Expressing a vector v as a linear combination of x and y 
Express the vector $v = \begin{bmatrix}49\\0\end{bmatrix}$ as a linear combination of $x = \begin{bmatrix}6\\5\end{bmatrix}$ and $y = \begin{bmatrix}-5\\4\end{bmatrix}$
$v = $ ____ $x + $ ______$y$

How exactly can I express this vector as a linear combination of these other two vectors? I've never seen a problem like this before. Nor do I know exactly what a Linear Combination is in "laymans terms"
 A: Just put:
$$v = ax+by $$
then solve for $a$ and $b$
\begin{cases}
6a-5b = 49\\
5a+4b = 0
\end{cases}
You will get: 
$$v = 4x-5y$$
A: Suppose that you have vectors $x_1,..,x_n \in V$, where $V$ is a vector space over a field $F$ (For example $\mathbb{R}^2$ over $\mathbb{R}$).  Then $z$ is said to be a linear combination of vectors $x_1,..,x_n$ if there exists scalars $a_1,..,a_n \in F$ such that:
\begin{equation}
z = a_1x_1+...+a_nx_n \in V
\end{equation}
In this case you have $V = \mathbb{R}^2$ and $F = \mathbb{R}$. You want to find scalars $a_1,a_2 \in \mathbb{R}$ such that $v = a_1x + a_2y$.
In order to solve this, start from this definition and plug in your information:
\begin{align}
    v = \begin{bmatrix}
           49 \\
           0
         \end{bmatrix} = a_1\begin{bmatrix}
           6 \\
          5
         \end{bmatrix} + a_2\begin{bmatrix}
           -5 \\
          4
         \end{bmatrix}
  \end{align}
Now you can use standard properties of vectors and multiply every component of $x$ by $a_1$, and multiply every component of $y$ by $a_2$. Hence you get a systems of two equations and two unknown, which is consistent.
you have
\begin{equation}
49 = 6a_1-5a_2 
\end{equation}
and:
\begin{equation}
0 = 5a_1+4a_2
\end{equation}
You can then solve for $a_1$ and $a_2$. From the second one you get $a_1 = -\frac{4}{5}a_2$, then you substitute into the first one and obtain $49 = -\frac{24}{5}a_2-5a_2$, which then becomes $49 = -\frac{49}{5}a_2$, so $a_2 = -5$, so $a_1 = 4$.
Finally you get $v = 4x-5y$, where $4,-5\in \mathbb{R}$
A: Let the equation be $v=lx+my$. where $l,n\in R$.So now you get two equations ie $6l-5n=49,5l+4n=0$. Now you can solve for $l,n$
