Solve 6/2x+1/3>4/5

$$\frac { 6 } { 2 } x + \frac { 1 } { 3 } > \frac { 4 } { 5 }$$

Solve for x

$$x>\frac{7}{45}$$

Solution Steps

Divide 6 by 2 to get 3.

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

Subtract \frac{1}{3} from both sides.

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

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

$$3x>\frac{12}{15}-\frac{5}{15}$$

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

$$3x>\frac{12-5}{15}$$

Subtract 5 from 12 to get 7.

$$3x>\frac{7}{15}$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x>\frac{\frac{7}{15}}{3}$$

Express \frac{\frac{7}{15}}{3} as a single fraction.

$$x>\frac{7}{15\times 3}$$

Multiply 15 and 3 to get 45.

$$x>\frac{7}{45}$$