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The solution to this system of equations will depend on the value of $a$. Let's consider two cases:

Case 1: $a = 0$

Substitute $a = 0$ into the second equation:

$y = 10$

Now, substitute $y = 10$ into the first equation:

$\left| x-100\right|+\left| 10-100\right|=25$

$\left| x-100\right| + 90 = 25$

This equation has no solution because the absolute value of a number is always non-negative, so it cannot equal a negative number.

Case 2: $a \neq 0$

Substitute $y = a\,x + 10a$ into the first equation:

$\left| x-100\right|+\left| a\,x+10a-100\right|=25$

Since we have two absolute values, we need to consider the different cases for each absolute value:

1. $x-100 \geq 0$ and $a\,x+10a-100 \geq 0$:

$x-100 + a\,x+10a-100 = 25$ $(1+a)x + 10a - 200 = 25$ $(1+a)x = 225 - 10a$ $x = \frac{225 - 10a}{1+a}$

2. $x-100 \geq 0$ and $a\,x+10a-100 < 0$:

$x-100 + 100 - a\,x-10a = 25$ $(1-a)x - 10a = 25$ $x = \frac{25 + 10a}{1-a}$

3. $x-100 < 0$ and $a\,x+10a-100 \geq 0$:

$100-x + a\,x+10a-100 = 25$ $(a-1)x + 10a = 25$ $x = \frac{25 - 10a}{1-a}$

4. $x-100 < 0$ and $a\,x+10a-100 < 0$:

$100-x + 100 - a\,x-10a = 25$ $(a-1)x - 10a = 25$ $x = \frac{25 + 10a}{1-a}$

These are the possible solutions for the system of equations given the value of $a$.