Given the following system:
[tex]\mleft\{\begin{aligned}x+5y=4 \\ 3x+15y=-1\end{aligned}\mright.[/tex]
We can solve for x on the first equation and then subsitute it's value on the other equation:
[tex]\begin{gathered} x+5y=4 \\ \Rightarrow x=4-5y \\ 3x+15y=-1 \\ \Rightarrow3(4-5y)+15y=-1 \\ \Rightarrow12-15y+15y=-1 \\ \Rightarrow12=-1 \end{gathered}[/tex]
We found an inconsistency on the system, since 12=-1 can never happen. Therefore, there is no solution to the system.
