A simple way to solve a system of three equations

Question

I would like to find a simple way to solve the system:

$mv=(m+M)v_1,\ \ \ \ \qquad(1)$

$2mv=(2m+M)v_2,\qquad(2)$

$3mv=(3m+M)v_3.\qquad(3)$

It will be nice to know some trick (of course, if it exists) to express $v_3=v_3(v_1,v_2)$ easy. The answer is:

$v_3=\dfrac{3v_1v_2}{4v_1-v_2}$.

My attempt:

All variables are positive: $m,M,v,v_1,v_2,v_3>0.$

Dividing (3) by (2) we obtain:

$\dfrac{3}{2}=\dfrac{3m+M}{2m+M}\dfrac{v_3}{v_2},$

and from here we get:

$v_3=\dfrac{3}{2}\dfrac{2m+M}{3m+M}v_2.\ \ \ \qquad(4)$

Also, dividing (2) by (1) we have:

$2=\dfrac{2m+M}{m+M}\dfrac{v_2}{v_1},$

and from here we obtain:

$2mv_2+Mv_2=2mv_1+2Mv_1,$

and from here:

$m=\dfrac{M(2v_1-v_2)}{2(v_2-v_1)}.$

Let's substitute this expression for $m$ into (4), we get:

$v_3=\dfrac{3}{2}v_2\dfrac{\dfrac{M(2v_1-v_2)}{v_2-v_1}+M}{\dfrac{3}{2}\dfrac{M(2v_1-v_2)}{v_2-v_1}+M}=\dfrac{3}{2}v_2\dfrac{2v_1-v_2+v_2-v_1}{\dfrac{3}{2}(2v_1-v_2)+v_2-v_1}=\dfrac{3}{2}\dfrac{v_1v_2}{2v_1-\dfrac{1}{2}v_2}=\dfrac{3v_1v_2}{4v_1-v_2}.$

So, the answer is:

$v_3=\dfrac{3v_1v_2}{4v_1-v_2}.$

And I am interested if there is an easier way to get this result for $v_3$.

Answer 1

Letting $$x:=\frac v{v_1},\;y:=\frac v{v_2},\;z:=\frac v{v_3},\;t:=\frac Mm,$$ the system becomes $$x=1+t,\;y=1+\frac t2,\;z=1+\frac t3$$ hence $$z=ax+(1-a)y$$ where $a$ is determined by $a+\frac{1-a}2=\frac13,$ i.e. $a=-\frac13.$ As a result, $$z=\frac{4y-x}3,$$ i.e. $$v_3=\frac{3v_1v_2}{4v_1-v_2}.$$