Solve -x/x+1>3/1-9/x+1

$$- \frac{ x }{ x+1 } > \frac{ 3 }{ 1 } - \frac{ 9 }{ x+1 }$$

Solve for x

$$x\in \left(-1,\frac{3}{2}\right)$$

Solution Steps

Anything divided by one gives itself.

$$-\frac{x}{x+1}>3-\frac{9}{x+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x+1}{x+1}.

$$-\frac{x}{x+1}>\frac{3\left(x+1\right)}{x+1}-\frac{9}{x+1}$$

Since \frac{3\left(x+1\right)}{x+1} and \frac{9}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{x}{x+1}>\frac{3\left(x+1\right)-9}{x+1}$$

Do the multiplications in 3\left(x+1\right)-9.

$$-\frac{x}{x+1}>\frac{3x+3-9}{x+1}$$

Combine like terms in 3x+3-9.

$$-\frac{x}{x+1}>\frac{3x-6}{x+1}$$

Subtract \frac{3x-6}{x+1} from both sides.

$$-\frac{x}{x+1}-\frac{3x-6}{x+1}>0$$

Since -\frac{x}{x+1} and \frac{3x-6}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-x-\left(3x-6\right)}{x+1}>0$$

Do the multiplications in -x-\left(3x-6\right).

$$\frac{-x-3x+6}{x+1}>0$$

Combine like terms in -x-3x+6.

$$\frac{-4x+6}{x+1}>0$$

For the quotient to be positive, 6-4x and x+1 have to be both negative or both positive. Consider the case when 6-4x and x+1 are both negative.

$$6-4x<0 x+1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 6-4x and x+1 are both positive.

$$x+1>0 6-4x>0$$

The solution satisfying both inequalities is x\in \left(-1,\frac{3}{2}\right).

$$x\in \left(-1,\frac{3}{2}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,\frac{3}{2}\right)$$