Find the number of ordered pairs (x, y) of real numbers such that
If you divide the first expression on the LHS(left hand side), you get (x^2+2x+4)(3y^2-6y+7) = 12
Putting the two expressions on the LHS in square form(by completing the square) you get that((x+1)^2+3)*(3(y-1)^2+4) = 12
Using this we know that the minimum for the two expressions are 3 and 4 respectively so that means that there is only 1 ordered pair satisfying this equation. Namely, (-1,1).