Given that x≠4 and we need to find the range of the solutions of the equation \(|14–x|=\frac{24}{(x−4)}\)

Now, in order to solve this equation we need to deal with the Absolute Value of |14-x|
We can do that by taking two cases

Case 1: 14 -x >= 0 or x <= 14
=> | 14-x | = 14 -x
=> \(14–x = \frac{24}{(x−4)}\)
=> (14-x) * ( x-4) = 24
We can either use Algebra (open the equation and solve) or can substitute values

Substitution

(14-x) * ( x-4) = 4 * 6
If we put x = 10 then we will get
(14-10) * (10-4) = 4*6
So, x = 10

(14-x) * ( x-4) = 6 * 4
If we put x = 8 then we will get
(14-8) * (8-4) = 6*4
So, x = 8

Algebra

(14-x) * ( x-4) = 24
=> 14x - 56 - \(x^2\) + 4x = 24
=> - \(x^2\) + 18x - 56 - 24 = 0
=> - \(x^2\) + 18x - 80 = 0
Multiply by -1 we get
=> \(x^2\) - 18x + 80 = 0
=> \(x^2\) - 8x - 10x + 80 = 0
=> x*(x-8) - 10*(x-8) = 0
=> (x-8) * (x-10) = 0
=> x = 8 or 10

Both values are <= 14
=> x can be 8 or 10

Case 2: 14 -x < 0 or x > 14
=> | 14-x | = -(14 -x) = x - 14
=> \(x-14 = \frac{24}{(x−4)}\)
=> (x-14) * ( x-4) = 24
We can either use Algebra (open the equation and solve) or can substitute values

Substitution

(x-14) * (x-4) = 2 * 12
If we put x = 16 then we will get
(16-14) * (16-4) = 2*12
So, x = 16
Since 16 > 14
=> x = 16 is a solution

Algebra

(x-14) * ( x-4) = 24
=> \(x^2\) - 4x -14x + 56 = 24
=> \(x^2\) - 18x + 56 - 24 = 0
=> \(x^2\) - 18x + 32 = 0
=> \(x^2\) - 2x - 16x + 32 = 0
=> x*(x-2) - 16*(x-2) = 0
=> (x-2) * (x-16) = 0
=> x = 2 or 16
But x > 14
=> x = 16 is a solution

Combining both the cases possible values of x are 8, 10, 16

Range = Highest value - Lowest value = 16-8 = 8

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Absolute Values