AmanGrg
AmanGrg AmanGrg
  • 17-03-2021
  • Mathematics
contestada

plz quick very fast
simplify​

plz quick very fastsimplify class=

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math3683 math3683
  • 17-03-2021
Factor both denominators resulting in
[1/(y+2)(y-2)] + [3/(y+2)(y+3)]
To get a common denominator
Multiply [1/(y+2)(y-2)] x [(y+3)/(y+3)] = (y+3)/[(y+2)(y-2)(y+3)]
Multiply [3/(y+2)(y+3)] x (y-2)/(y-2) = [3(y-2)]/[(y+2)(y-2)(y+3)]
Add (y+3)/[(y+2)(y-2)(y+3)] + [3(y-2)]/[(y+2)(y-2)(y+3)] = [3(y-2) + (y+3)]/[(y+2)(y-2)(y+3)]
Distribute the numerator
[3y - 6 + y+3]/[(y+2)(y-2)(y+3)]
Combine like terms
[4y - 3]/[(y+2)(y-2)(y+3)]
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