Solve x/8-3/16geqx/24*2/3

$$\frac { x } { 8 } - \frac { 3 } { 16 } \geq \frac { x } { 24 } \cdot \frac { 2 } { 3 }$$

Solve for x

$$x\geq \frac{27}{14}$$

Solution Steps

Multiply both sides of the equation by 48, the least common multiple of 8,16,24,3. Since 48 is positive, the inequality direction remains the same.

$$6x-9\geq 2x\times \frac{2}{3}$$

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

$$6x-9\geq \frac{2\times 2}{3}x$$

Multiply 2 and 2 to get 4.

$$6x-9\geq \frac{4}{3}x$$

Subtract \frac{4}{3}x from both sides.

$$6x-9-\frac{4}{3}x\geq 0$$

Combine 6x and -\frac{4}{3}x to get \frac{14}{3}x.

$$\frac{14}{3}x-9\geq 0$$

Add 9 to both sides. Anything plus zero gives itself.

$$\frac{14}{3}x\geq 9$$

Multiply both sides by \frac{3}{14}, the reciprocal of \frac{14}{3}. Since \frac{14}{3} is positive, the inequality direction remains the same.

$$x\geq 9\times \frac{3}{14}$$

Express 9\times \frac{3}{14} as a single fraction.

$$x\geq \frac{9\times 3}{14}$$

Multiply 9 and 3 to get 27.

$$x\geq \frac{27}{14}$$