# How do you solve (2a-5)/(a-9)+a/(a+9)=-6/(a^2-81)?

Jul 8, 2017

Solution : $a = 3 \mathmr{and} a = - \frac{13}{3}$

#### Explanation:

 (2a-5)/(a-9) +a /(a+9) = - 6/((a^2-81) or

 (2a-5)/(a-9) +a /(a+9) = - 6/((a+9(a-9) Multiplying by (a+9(a-9) on boh sides we get,

$\left(2 a - 5\right) \left(a + 9\right) + a \left(a - 9\right) = - 6$ or

$2 {a}^{2} + 13 a - 45 + {a}^{2} - 9 a + 6 = 0$ or

$3 {a}^{2} + 4 a - 39 = 0 \mathmr{and} 3 {a}^{2} + 13 a - 9 a - 39 = 0$ or

$a \left(3 a + 13\right) - 3 \left(3 a + 13\right) = 0$ or

$\left(3 a + 13\right) \left(a - 3\right) = 0$, either $3 a + 13 = 0 \therefore a = - \frac{13}{3}$ or

$a - 3 = 0 \therefore a = 3$

Solution: $a = 3 \mathmr{and} a = - \frac{13}{3}$ [Ans]

Jul 8, 2017

No real solutions and $x = - \frac{13}{3} , x = 3$ extraneous solutions

#### Explanation:

$\frac{2 a - 5}{a - 9} + \frac{a}{a + 9} = - \frac{6}{{a}^{2} - 81}$

$\frac{2 a - 5}{a - 9} + \frac{a}{a + 9} + \frac{6}{\left(a - 9\right) \left(a + 9\right)} = 0$

$\frac{\left(a + 9\right) \left(2 a - 5\right) + a \left(a - 9\right) + 6 \left(a + 9\right)}{\left(a - 9\right) \left(a + 9\right)} = 0$

Multiply both sides by $\left(a - 9\right) \left(a + 9\right)$

$2 {a}^{2} + 13 a - 45 + {a}^{2} - 9 a + 6 a + 54 = 0$

$3 {a}^{2} + 10 a + 9 = 0$

$a = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = \frac{- \left(10\right) \pm \sqrt{{\left(10\right)}^{2} - 4 \left(3\right) \left(9\right)}}{2 \left(3\right)}$

$a = \frac{- 10 \pm \sqrt{100 - 108}}{2 \left(3\right)}$

$a = \frac{- 10 \pm \sqrt{- 8}}{6}$

No real solution because of $\sqrt{- 8}$

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Alternate method:

$\frac{2 a - 5}{a - 9} + \frac{a}{a + 9} = - \frac{6}{{a}^{2} - 81}$

$\frac{2 a - 5}{a - 9} + \frac{a}{a + 9} = - \frac{6}{\left(a - 9\right) \left(a + 9\right)}$

$\frac{\left(a + 9\right) \left(2 a - 5\right) + \left(a - 9\right) a = - 6}{\left(a - 9\right) \left(a + 9\right)}$

Multiply both sides by $\left(a - 9\right) \left(a + 9\right)$

$\left(a + 9\right) \left(2 a - 5\right) + \left(a - 9\right) a = - 6$

$2 {a}^{2} + 13 a - 45 + {a}^{2} - 9 a = - 6$

$3 {a}^{2} + 4 a - 39 = 0$

$\left(3 a + 13\right) \left(a - 3\right) = 0$

$3 a = - 13 \mathmr{and} a = 3$

$a = - \frac{13}{3} \mathmr{and} a = 3$

extraneous solutions