Solve 4x+7/x-3+5>0

$$\frac { 4 x + 7 } { x - 3 } + 5 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{8}{9}\right)\cup \left(3,\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{x-3}{x-3}.

$$\frac{4x+7}{x-3}+\frac{5\left(x-3\right)}{x-3}>0$$

Since \frac{4x+7}{x-3} and \frac{5\left(x-3\right)}{x-3} have the same denominator, add them by adding their numerators.

$$\frac{4x+7+5\left(x-3\right)}{x-3}>0$$

Do the multiplications in 4x+7+5\left(x-3\right).

$$\frac{4x+7+5x-15}{x-3}>0$$

Combine like terms in 4x+7+5x-15.

$$\frac{9x-8}{x-3}>0$$

For the quotient to be positive, 9x-8 and x-3 have to be both negative or both positive. Consider the case when 9x-8 and x-3 are both negative.

$$9x-8<0 x-3<0$$

The solution satisfying both inequalities is x<\frac{8}{9}.

$$x<\frac{8}{9}$$

Consider the case when 9x-8 and x-3 are both positive.

$$x-3>0 9x-8>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<\frac{8}{9}\text{; }x>3$$