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Dmitry_Shevchenko [17]
2 years ago
7

Solve for x and y in the system of equation3x²+4y=17 2x²+5y=2I'll mark brainliest​

Mathematics
1 answer:
Novay_Z [31]2 years ago
7 0

Answer:

x=\pm \sqrt{11} \\y=-4.

Step-by-step explanation:

\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}.

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