Question

I need help with this problem. I started it but I got to a point where it seemed wrong so I just can't figure out what I did wrong.

Answer 1

Given the following system:

`\mleft\{\begin{aligned}x+5y=4 \\ 3x+15y=-1\end{aligned}\mright.`

We can solve for x on the first equation and then subsitute it's value on the other equation:

`\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}`

We found an inconsistency on the system, since 12=-1 can never happen. Therefore, there is no solution to the system.