Solve 1/4-3/2xleq2/3x

$$\frac { 1 } { 4 } - \frac { 3 } { 2 x } \leq \frac { 2 } { 3 x }$$

Solve for x

$$x\in (0,\frac{26}{3}]$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2x is 4x. Multiply \frac{1}{4} times \frac{x}{x}. Multiply \frac{3}{2x} times \frac{2}{2}.

$$\frac{x}{4x}-\frac{3\times 2}{4x}\leq \frac{2}{3x}$$

Since \frac{x}{4x} and \frac{3\times 2}{4x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x-3\times 2}{4x}\leq \frac{2}{3x}$$

Do the multiplications in x-3\times 2.

$$\frac{x-6}{4x}\leq \frac{2}{3x}$$

Subtract \frac{2}{3x} from both sides.

$$\frac{x-6}{4x}-\frac{2}{3x}\leq 0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4x and 3x is 12x. Multiply \frac{x-6}{4x} times \frac{3}{3}. Multiply \frac{2}{3x} times \frac{4}{4}.

$$\frac{3\left(x-6\right)}{12x}-\frac{2\times 4}{12x}\leq 0$$

Since \frac{3\left(x-6\right)}{12x} and \frac{2\times 4}{12x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(x-6\right)-2\times 4}{12x}\leq 0$$

Do the multiplications in 3\left(x-6\right)-2\times 4.

$$\frac{3x-18-8}{12x}\leq 0$$

Combine like terms in 3x-18-8.

$$\frac{3x-26}{12x}\leq 0$$

For the quotient to be ≤0, one of the values 3x-26 and 12x has to be ≥0, the other has to be ≤0, and 12x cannot be zero. Consider the case when 3x-26\geq 0 and 12x is negative.

$$3x-26\geq 0 12x<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 3x-26\leq 0 and 12x is positive.

$$3x-26\leq 0 12x>0$$

The solution satisfying both inequalities is x\in \left(0,\frac{26}{3}\right].

$$x\in (0,\frac{26}{3}]$$

The final solution is the union of the obtained solutions.

$$x\in (0,\frac{26}{3}]$$