The given set of equations are:
[tex]\begin{gathered} 3x-8y=19\ldots\ldots\text{.}(1) \\ 3x+5y=-46\ldots\ldots(2) \end{gathered}[/tex]Subtract equation 1 from equation 2, we get,
[tex]\begin{gathered} 3x-3x+5y+8y=-46-19 \\ 13y=-65 \\ y=-\frac{65}{13}=-5 \end{gathered}[/tex]Now, substitute -5 for y in equation 1, we get,
[tex]\begin{gathered} 3x-(8\times-5)=19 \\ 3x+40=19 \\ 3x=19-40=-21 \\ x=-\frac{21}{3}=-7 \end{gathered}[/tex]Thus, x = -7 and y = -5
