Solution
Solve the system by any method.
[tex]\begin{gathered} 3x-10y=-58 \\ 9x+10y=-14 \end{gathered}[/tex]Using substituttion method:
[tex]\begin{gathered} 3x-10y=-58\ldots\ldots.....(1) \\ 9x+10y=-14\ldots\ldots\ldots\text{.}(2) \\ \text{from equ(1)} \\ 3x-10y=-58 \\ 3x=-58+10y \\ \frac{3x}{3}=\frac{-58+10y}{4} \\ \mathrm{Substitute\: }x=\frac{-58+10y}{3} \end{gathered}[/tex][tex]\begin{gathered} \text{substitute x} \\ \begin{bmatrix}9\cdot\frac{-58+10y}{3}+10y=-14\end{bmatrix} \\ \begin{bmatrix}40y-174=-14\end{bmatrix} \\ 40y=-14+174 \\ 40y=160 \\ y=\frac{160}{40} \\ y=4 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-58+10\cdot\:4}{3} \\ x=x=\frac{-58+40}{3} \\ x=-\frac{18}{3} \\ x=-6 \end{gathered}[/tex]Therefore the solution to the system of equation is
[tex]x,y=-6,4[/tex]