The teacher will be very surprised to see your correct solution 😉
The solution set is obviously symmetric with respect to the y-axis. Therefore we may assume x≥0. In the domain {(x,y)∈R2∣x≥0} the equation is equivalent with\nx2+y2−1=x2/3y,\nwhich can easily be solved for y:\ny=21(x2/3±x4/3+4(1−x2)).\nNow plot this, taking both branches of the square root into account. You might have to numerically solve the equation x4/3+4(1−x2)=0 in order to get the exact x-interval.
Move free summands (without y3) from left part to right part, we given: −x2y3+3x2+3y2=3 Move the summands with the other variables from left part to right part, we given: −x2y3+3y2=−3x2+3 Divide both parts of the equation by (3*y2 - x2*y3)/y3
y3 = 3 - 3*x2 / ((3*y2 - x2*y3)/y3)
We get the answer: y3 = 3*(-1 + x2 + y2)/x2
The solution of the parametric equation
Given the equation with a parameter: −x2y3+3x2+3y2−3=0 The coefficient at y3 is equal to −x2 then possible cases for x2 : x2<0 x2=0 Consider all cases in more detail: With x2<0 the equation 3y2+y3−6=0 its solution y3=−3y2+6 With x2=0 the equation 3y2−3=0 its solution