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\(x-2y=5 \\ 3x-5y=8 \\ \\ \hbox{solve the first equation for x:} \\ x-2y=5 \ \ \ |+2y \\ x=2y+5 \\ \\ \hbox{substitute 2y+5 for x in the second equation:} \\ 3(2y+5)-5y=8 \\ 6y+15-5y=8 \\ y+15=8 \ \ \ |-15 \\ y=-7 \\ \\ x=2y+5=2 \times (-7)+5=-14+5=-9 \\ \\ \boxed{(x,y)=(-9,-7)}\)

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