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This question has been asked before, but I want it specifically answered without the use of matrices. The purpose of this question is to answer it without matrices, just using the definition/knowledge of a linear combination.

My approach was to let $r$, $s$ be some scalars, so I would get

$$r(1,5) + s(3,c) = (13,-15) =>(r,5r) + (3s,cs) - (13,-15)$$ $$=> r+3s=13 \text{ and } 5r+cs=-15$$ I tried to solve this by isolating variables, but it didn't work out for me. I know the answer is all $c$, $c$ cannot be $15$. Help Please.

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If you want to do it without matrices too, the solution is straightforward. Just eliminate one of the variables, say r(because it doesn't have c in it). Then we have, $$5*13+(c-15)*s = -15 \implies s*(c-15) = -80$$. Now observe that $s$ has always a unique solution when $c\neq15.$When, $c=15,$ LHS=0, while RHS=-80, so there is no solution for s and so no solution for c. In other terms, $x^{-1}$ exists when $x\neq0 (x\in \mathbb R)$, and $\mathbb R$ is a field. Now try to generalize this statement to vector spaces and you get the matrix solution too. (The notion of invertibility of a matrix)

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\begin{align} r + 3s &= 13\\ 5r + cs &= -15 \\ \\ 5r + 15s &= 65 & \text{multiply first row by $5$}\\ 5r + cs &= -15\\ \\ 5r + 15s &= 65 & \\ 0r + (c-15)s &= -80& \text{subtract first from second}\\\\ \\ 5r + 15s &= 65 & \\ 0r + s &= \frac{-80}{c-15}& \text{divide through}\\\\ \end{align} and from this, you can substitute for $s$ in the first equation to get $r$ and you're done.