solve system to a variable: (v1+v2)/t1 = s, (v1-v2)/t2 = s

[xoninhas]

I don't know if this belongs here but here it goes:

solve the system in order to v1 without v2 showing up.
(v1+v2)/t1 = s and (v1 - v2)/t2 = s

can anyone get to v1 = (s/2) * (1/t1 + 1/t2)

if so, can you post the steps?

first I get

v1 = s.t1 - v2
=> s.t1 -2.v2 = s.t2
=> v2 = (s/2)*(t1 - t2)
=> v1 = s.t1 - (s/2)*(t1-t2)
=> v1 = (s/2)*(t1+t2)

 

[soroban]

Re: solve system to a variable

Hello, xoninhas!

Solve the system in order to \(\displaystyle v_1\) without \(\displaystyle v_2\) showing up.

. . \(\displaystyle \frac{v_1+v_2}{t_1} \:=\: s\;\text{ and }\;\frac{v_1 - v_2}{t_2} \:=\: s\)

\(\displaystyle \text{Can anyone get to: }\;v_1 \:= \:\frac{s}{2}\left(\frac{1}{t_1} + \frac{1}{t_2}}\right)\) . . . . no

\(\displaystyle \text{We have: }\;\begin{array}{cccccccc}\dfrac{v_1 + v_2}{t_1} \:=\: s & \Rightarrow & v_1 + v_2 \:=\: st_1 & {\bf[1]} \\ \\[-3mm] \dfrac{v_1-v_2}{t_2} \:=\: s & \Rightarrow & v_1-v_2 \:=\: st_2 & {\bf[2]}\end{array}\)

\(\displaystyle \text{Add {\bf[1]} and {\bf[2]}: }\;2v_1 \:=\:st_1 + st_2\quad\Rightarrow\quad\boxed{v_1 \:=\:\frac{s}{2}(t_1 + t_2)}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

There is a typo somewhere . . .

\(\displaystyle \text{If the original equations were: }\;\begin{array}{c}(v_1+v_2)t_1 \:=\:s \\ (v_1-v_2)t_2 \:=\:s\end{array}\)
. \(\displaystyle \text{their answers would be correct.}\)

 

[xoninhas]

Yep, exactly what I thought, and exactly what I got. Thanks for confirming it... it seems I always need another set of eyes to take a look at things! :S Thanks!