Answer:
[tex]x=\pm \sqrt{11} \\y=-4.[/tex]
Step-by-step explanation:
[tex]\left \{ {{3x^2+4y=17} \atop {2x^2+5y=2}} \right. \\\\Then\\\\\left \{ {{2 \times (3x^2+4y)=2 \times 17} \atop {3\times(2x^2+5y)=3\times2}} \right. \\\\\left \{ {{6x^2+8y=34 \quad (1)} \atop {6x^2+15y=6\quad (2)}} \right.\\\\(1)-(2) \Rightarrow -7y=28 \Rightarrow y =-4.\\Since \: \: 2x^2+5y=2 \Rightarrow 2x^2=2-5y\\\Rightarrow 2x^2=2+20=22\\\Rightarrow x^2=22:2=11\\\Rightarrow x=\pm \sqrt{11}.[/tex]
