I understand Your reply fully, thanks:-)
But since there is an n on both sides the correct equation:
nLambda 1=(n+1)Lambda 2
n=lambda 2/(lavmbda 1-lambda 2) has a different technique which I cant figure out. If You know how I would be very grateful
Still trying to get n by itself, subtract from both sides:
Now use the "distributive" property again:
and, finally, divide both sides by .
That's essentially just what Plato did before.