Verified answer

[tex]\displaystyle\bf\\\left \{ {{2x^{2} -3xy+5y-5=0} \atop {(x-2)(y-1)=0}} \right. \\\\\\(x-2)(y-1)=0\\\\\\\left[\begin{array}{ccc}x-2=0\\y-1=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=2\\y=1\end{array}\right \\\\\\1)\\\\\left \{ {{x=2} \atop {2x^{2}-3xy+5y-5=0 }} \right. \\\\\\\left \{ {{x=2} \atop {2\cdot 2^{2} -3y\cdot 2+5y-5=0}} \right.\\\\\\\left \{ {{x=2} \atop {8-6y+5y-5=0} \right. \\\\\\\left \{ {{x=2} \atop {-y=-3}} \right. \\\\\\\left \{ {{x=2} \atop {y=3}} \right.[/tex]

[tex]\displaystyle\bf\\2)\\\\\left \{ {{y=1} \atop {2x^{2} -3xy+5y-5=0}} \right.\\\\\\\left \{ {{y=1} \atop {2x^{2} -3x\cdot 1+5\cdot 1-5=0}} \right. \\\\\\\left \{ {{y=1} \atop {2x^{2} -3x=0}} \right. \\\\\\\left \{ {{y=1} \atop {x(2x-3)=0}} \right. \\\\\\\left \{ {{y=1} \atop {\left[\begin{array}{ccc}x=0\\2x-3=0\end{array}\right}} \right. \\\\\\\left \{ {{y=1} \atop {\left[\begin{array}{ccc}x_{1} =0\\x_{2} =1,5\end{array}\right}} \right.[/tex]

[tex]\displaystyle\bf\\Otvet: \ \Big(2 \ ; \ 3\Big) \ \ , \ \ \Big(0 \ ; \ 1\Big) \ \ , \ \ \Big(1,5 \ ; \ 1\Big)[/tex]