# How do you add \frac { x - 4} { 2x ^ { 2} + 9x - 5} + \frac { x + 3} { x ^ { 2} + 5x }?

Dec 7, 2016

$\frac{3 {x}^{2} + x - 3}{\left(x\right) \left(x + 5\right) \left(2 x - 1\right)}$

#### Explanation:

Factor the denominators to discover the least common denominator (LCD).

$2 {x}^{2} + 9 x - 5 = 2 {x}^{2} + 10 x - x - 5 = 2 x \left(x + 5\right) - \left(x + 5\right) = \left(2 x - 1\right) \left(x + 5\right)$

${x}^{2} + 5 x = x \left(x + 5\right)$

Therefore, the LCD is $\textcolor{g r e e n}{\left(x\right) \left(x + 5\right) \left(2 x - 1\right)}$.

$\implies \frac{x \left(x - 4\right)}{\left(x\right) \left(x + 5\right) \left(2 x - 1\right)} + \frac{\left(x + 3\right) \left(2 x - 1\right)}{\left(x\right) \left(x + 5\right) \left(2 x - 1\right)}$

$\implies \frac{{x}^{2} - 4 x + 2 {x}^{2} + 6 x - x - 3}{\left(x\right) \left(x + 5\right) \left(2 x - 1\right)}$

$\implies \frac{3 {x}^{2} + x - 3}{\left(x\right) \left(x + 5\right) \left(2 x - 1\right)}$

Hopefully this helps!