"Get $a$ and $B$ in terms of $m_1$, $m_2$": where $m_1 = aB / (a - 1)$ , $m_2 = ( aB^2 ) / (a - 2)$

Question

I have no idea how to solve this.
- I tried solving for B in terms of a and m1 - Then I plugged B into m2, but was not able to solve for a.

How do you do this ?

Thanks !

Answer 1

This is the system of equations:

$\begin{align*} m_1 (a - 1) &= a B \\ m_2 (a - 2) &= a B^2 \end{align*}$

Get $B$ from the first equation, replace in the second:

$\begin{align*} B &= \frac{m_1 (a - 1)}{a} \\ m_2 (a - 2) &= a \left(\frac{m_1 (a - 1)}{a}\right)^2 \\ m_2 a (a - 2) &= m_1^2 (a - 1)^2 \\ m_2 a^2 - 2 m_2 a &= m_1^2 a^2 - 2 m_1^2 + m_1^2 \end{align*}$

This results in the quadratic:

$\begin{align*} (m_2 - m_1^2) a^2 - 2 (m_2 - m_1^2) a - m_1^2 &= 0 \end{align*}$

Solve that one, replace in the expression for $B$.