Solve 5x/x-2>3/x-5

$$\frac{ 5x }{ x } -2 > \frac{ 3 }{ x } -5$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{3}{8},\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$\frac{5x}{x}-\frac{2x}{x}>\frac{3}{x}-5$$

Since \frac{5x}{x} and \frac{2x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5x-2x}{x}>\frac{3}{x}-5$$

Combine like terms in 5x-2x.

$$\frac{3x}{x}>\frac{3}{x}-5$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{x}{x}.

$$\frac{3x}{x}>\frac{3}{x}-\frac{5x}{x}$$

Since \frac{3}{x} and \frac{5x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x}{x}>\frac{3-5x}{x}$$

Subtract \frac{3-5x}{x} from both sides.

$$\frac{3x}{x}-\frac{3-5x}{x}>0$$

Since \frac{3x}{x} and \frac{3-5x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x-\left(3-5x\right)}{x}>0$$

Do the multiplications in 3x-\left(3-5x\right).

$$\frac{3x-3+5x}{x}>0$$

Combine like terms in 3x-3+5x.

$$\frac{8x-3}{x}>0$$

For the quotient to be positive, 8x-3 and x have to be both negative or both positive. Consider the case when 8x-3 and x are both negative.

$$8x-3<0 x<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when 8x-3 and x are both positive.

$$x>0 8x-3>0$$

The solution satisfying both inequalities is x>\frac{3}{8}.

$$x>\frac{3}{8}$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{3}{8}$$