# How do you solve (2a-5)/(a-9)-(a-3)/(3a+2)=5/(3a^2-25a-18)?

Oct 20, 2016

$a = - 3$ OR $a = \frac{14}{5}$

#### Explanation:

$\frac{2 a - 5}{a - 9} - \frac{a - 3}{3 a + 2} = \frac{5}{3 {a}^{2} - 25 a - 18}$

$\Rightarrow \frac{\left(2 a - 5\right) \textcolor{b l u e}{\left(3 a + 2\right)} - \left(a - 3\right) \textcolor{b l u e}{\left(a - 9\right)}}{\textcolor{b l u e}{\left(a - 9\right) \left(3 a + 2\right)}} = \frac{5}{3 {a}^{2} - 25 a - 18}$

$\Rightarrow \frac{\left(6 {a}^{2} + 4 a - 15 a - 10\right) - \left({a}^{2} - 9 a - 3 a + 27\right)}{\textcolor{b l u e}{\left(a - 9\right) \left(3 a + 2\right)}} = \frac{5}{3 {a}^{2} - 25 a - 18}$

$\Rightarrow \frac{6 {a}^{2} + 4 a - 15 a - 10 - {a}^{2} + 9 a + 3 a - 27}{3 {a}^{2} + 2 a - 27 a - 18} = \frac{5}{3 {a}^{2} - 25 a - 18}$

$\Rightarrow \frac{6 {a}^{2} - {a}^{2} + 4 a - 15 a + 9 a + 3 a - 10 - 27}{3 {a}^{2} - 25 a - 18} = \frac{5}{3 {a}^{2} - 25 a - 18}$

$\Rightarrow \frac{5 {a}^{2} + a - 37}{3 {a}^{2} - 25 a - 18} = \frac{5}{3 {a}^{2} - 25 a - 18}$

$\Rightarrow 5 {a}^{2} + a - 37 = 5$
$\Rightarrow 5 {a}^{2} + a - 37 - 5 = 0$
$\textcolor{b r o w n}{\Rightarrow 5 {a}^{2} + a - 42 = 0}$

Let us compute the quadratic roots based on the property:
$\textcolor{red}{\delta = {b}^{2} - 4 a c}$
Roots are:
color(red)(a_1=(-b-sqrtdelta)/(2*a)
$\textcolor{red}{{a}_{2} = \frac{- b + \sqrt{\delta}}{2 \cdot a}}$

$\delta = {b}^{2} - 4 a c = {1}^{2} - 4 \left(5\right) \left(- 42\right) = 1 + 840 = 841$

roots are:
${a}_{1} = \frac{- 1 - \sqrt{841}}{2 \cdot 5} = \frac{- 1 - 29}{10} = \left(- \frac{30}{10}\right) = - 3$
${a}_{2} = \frac{- 1 + \sqrt{841}}{2 \cdot 5} = \frac{- 1 + 29}{10} = \left(\frac{28}{10}\right) = \frac{14}{5}$

$\textcolor{b r o w n}{\Rightarrow 5 {a}^{2} + a - 42 = 0}$
$\Rightarrow \left(a - \left(\textcolor{red}{-} 3\right)\right) \left(a - \textcolor{red}{\frac{14}{5}}\right) = 0$
$\Rightarrow \left(a + 3\right) \left(a - \frac{14}{5}\right) = 0$
Therefore,

$a + 3 = 0 \Rightarrow a = - 3$
OR
$a - \frac{14}{5} = 0 \Rightarrow a = \frac{14}{5}$