Solve 8+3/y>19/y

$$8 + \frac { 3 } { y } > \frac { 19 } { y }$$

Solve for y

$$y\in \left(-\infty,0\right)\cup \left(2,\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 8 times \frac{y}{y}.

$$\frac{8y}{y}+\frac{3}{y}>\frac{19}{y}$$

Since \frac{8y}{y} and \frac{3}{y} have the same denominator, add them by adding their numerators.

$$\frac{8y+3}{y}>\frac{19}{y}$$

Subtract \frac{19}{y} from both sides.

$$\frac{8y+3}{y}-\frac{19}{y}>0$$

Since \frac{8y+3}{y} and \frac{19}{y} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8y+3-19}{y}>0$$

Combine like terms in 8y+3-19.

$$\frac{8y-16}{y}>0$$

For the quotient to be positive, 8y-16 and y have to be both negative or both positive. Consider the case when 8y-16 and y are both negative.

$$8y-16<0 y<0$$

The solution satisfying both inequalities is y<0.

$$y<0$$

Consider the case when 8y-16 and y are both positive.

$$y>0 8y-16>0$$

The solution satisfying both inequalities is y>2.

$$y>2$$

The final solution is the union of the obtained solutions.

$$y<0\text{; }y>2$$