Question

$\dfrac{24x^2+25x-47}{ax-2}=-8x-3-\dfrac{53}{ax-2}$
Here, x≠a/2
Find the value of a.

Answer 1

ANSWER:

Given:

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

To Find:

• Value of a.

Solution:

$:\longrightarrow\dfrac{24x^2+25x-47}{ax-2}=-8x-3-\dfrac{53}{ax-2}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{24x^2+25x-47}{ax-2}=\dfrac{-8x(ax-2)-3(ax-2)-53}{ax-2}\\\\:\implies\dfrac{24x^2+25x-47}{ax-2}=\dfrac{-8ax^2+16x-3ax+6-53}{ax-2}\\\\\text{Cancelling (ax-2) on both the fractions' denominators.}\\\\:\implies 24x^2+25x-47= -8ax^2+16x-3ax+6-53\\\\:\implies 24x^2+25x-47= -8ax^2+16x-3ax-47\\\\:\implies 24x^2+25x-47= -8ax^2+(16-3a)x-47 \\\\\text{As both the equations are quadratic,}\\\\\text{We compare the coefficients of x ^2 and x}\\\\:\implies24=(-8a) \:\:\:\:AND\:\:\:\:25=(16-3a)\\\\\text{So,}\\\\:\implies a=-3\:\:\:\:AND\:\:\:\:25-16=-3a\\\\:\implies a=-3\:\:\:\:AND\:\:\:\:9=-3a\\\\:\implies a=-3\:\:\:\:AND\:\:\:\:a=-3\\\\\text{\bf{So, a = -3}}$