Question 469553
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Hi,
{{{(y^3-3y^2)/(y^2+2y-15)=(13y-40)/(y+5)}}}
{{{y^2cross((y-3))/((y+5)cross((y-3)))=(13y-40)/(y+5)}}}
  y^2 = 13y - 40
  y^2 - 3y + 40 = 0
  factoring
 (y-8)(y-5) = 0
 (y-8)= 0   y = 8
 (y-5)= 0   y = 5