# Thread: Trick 4 solving 4 n

1. ## Trick 4 solving 4 n

In the equation:

Lambda1=(n+1)Lambda2

What is the trick 2 get the solution:

n=Lambda2/(Lambda1-Lambda2)

2. Originally Posted by Jaxwell
In the equation:
Lambda1=(n+1)Lambda2
What is the trick 2 get the solution:
n=Lambda2/(Lambda1-Lambda2)
That is actually wrong.

$\begin{gathered}
\lambda _1 = \left( {n + 1} \right)\lambda _2 \hfill \\
\lambda _1 = n\lambda _2 + \lambda _2 \hfill \\
n\lambda _2 = \lambda _1 - \lambda _2 \hfill \\
\end{gathered}$

3. Thank You,
You are of course right.

I meant: n*Lambda 1 =(n+1)Lambda 2

with above solution?

4. Well, you see how things are done from my reply.

5. I understand Your reply fully, thanks:-)
But since there is an n on both sides the correct equation:
nLambda 1=(n+1)Lambda 2
with solution:
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

6. Do you realize how difficult it is to read your postings?
Why not learn to post in symbols? You can use LaTeX tags
$$n\lambda_1=(n+1)\lambda_2$$ gives $n\lambda_1=(n+1)\lambda_2$.

7. Thank You, I will try that. The equation:

$n\lambda_1=(n+1)\lambda_2$

has the solution:

$n=\lambda_2/(\lambda_1-\lambda_2)$

If You know the technique, I would be highly grateful

8. Originally Posted by Jaxwell
Thank You, I will try that. The equation:

$n\lambda_1=(n+1)\lambda_2$

has the solution:

$n=\lambda_2/(\lambda_1-\lambda_2)$

If You know the technique, I would be highly grateful
Just change slightly what Plato did for your original equation:

$n\lambda_1 = n \lambda_2 + \lambda_2$

$\Rightarrow n (\lambda_1 - \lambda_2) = \lambda_2$

and I think the answer is straight ahead.

9. Originally Posted by Jaxwell
Thank You, I will try that. The equation:

$n\lambda_1=(n+1)\lambda_2$

has the solution:

$n=\lambda_2/(\lambda_1-\lambda_2)$

If You know the technique, I would be highly grateful
It's pretty much basic algebra- you have to get n by itself so first you have to "free" it from the right side by using the "distributive" property: $(n+1)\lambda_2= n\lambda_2+ \lambda_2$. At that point you have
$n= n\lambda2+ \lambda2$
Still trying to get n by itself, subtract $n\lambda$ from both sides:
$n\lambda_1- n\lambda_2= \lambda_2$

Now use the "distributive" property again:
$n(\lambda_1- \lambda_2)= \lambda_2$
and, finally, divide both sides by $(\lambda_1- \lambda_2)$.

That's essentially just what Plato did before.

10. Originally Posted by Jaxwell
I meant: n*Lambda 1 =(n+1)Lambda 2
Why don't you make your life a bit easier; use x and y for them Lambdas:
nx = (n + 1)y
nx = ny + y
nx - ny = y
Take over, Rover...