How do you graph 3x^2+7y^2=71

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It's an ellipse.  You need to put this in the form (x/a)^2 + (y/b)^2 = 1 and find, a, b and c = SQR (a^2 - b^2).If a is > than b then it is the semimajor axis length and b is the semiminor length.  c is the focal length along the semimajor axis whether it is y or x.  Divide all terms by 71Put the coefficients of the left side in the denominator (x^2) / (71 / 3) + (y^2) / (71 / 7) = 1a^2 = 71 / 3 and  b^2 = 71 / 7a^2 = b^2 + c^2 so you can find c^2 = 71/3 -71/7c^2 = 124 / 21 a is the semimajor axis length.  The ellipse is horizontal and has vertices of +/-  SQR(71/3) along the x axis.The semiminor axis length is b = +/- SQR(71/7)The focal points are along the x axis at distances of c = +/- SQR(124/21)