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The equations $\frac{24 x^{2}+25 x-47}{a x-2}=-8 x-3-\frac{53}{a x-2}$ is true for all values of $x \neq \frac{2}{a}$, where $a$ is a constant. What is the value of $a$ ?

Step 1 ：First, we can rewrite the equation as \(\frac{24 x^{2}+25 x-47+53}{a x-2}=-8 x-3\).

Step 2 ：Then, we can multiply both sides by \(a x-2\) to get rid of the denominator, which gives us \(24 x^{2}+25 x-47+53=-8 x(a x-2)-3(a x-2)\).

Step 3 ：Simplify the equation to get \(24 x^{2}+25 x+6=-8 a x^{2}+16 x-3 a x+6\).

Step 4 ：By comparing the coefficients on both sides, we can get two equations: \(24=-8 a\) and \(25=16-3 a\).

Step 5 ：Solving the first equation gives us \(a=-3\).

Step 6 ：Substitute \(a=-3\) into the second equation, we get \(25=16+9\), which is true.

Step 7 ：So, the value of \(a\) is \(\boxed{-3}\).

Step 8 ：We can check our answer by substituting \(a=-3\) back into the original equation, and it holds true for all values of \(x \neq \frac{2}{a}\).