How do you solve (2x+3)/(x^2-9) + x/(x-3)?

Apr 3, 2018

$\frac{{x}^{2} + 5 x + 3}{{x}^{2} - 9}$

Explanation:

Add them like fractions (make sure their denominators are the same)

The two denominators are ${x}^{2} - 9$ are $x - 3$. Find the least common denominator.

${x}^{2} - 9 = \left(x - 3\right) \left(x + 3\right) \rightarrow$ One of the denominators is a factor of the other

Multiply the second fraction's numerator and denominator by $x + 3$ to get the same denominator as the first fraction

$\frac{x}{x - 3} \cdot \frac{x + 3}{x + 3}$

$\frac{{x}^{2} + 3 x}{{x}^{2} - 9}$

$\frac{2 x + 3 + {x}^{2} + 3 x}{{x}^{2} - 9}$
$\frac{{x}^{2} + 5 x + 3}{{x}^{2} - 9}$