subtract 12 from 5 to get -7.

$$x\geq -\frac{7}{15}$$