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

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

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

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

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

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 $\boxed{{\displaystyle -3}}$.

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