<s>[INST] Find basis for subspace given by $x_1/3=x_2/4=x_3/2$ Find a basis for the subspace W: $$
W = \left\{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix} \in \mathbb{R}^3 \left| \dfrac{x_{1}}{3}=\dfrac{x_{2}}{4}=\dfrac{x_{3}}{2} \right.\right\}
$$
The two equation signs are really throwing me off. If it was a linear equation with one equation sign then I could easily find the basis by putting the equation in a matrix and solve for each variable but I cannot do this for this equation. Can anyone help me out? Thanks.
 [/INST] What you have here are two equation:
$$
\frac{x_{1}}{3}=\frac{x_{2}}{4}
$$
$$
\frac{x_{2}}{4}=\frac{x_{3}}{2}
$$
from the first one you get $x_{2}=\frac{4}{3}x_{1}$ and from the
second one you get $x_{3}=\frac{1}{2}x_{2}$, now using $x_{2}=\frac{4}{3}x_{1}$
we get $x_{3}=\frac{1}{2}\cdot\frac{4}{3}x_{1}=\frac{2}{3}x_{1}$
so 
$$
W=\{\begin{pmatrix}x_{1}\\
\frac{4}{3}x_{1}\\
\frac{2}{3}x_{1}
\end{pmatrix}\mid x_{1}\in\mathbb{R}\}=\{\begin{pmatrix}1\\
\frac{4}{3}\\
\frac{2}{3}
\end{pmatrix}x_{1}\mid x_{1}\in\mathbb{R}\}=sp\{\begin{pmatrix}1\\
\frac{4}{3}\\
\frac{2}{3}
\end{pmatrix}\}
$$
A small thing to note: we have two independent equations and so the
dimension of $W$ is $3-2=1$ as expected
</s>