Solve 4x+1/3<1/6+2

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

Solve for x

$$x<\frac{11}{24}$$

Solution Steps

Convert 2 to fraction \frac{12}{6}.

$$4x+\frac{1}{3}<\frac{1}{6}+\frac{12}{6}$$

Since \frac{1}{6} and \frac{12}{6} have the same denominator, add them by adding their numerators.

$$4x+\frac{1}{3}<\frac{1+12}{6}$$

Add 1 and 12 to get 13.

$$4x+\frac{1}{3}<\frac{13}{6}$$

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

$$4x<\frac{13}{6}-\frac{1}{3}$$

Least common multiple of 6 and 3 is 6. Convert \frac{13}{6} and \frac{1}{3} to fractions with denominator 6.

$$4x<\frac{13}{6}-\frac{2}{6}$$

Since \frac{13}{6} and \frac{2}{6} have the same denominator, subtract them by subtracting their numerators.

$$4x<\frac{13-2}{6}$$

Subtract 2 from 13 to get 11.

$$4x<\frac{11}{6}$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x<\frac{\frac{11}{6}}{4}$$

Express \frac{\frac{11}{6}}{4} as a single fraction.

$$x<\frac{11}{6\times 4}$$

Multiply 6 and 4 to get 24.

$$x<\frac{11}{24}$$