Question 646728
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi, Previously Posted
{{{3/(3x)-1/(x+1)= 1}}} |Multiplying thru by 3x(x+1) so as all denominators = 1
   3(x+1) - 3x = 3x(x+1)
    3x^2+3x - 3 = 0
     x^2 + x - 1 = 0    || y = ax^2 + bx + c = 0
Perhaps easier if one started with:
{{{1/x - 1/(x+1) = 1}}} |Multiplying thru by x(x+1) so as all denominators = 1
{{{cross(x)(x+1)/cross(x) - 1*x*(cross(x+1))/(cross(x+1)) = 1*x(x+1)}}}
  1 = x^2+x
  x^2 + x-1 = 0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-1 +- sqrt( 5 ))/(2) }}}