Respuesta :

carlosego
You have the following expressions given in the problem above:

 (y^2/y-3)(y^2-y-6/y^2+y)
 
 By applying the exponents properties, you can simplify it, as it shown below:

 (y^2/y-3)(y^2-y-6/y^2+y)
 (y^4-y^3-6y^2)/(y^3+y^2-3y2-3y)
 (y^4-y^3-6y^2)/(y^3-2y2-3y)

 Then, you have:
 y^2(y^2-y-6)/y(y^2-2y-3)
 (y^2-y-6)/(y^2-2y-3)

 The answer is: (y^2-y-6)/(y^2-2y-3)

 


 
Matheng
The first term = [tex] \frac{y^2}{y-3} [/tex]
The second term = [tex] \frac{y^2-y-6}{y^2+y} = \frac{(y-3)(y+2)}{y(y+1)} [/tex] ⇒ factoring the numerator and denominator

multiplying the both terms:
[tex] \frac{y^2}{y-3} * \frac{y^2-y-6}{y^2+y} = \frac{y^2}{y-3} * \frac{(y-3)(y+2)}{y(y+1)}[/tex] 

= [tex] \frac{y^2}{y} * \frac{(y-3)(y+2)}{(y-3)(y+1)} = \frac{y(y+2)}{y+1} [/tex]






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