Linear combination problem

Question

Show that $v$ can be written as the linear combination:

$$v = \frac{1}{3}s + \frac{1}{6}t$$

I don't see it. When I combine the linear combinations for $s$ and $t$ expressed by $u$ and $v$, I get $s+t=u+5v$, but as far as I understand, $u$ should be eliminated in the equation. Here are the equations for $s$ and $t$ (the equation for $s$ is right, but I'm not sure about $t$):

$$s=2v-u$$

$$t=2u+3v$$

Answer 1

First note that $t = 2u + 2v$, and not $t = 2u + 3v$.

Then all it takes is to substitute $s$ and $t$ in the first equation:

$$\frac13 s + \frac16 t =\\ \frac13 (2v - u) + \frac16 (2u + 2v) =\\ \frac23v - \frac{1}{3}u + \frac26u + \frac26v =\\ \frac23v + \frac13v - \frac13u + \frac13u =\\ \frac33v + \frac03u = v$$

Answer 2

Hint (using the equations as posted, not as indicated by the figure):

$$u=2v-s$$

$$3v=t-2u=t-2(2v-s)=-4v+2s+t$$

$$7v=2s+t$$

If the second equation $t=2u+3v$ is replaced with $t=2u+2v$ then you obtain the given relation.

Answer 3

Using vector notation $\left<1,1\right>=u+v$: $$\frac13*\left<-1,2\right>+\frac16\left<2,2\right>=\left<-\frac13,\frac23\right>+\left<\frac13,\frac13\right>=\left<-\frac13+\frac13,\frac23+\frac13\right>=\left<0,1\right>$$