Solve 4/5x+1/10<3/2x

$$\frac{ 4 }{ 5x } + \frac{ 1 }{ 10 } < \frac{ 3 }{ 2x }$$

Solve for x

$$x\in \left(0,7\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5x and 10 is 10x. Multiply \frac{4}{5x} times \frac{2}{2}. Multiply \frac{1}{10} times \frac{x}{x}.

$$\frac{4\times 2}{10x}+\frac{x}{10x}<\frac{3}{2x}$$

Since \frac{4\times 2}{10x} and \frac{x}{10x} have the same denominator, add them by adding their numerators.

$$\frac{4\times 2+x}{10x}<\frac{3}{2x}$$

Do the multiplications in 4\times 2+x.

$$\frac{8+x}{10x}<\frac{3}{2x}$$

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

$$\frac{8+x}{10x}-\frac{3}{2x}<0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 10x and 2x is 10x. Multiply \frac{3}{2x} times \frac{5}{5}.

$$\frac{8+x}{10x}-\frac{3\times 5}{10x}<0$$

Since \frac{8+x}{10x} and \frac{3\times 5}{10x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8+x-3\times 5}{10x}<0$$

Do the multiplications in 8+x-3\times 5.

$$\frac{8+x-15}{10x}<0$$

Combine like terms in 8+x-15.

$$\frac{-7+x}{10x}<0$$

For the quotient to be negative, x-7 and 10x have to be of the opposite signs. Consider the case when x-7 is positive and 10x is negative.

$$x-7>0 10x<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 10x is positive and x-7 is negative.

$$10x>0 x-7<0$$

The solution satisfying both inequalities is x\in \left(0,7\right).

$$x\in \left(0,7\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(0,7\right)$$