Linear combinations problem The vectors $\dbinom{3}{2}$ and $\dbinom{-4}{1}$ can be written as linear combinations of $\mathbf{u}$ and $\mathbf{w}$:
\begin{align*}
 \dbinom{3}{2} &= 5\mathbf{u}+8\mathbf{w} \\
 \dbinom{-4}{1} &= -3\mathbf{u}+\mathbf{w} . 
\end{align*}The vector $\dbinom{5}{-2}$ can be written as the linear combination $a\mathbf{u}+b\mathbf{w}$. Find the ordered pair $(a,b)$.
I've tried to eliminate $\mathbf{u}$ by multiplying the first equation by 3, the second equation by 5, then adding, but it only leads to $\mathbf{w}=\dbinom{-\frac{11}{29}}{\frac{11}{29}}$. I feel like the algebra from here would be too complicated for what the people who wrote the problem were intending, so perhaps I'm going down the wrong path. Would there instead be a convenient way to manipulate the terms to eventually get $\dbinom{5}{-2}$ on the LHS?
 A: You can certainly solve for $\mathbf u$ and $\mathbf w$ via the system of equations that you’ve got. If you expand component by component, there are four linear equations in four unknowns, so there’s a unique solution. Once you have $\mathbf u$ and $\mathbf w$, you can then set up another system of equations to compute $a$ and $b$.  
I’m pretty sure that the intended method is instead to write $(5,-2)^T$ as a linear combination of $(3,2)^T$ and $(-4,1)^T$ and then use linearity. That is, \begin{align} \pmatrix{5\\-2} &= c\pmatrix{3\\2}+d\pmatrix{-4\\1} \\ &= c(5\mathbf u+8\mathbf w)+d(-3\mathbf u+\mathbf w) \\ &=(5c-3d)\mathbf u + (8c+d)\mathbf w. \end{align}
A: Here is how I would do it.
$\pmatrix{3\\2} - 2\pmatrix{-4\\1} = \pmatrix{11\\0}$
Allowing us to find one of the princicipal component vectors of the standard basis in terms of the $\{u,w\}$ basis.
$5u + 8w - 2(-3u+w) = \pmatrix{11\\0}\\
11u + 6w = \pmatrix{11\\0}\\
\pmatrix{1\\0} = u + \frac {6}{11} w$
With that we can say:
$\pmatrix{5\\-2} = 8\pmatrix{1\\0}-\pmatrix{3\\2}$
And write our vector in terms of this basis.
$\pmatrix{5\\-2} = 8(u+\frac 6{11}w) - (5u+8w) = 3u- \frac {40}{11}w$
