Solve 4divx-3>2divx-7

$$4 \div x-3 > 2 \div x-7$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{2}\right)\cup \left(0,\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x}{x}.

$$\frac{4}{x}-\frac{3x}{x}>\frac{2}{x}-7$$

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

$$\frac{4-3x}{x}>\frac{2}{x}-7$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{x}{x}.

$$\frac{4-3x}{x}>\frac{2}{x}-\frac{7x}{x}$$

Since \frac{2}{x} and \frac{7x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4-3x}{x}>\frac{2-7x}{x}$$

Subtract \frac{2-7x}{x} from both sides.

$$\frac{4-3x}{x}-\frac{2-7x}{x}>0$$

Since \frac{4-3x}{x} and \frac{2-7x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4-3x-\left(2-7x\right)}{x}>0$$

Do the multiplications in 4-3x-\left(2-7x\right).

$$\frac{4-3x-2+7x}{x}>0$$

Combine like terms in 4-3x-2+7x.

$$\frac{2+4x}{x}>0$$

For the quotient to be positive, 4x+2 and x have to be both negative or both positive. Consider the case when 4x+2 and x are both negative.

$$4x+2<0 x<0$$

The solution satisfying both inequalities is x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

Consider the case when 4x+2 and x are both positive.

$$x>0 4x+2>0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{2}\text{; }x>0$$