Solve x-3/3x+x/6leqx^2+9/6x-x+3/x

$$\frac { x - 3 } { 3 x } + \frac { x } { 6 } \leq \frac { x ^ { 2 } + 9 } { 6 x } - \frac { x + 3 } { x }$$

Solve for x

$$x\in [-\frac{3}{8},0)$$

Solution Steps

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

$$\frac{2\left(x-3\right)}{6x}+\frac{xx}{6x}\leq \frac{x^{2}+9}{6x}-\frac{x+3}{x}$$

Since \frac{2\left(x-3\right)}{6x} and \frac{xx}{6x} have the same denominator, add them by adding their numerators.

$$\frac{2\left(x-3\right)+xx}{6x}\leq \frac{x^{2}+9}{6x}-\frac{x+3}{x}$$

Do the multiplications in 2\left(x-3\right)+xx.

$$\frac{2x-6+x^{2}}{6x}\leq \frac{x^{2}+9}{6x}-\frac{x+3}{x}$$

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

$$\frac{2x-6+x^{2}}{6x}\leq \frac{x^{2}+9}{6x}-\frac{6\left(x+3\right)}{6x}$$

Since \frac{x^{2}+9}{6x} and \frac{6\left(x+3\right)}{6x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x-6+x^{2}}{6x}\leq \frac{x^{2}+9-6\left(x+3\right)}{6x}$$

Do the multiplications in x^{2}+9-6\left(x+3\right).

$$\frac{2x-6+x^{2}}{6x}\leq \frac{x^{2}+9-6x-18}{6x}$$

Combine like terms in x^{2}+9-6x-18.

$$\frac{2x-6+x^{2}}{6x}\leq \frac{x^{2}-9-6x}{6x}$$

Subtract \frac{x^{2}-9-6x}{6x} from both sides.

$$\frac{2x-6+x^{2}}{6x}-\frac{x^{2}-9-6x}{6x}\leq 0$$

Since \frac{2x-6+x^{2}}{6x} and \frac{x^{2}-9-6x}{6x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x-6+x^{2}-\left(x^{2}-9-6x\right)}{6x}\leq 0$$

Do the multiplications in 2x-6+x^{2}-\left(x^{2}-9-6x\right).

$$\frac{2x-6+x^{2}-x^{2}+9+6x}{6x}\leq 0$$

Combine like terms in 2x-6+x^{2}-x^{2}+9+6x.

$$\frac{8x+3}{6x}\leq 0$$

For the quotient to be ≤0, one of the values 8x+3 and 6x has to be ≥0, the other has to be ≤0, and 6x cannot be zero. Consider the case when 8x+3\geq 0 and 6x is negative.

$$8x+3\geq 0 6x<0$$

The solution satisfying both inequalities is x\in \left[-\frac{3}{8},0\right).

$$x\in [-\frac{3}{8},0)$$

Consider the case when 8x+3\leq 0 and 6x is positive.

$$8x+3\leq 0 6x>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{3}{8},0)$$