Solve 2n/n+1>9/5

$$\frac { 2 n } { n + 1 } > \frac { 9 } { 5 }$$

Solve for n

$$n\in \left(-\infty,-1\right)\cup \left(9,\infty\right)$$

Solution Steps

Denominator n+1 cannot be zero since division by zero is not defined. There are two cases.

$$n+1>0 n+1<0$$

Consider the case when n+1 is positive. Move 1 to the right hand side.

$$n>-1$$

The initial inequality does not change the direction when multiplied by n+1 for n+1>0.

$$2n>\frac{9}{5}\left(n+1\right)$$

Multiply out the right hand side.

$$2n>\frac{9}{5}n+\frac{9}{5}$$

Move the terms containing n to the left hand side and all other terms to the right hand side.

$$2n-\frac{9}{5}n>\frac{9}{5}$$

Combine like terms.

$$\frac{1}{5}n>\frac{9}{5}$$

Divide both sides by \frac{1}{5}. Since \frac{1}{5} is positive, the inequality direction remains the same.

$$n>9$$

Consider condition n>-1 specified above. The result remains the same.

$$n>9$$

Now consider the case when n+1 is negative. Move 1 to the right hand side.

$$n<-1$$

The initial inequality changes the direction when multiplied by n+1 for n+1<0.

$$2n<\frac{9}{5}\left(n+1\right)$$

Multiply out the right hand side.

$$2n<\frac{9}{5}n+\frac{9}{5}$$

Move the terms containing n to the left hand side and all other terms to the right hand side.

$$2n-\frac{9}{5}n<\frac{9}{5}$$

Combine like terms.

$$\frac{1}{5}n<\frac{9}{5}$$

Divide both sides by \frac{1}{5}. Since \frac{1}{5} is positive, the inequality direction remains the same.

$$n<9$$

Consider condition n<-1 specified above.

$$n<-1$$

The final solution is the union of the obtained solutions.

$$n\in \left(-\infty,-1\right)\cup \left(9,\infty\right)$$