Solve 7y+4/y+2>-4/3

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

Solve for y

$$y\in \left(-\infty,-2\right)\cup \left(-\frac{4}{5},\infty\right)$$

Solution Steps

Denominator y+2 cannot be zero since division by zero is not defined. There are two cases.

$$y+2>0 y+2<0$$

Consider the case when y+2 is positive. Move 2 to the right hand side.

$$y>-2$$

The initial inequality does not change the direction when multiplied by y+2 for y+2>0.

$$7y+4>-\frac{4}{3}\left(y+2\right)$$

Multiply out the right hand side.

$$7y+4>-\frac{4}{3}y-\frac{8}{3}$$

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

$$7y+\frac{4}{3}y>-4-\frac{8}{3}$$

Combine like terms.

$$\frac{25}{3}y>-\frac{20}{3}$$

Divide both sides by \frac{25}{3}. Since \frac{25}{3} is positive, the inequality direction remains the same.

$$y>-\frac{4}{5}$$

Consider condition y>-2 specified above. The result remains the same.

$$y>-\frac{4}{5}$$

Now consider the case when y+2 is negative. Move 2 to the right hand side.

$$y<-2$$

The initial inequality changes the direction when multiplied by y+2 for y+2<0.

$$7y+4<-\frac{4}{3}\left(y+2\right)$$

Multiply out the right hand side.

$$7y+4<-\frac{4}{3}y-\frac{8}{3}$$

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

$$7y+\frac{4}{3}y<-4-\frac{8}{3}$$

Combine like terms.

$$\frac{25}{3}y<-\frac{20}{3}$$

Divide both sides by \frac{25}{3}. Since \frac{25}{3} is positive, the inequality direction remains the same.

$$y<-\frac{4}{5}$$

Consider condition y<-2 specified above.

$$y<-2$$

The final solution is the union of the obtained solutions.

$$y\in \left(-\infty,-2\right)\cup \left(-\frac{4}{5},\infty\right)$$