Question 465729
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Hi,
Multiplying thru by term such as ALL denominators = 1
{{{6/(y+3)+ 2/y = (5y-3)/(y^2-9)}}}  |multiplying thru by y(y^2-9)
   6y(y-3) + 2(y^2-9)= (5y-3)y
  6y^2 - 18y + 2y^2 - 18 = 5y^2 - 3y
     3y^2 - 15y - 18 = 0
      y^2 - 5y - 6 = 0 
 factoring
   (y-6)(y+1) = 0    y = 6 and y = -1 are the solutions

 2/(x+5) + 1/(x-5) = 16/(x^2-25)  |multiplying thru by x^2-25 = (x-5)(x+5)
  2(x-5) + (x+5) = 16
      3x = 21
       x = 7 is the solution