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