Hello, blackjack21!

From [1]: .

From [2]: .

Equate [4] and [5]: .

. . .

Substitute into [3]: .

Divide by

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

After dividing by , I had to consider: what if

Then equation [6] says:

And equation [3] becomes: .

. . which is true forvalue ofany

I suppose that counts as another solution.