How do you solve $a/(2a+1) - (2a^2+5)/ (2a^2-5a-3) =3/(a-3)$ ?

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Answer 1 · 2015-05-26T11:53:10

Given $a/(2a+1)-(2a^2+5)/(2a^2-5a-3) = 3/(a-3)$

Noting that $2a^2-5a-3 = (2a+1)(a-3)$
we can clear the denominators from the given equation by multiplying by $2a^2-5a-3$

$a(a-3) -(2a^2+5) = 3(2a+1)$

$a^2 -3a -2a^2 -5 = 6a +3$

$a^2+9a+8 =0$

Factor the left side
$(a+1)(a+9) = 0$

State the solutions
$a=-1$ or $a=-9$

How do you solve a/(2a+1) - (2a^2+5)/ (2a^2-5a-3) =3/(a-3)a/(2a+1) - (2a^2+5)/ (2a^2-5a-3) =3/(a-3)?