Real Algebra

$\begin{aligned} \frac {24x^2+25x-47}{ax-2} & = - 8x - 3 - \frac {53}{ax-2} \\ \frac {24x^2+25x+6- 53}{ax-2} & = - 8x - 3 - \frac {53}{ax-2} \\ \frac {24x^2+25x+6}{ax-2} -\frac {53}{ax-2} & = - 8x - 3 - \frac {53}{ax-2} \\ \implies 24x^2+25x+6 & = (-8x-3)(ax-2) \\ \implies a & = \boxed {-3} \end{aligned}$

$\dfrac{24x^2+25x-47}{ax-2} = -8x -3 - \dfrac{53}{ax-2}$

$\implies 24x^2+25x-47 = -8x(ax-2) -3(ax-2) - 53$

$\implies 24x^2+25x-47 = -8ax^2+16x -3ax +6 - 53$

$\implies 24x^2+25x-47 = -8ax^2+16x -3ax -47$

$\implies 24x^2+25x = -8ax^2+16x -3ax$

$\implies x(24x+25) = x(-8ax+16-3a)$

$\implies (24x+25) = (-8ax+16-3a)$

$\implies 24x+9 = -8ax-3a = -a(8x+3)$

$\implies 3(8x+3) = -a(8x+3)$

$\implies 3 = -a$

$\implies a=\boxed{-3}$

There are two ways to solve this question. The faster way is to multiply each side of the given equation by ax−2 (so you can get rid of the fraction). When you multiply each side by ax−2, you should have: $24{ x }^{ 2 }+25x-47$ ==(−8x−3)(ax−2)−53 You should then multiply (−8x−3) and (ax−2) . $24{ x }^{ 2 }+25x-47$ =−8ax^2−3ax+16x+6−53 Then, reduce on the right side of the equation $24{ x }^{ 2 }+25x-47$ =−8ax^2−3ax+16x−47 Since the coefficients of the x^2-term have to be equal on both sides of the equation, −8a=24, or a=−3.

Comparing coefficients of $x^2$ on both sides of the equation we get $-8a=24\implies a=\boxed {-3}$ .