given the following statement: the sum of 2 numbers namely 'a' and 'b' is equal to 'm' times their difference, and the product of those 2 numbers is 'n' times their difference.
we get the system:
a+b=m(a-b)
a*b=n(a-b)
and I'm able to decompose it like:
a+b=ma-mb
a*b=na-nb
.........
a=m(a-b)-b
b=m(a-b)-a
a=n(a-b)/b
b=n(a-b)/a
now, I know the answer is of the form:
a = 2n/m-1 and b = 2n/m+1
how is the process to solve this system of equations?
Hint: set $b=ka$ then the first equation gives you $k=\frac{m-1}{m+1}$ and proceed to solve the second equation.
When you have some equations like this, it is often useful to test things like $b=ka$ our $b=a+k$ and see if you can get something out of it.
\begin{align} 4ab &=(a+b)^2-(a-b)^2\\ 4n(a-b) & =(a+b)^2-(a-b)^2\\ &= m^2(a-b)^2-(a-b)^2\\ &= (m^2-1)(a-b)^2\\ \implies (a-b) &=\dfrac{4n}{m^2-1} \text{, as obviously $a \ne b$}\\ \end{align}
Can you proceed from here?
