Subtract 5 from both sides

2x=3kx

Divide both sides by 3x, which can only be
done for non-zero values of x.





That's no doubt the desired value of k, but we must continue
on to show that it is also true when x=0 and k=.

So we investigate to see if it also holds true for x=0:

When we replace k by  in 



we get 



and when we substitute x=0







Now since we have shown that it also holds for x=0, 
it holds for ALL values of x.

[To have a rigorous proof we needed to show that
it also held for x=0]

Edwin