Show that no choice of numbers $a$ and $b$ can make $ax + by = (3,0,0)$ 
Show that no choice of numbers $a$ and $b$ can make $ax + by = (3,0,0)$ when $x = (3,-1,0)$ and $y = (0,1,5)$.

The only materials in the chapter talked about are:


*

*Vector Space Operations

*Standard Basis

*Coordinates of a vector $x$

*Components of a vector $x$
I don't think that the vector $v=(3,0,0)$ has a unique linear combination of coordinates with respect to the standard basis with how $x$ and $y$ are defined. I also don't think stating this is "rigorous" enough of an answer. Is there a better (clearer) way I can show that the above statement is true?
How can I show that the statement is true from a geometric point of view?

You may not have learned this yet, but
$n$ vectors in $ \mathbb {R} ^{n}$ are linearly independent if and only if 
the determinant of the matrix formed by taking the vectors as its columns is non-zero.   
In your particular case, that matrix has a form (upper triangular) that makes it very easy to compute 
the determinant (simply multiply the diagonal entries):
$\begin{vmatrix} 3 & 3 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 5 \end{vmatrix}=-15\ne0,$ so they're linearly independent.
