Solve 1/2x+3geq1/3

$$\frac { 1 } { 2 } x + 3 \geq \frac { 1 } { 3 }$$

Solve for x

$$x\geq -\frac{16}{3}$$

Solution Steps

Subtract 3 from both sides.

$$\frac{1}{2}x\geq \frac{1}{3}-3$$

Convert 3 to fraction \frac{9}{3}.

$$\frac{1}{2}x\geq \frac{1}{3}-\frac{9}{3}$$

Since \frac{1}{3} and \frac{9}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}x\geq \frac{1-9}{3}$$

Subtract 9 from 1 to get -8.

$$\frac{1}{2}x\geq -\frac{8}{3}$$

Multiply both sides by 2, the reciprocal of \frac{1}{2}. Since \frac{1}{2} is positive, the inequality direction remains the same.

$$x\geq -\frac{8}{3}\times 2$$

Express -\frac{8}{3}\times 2 as a single fraction.

$$x\geq \frac{-8\times 2}{3}$$

Multiply -8 and 2 to get -16.

$$x\geq \frac{-16}{3}$$

Fraction \frac{-16}{3} can be rewritten as -\frac{16}{3} by extracting the negative sign.

$$x\geq -\frac{16}{3}$$