Let's v1 = (1,4) and v2 = (0,3). Now find a and b such that
(2,3) = a*v1 + b*v2
(2,3) = a*(1,4) + b*(0,3)
From here we have linear system
"\\begin{cases}\n a*1 + b*0 = 2\\\\\n a*4 + b*3 = 3\n\\end{cases}"
From first equation we find a = 2 and substitute it to second we find b = -5/3. Exist a, b in R such that (2,3) = a*v1 + b*v2, where v1, v2 in S
Hence (2,3) belongs to L(S) - the linear span of S.
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