How do you add the rational expressions (2x-1)/((x-3)(x+2))+(x-4)/(x-3)?

May 30, 2018

$\frac{x + 3}{x + 2} , x \ne 3$

Explanation:

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

$\frac{2 \cdot x - 1 + {x}^{2} - 4 x + 2 x - 8}{\left(x - 3\right) \left(x + 2\right)} =$

$\frac{{x}^{2} - 9}{\left(x - 3\right) \left(x + 2\right)} =$

$\frac{\left(x - 3\right) \left(x + 3\right)}{\left(x - 3\right) \left(x + 2\right)} =$

$\frac{x + 3}{x + 2} , x \ne 3$

May 30, 2018

$\frac{x + 3}{x + 2} , x \ne 3$

Explanation:

1st way:

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

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

=$\frac{2 x - 1 + {x}^{2} - 2 x - 8}{\left(x - 3\right) \left(x + 2\right)}$

=$\frac{{x}^{2} - 9}{\left(x - 3\right) \left(x + 2\right)}$

=$\frac{\left(x + 3\right) \left(x - 3\right)}{\left(x - 3\right) \left(x + 2\right)}$

=$\frac{x + 3}{x + 2} , x \ne 3$

2nd way:

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

=$\frac{\left(x - 3\right) + \left(x + 2\right)}{\left(x - 3\right) \cdot \left(x + 2\right)} + \frac{x - 4}{x - 3}$

=$\frac{1}{x - 3} + \frac{1}{x + 2} + \frac{x - 4}{x - 3}$

=$\frac{1}{x + 2} + \frac{x - 3}{x - 3}$

=$\frac{1}{x + 2} + 1 , x \ne 3$

=$\frac{x + 3}{x + 2} , x \ne 3$