Work Step by Step
Adding the first and the second equation we get: $7x^2-28=0\\x^2=4\\x=\pm2$ $y^2=13-x^2=9$, thus $y=\pm3$ Thus the solutions are: $(\pm2,\pm3)$
You can help us out by revising, improving and updating this answer.Update this answer
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.