# A simple way to solve a system of three equations

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:

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

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

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

• That system is not linear. Commented Nov 20, 2023 at 18:47
• The answer is not correct - consider, say, $m=1$ and $M=-3$. Then your formula gives $v_3=\frac{0}{0}$, which is not defined. Commented Nov 20, 2023 at 19:27
• Where do $m,M,v$ live? They could be for instance the indeterminates of a polynomial ring like $\Bbb Z[m,M,v]$. And the solutions $(v_1,v_2,v_3)$ (of the linear system) could then belong to the field $\Bbb Q(m,M,v)$. Commented Nov 20, 2023 at 19:37
• @AnneBauval, thanks and sorry please, Now I've done this. Commented Nov 21, 2023 at 21:22

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. $$v_3=\frac{3v_1v_2}{4v_1-v_2}.$$
• If $a=-\frac{1}{3}$ then why did you write $z$ with respect to $a$ first?
• I solved $1+\frac t3=a(1+t)+(1-a)\left(1+\frac t2\right)$ for $a$ and found $a=-\frac13.$ @EtackSxchange Commented Nov 21, 2023 at 22:34
• @AnneBauval, thanks. I've understood. Could you tell please how did you know that we can write $z=ax+(1-a)y$ and then find $a$ from here? We can write $z=ax+by$, but how did you know that b=$1-a$? Or you just found this through calculations? Commented Nov 22, 2023 at 17:57
• I noticed the $1$ in each of $z,x,y$, which made me immediately find $b=1-a.$ Commented Nov 22, 2023 at 19:37
• @AnneBauval, thanks! I understood. It follows from $1=a\cdot1+b\cdot1$. Commented Nov 22, 2023 at 21:34