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The equation $$\frac{24x^2+25x-47}{ax-2} = -8x-3-\frac{53}{ax-2} $$ is true for all values of $$ x \neq \frac{2}{a} $$ where a is a constant.

What is the value of a?

Please someone help me solve this question? This is a question from the RD Sharma Book which I was not able to solve. Actually I solved but I want a more better way to solve it. I putted x as 0 and then got the value of a as -3.

Please note this has been edited so the equation in the question accords with the accepted answer.

Mark Bennet
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    I have edited your equation (using \frac to format fractions) as I think you intended. For large $x$ the left-hand side is large and positive while the right-hand side is smaller (in absolute value) but negative. Could you please check that you have got the right equation (and that my edit is correct). – Mark Bennet Jan 27 '21 at 08:19
  • Thanks @MarkBennet your edit and equation is correct –  Jan 27 '21 at 09:50
  • There is nothing wrong with taking a specific value of $x$. However, taking $x=0$ eliminates every occurrence of $a$ and can't be used to solve for $a$ – Mark Bennet Jan 27 '21 at 10:28

1 Answers1

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It appears the initial equation has been edited several times, I believe this might be what you intended originally:

$$ \frac{24x^2+25x-47}{ax-2} = -8x-3-\frac{53}{ax-2} , \ x \neq \frac{2}{a} . $$

If that's the case, then we have

$$ \frac{24x^2+25x-47}{ax-2} = -8x-3-\frac{53}{ax-2} , \ x \neq \frac{2}{a} $$

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

$$ \Rightarrow \frac{24x^2+25x+6}{ax-2} = -8x-3 $$

$$ \Rightarrow \frac{(8x+3)(3x+2)}{ax-2} = -8x-3 $$

$$ \Rightarrow (8x+3)(3x+2) = -(8x+3)(ax-2) $$

$$ \Rightarrow 3x+2 = -(ax-2) $$

$$ \Rightarrow a = -3 $$

Amaan M
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