Solve 3x^2+2x/x+1<3x

$$\frac { 3 x ^ { 2 } + 2 x } { x + 1 } < 3 x$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(0,\infty\right)$$

Solution Steps

Subtract 3x from both sides.

$$\frac{3x^{2}+2x}{x+1}-3x<0$$

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

$$\frac{3x^{2}+2x}{x+1}+\frac{-3x\left(x+1\right)}{x+1}<0$$

Since \frac{3x^{2}+2x}{x+1} and \frac{-3x\left(x+1\right)}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{3x^{2}+2x-3x\left(x+1\right)}{x+1}<0$$

Do the multiplications in 3x^{2}+2x-3x\left(x+1\right).

$$\frac{3x^{2}+2x-3x^{2}-3x}{x+1}<0$$

Combine like terms in 3x^{2}+2x-3x^{2}-3x.

$$\frac{-x}{x+1}<0$$

For the quotient to be negative, -x and x+1 have to be of the opposite signs. Consider the case when -x is positive and x+1 is negative.

$$-x>0 x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x+1 is positive and -x is negative.

$$x+1>0 -x<0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

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