# How do you simplify (2x-1)/((x-3)(x+2)) + (x-4)/(x-3)?

Aug 2, 2018

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

#### Explanation:

$\text{We require the fractions to have common denominators}$

$\text{multiply the numerator/denominator of}$

$\frac{x - 4}{x - 3} \text{ by } \left(x + 2\right)$

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

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

$\text{add terms on numerators leaving the denominator}$

$= \frac{{x}^{2} - 9}{\left(x - 3\right) \left(x + 2\right)} \leftarrow \textcolor{b l u e}{\text{difference of squares}}$

$= \frac{\cancel{\left(x - 3\right)} \left(x + 3\right)}{\cancel{\left(x - 3\right)} \left(x + 2\right)} \leftarrow \textcolor{b l u e}{\text{cancel common factor}}$

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

$\text{with restriction } x \ne 3$

Aug 2, 2018

color(brown)(=> (x+ 3) / (x+2)

#### Explanation:

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

$\implies \frac{\left(2 x - 1\right) + \left(\left(x - 4\right) \left(x + 2\right)\right)}{\left(x - 3\right) \left(x + 2\right)}$, taking L C M.

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

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

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

color(brown)(=> (x+ 3) / (x+2)