Solve 3/x-1/4>1/x+1

$$\frac { 3 } { x } - \frac { 1 } { 4 } > \frac { 1 } { x } + 1$$

Solve for x

$$x\in \left(0,\frac{8}{5}\right)$$

Solution Steps

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

$$\frac{3\times 4}{4x}-\frac{x}{4x}>\frac{1}{x}+1$$

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

$$\frac{3\times 4-x}{4x}>\frac{1}{x}+1$$

Do the multiplications in 3\times 4-x.

$$\frac{12-x}{4x}>\frac{1}{x}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$\frac{12-x}{4x}>\frac{1}{x}+\frac{x}{x}$$

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

$$\frac{12-x}{4x}>\frac{1+x}{x}$$

Subtract \frac{1+x}{x} from both sides.

$$\frac{12-x}{4x}-\frac{1+x}{x}>0$$

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

$$\frac{12-x}{4x}-\frac{4\left(1+x\right)}{4x}>0$$

Since \frac{12-x}{4x} and \frac{4\left(1+x\right)}{4x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12-x-4\left(1+x\right)}{4x}>0$$

Do the multiplications in 12-x-4\left(1+x\right).

$$\frac{12-x-4-4x}{4x}>0$$

Combine like terms in 12-x-4-4x.

$$\frac{8-5x}{4x}>0$$

For the quotient to be positive, 8-5x and 4x have to be both negative or both positive. Consider the case when 8-5x and 4x are both negative.

$$8-5x<0 4x<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 8-5x and 4x are both positive.

$$4x>0 8-5x>0$$

The solution satisfying both inequalities is x\in \left(0,\frac{8}{5}\right).

$$x\in \left(0,\frac{8}{5}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(0,\frac{8}{5}\right)$$