The Three-Number Problem I am currently working on a little extra credit for my 9th grade math class and I am stuck on the one of the problems.

The Three-Number Problem
I have chosen three numbers. The second is twice the first, and the third is three times the second. The sum of the first two when multiplied by the sum of the last two happens to be the same as the first number multiplied by the quare of the second number.

I decided to approach this by first starting with 3 variables.
f, s, and t.
f standing for the first number.
s standing for the second number.
t standing for the third number.
f = #
s = 2f
t = 3s
The equation I came up with was:
(f + s) * (s + t) = fs2
Which can be simplified to:
fs + ft + s2 = fs2
However, from there on, I am stuck. I have no idea how to solve for f. However, i'm sure that i'm in the right place and that somebody here can help me out. Thank you!
 A: So you have $f$; $s = 2f$; and $t = 3s = 6f$.
Furthermore, $(f+s)(s+t) = fs^2$.
Rewriting the latter equation with only $f$, this would mean:
$$(f+2f)(2f + 6f) = f(4f^2)$$
Can you now solve for $f$?

Since the other answer gives full details:
Either $f = 0$ and we are done, or divide both sides by $f^2$ to obtain $24 = 4f$.
Then $f = 6$. QED
A: Welcome to Math.SE. 
Your approach is quite correct so far, though you made a mistake. You wrote that:

$(f + s) \cdot (s + t) = fs^{2}$
Which can be simplified to:
$fs + ft + s^{2} = fs^{2}$

Whereas it really simplifies to:
$fs + ft + s^{2} + st = fs^{2}$.
Let's use your definition of the variables to do some substitution.
$fs + ft + s^{2} + st = fs^{2}$
Substitute $t$ for $3s$.
$fs + f3s + s^{2} + s(3s) = fs^{2}$.
$fs + f3s + s^{2} + 3s^{2} = fs^{2}$.
$fs + f3s + 4s^{2} = fs^{2}$.
Substitute $s$ for $2f$.
$f(2f) + f(6f) + 4(2f)^{2} = f(2f)^{2}$
$2f^{2} + 6f^{2} + 4(4f^{2}) = f(4f^{2})$
$2f^{2} + 6f^{2} + 16f^{2} = f(4f^{2})$
$24f^{2} = 4f^{3}$.
$6f^{2} = f^{3}$
Divide both sides by $f^{2}$.
$6 = f$
