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Algebraic Manipulation

  • Thread starter Schniz2
  • Start date
  • #1
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Homework Statement



As part of a calculus question, the solutions manual takes [tex]\frac{2y^{2}}{4+y^{2}}[/tex]
And somehow turns it into [tex]\left(2-\frac{8}{4+y^{2}}\right)[/tex]

Ive scribbled all the things i can thinkof on paper and still cant seem to get from one to the other, its driving me nuts!

Any help would be much appreciated :D.

Cheers.
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
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72
They appear to add 8 and subtract 8 from the numerator. That is,

[tex]\frac{2y^2}{4+y^2}=\frac{2y^2+8-8}{4+y^2}=\frac{2(4+y^2)-8}{4+y^2}=2-\frac{8}{4+y^2}[/tex]
 
  • #3
19
0
They appear to add 8 and subtract 8 from the numerator. That is,

[tex]\frac{2y^2}{4+y^2}=\frac{2y^2+8-8}{4+y^2}=\frac{2(4+y^2)-8}{4+y^2}=2-\frac{8}{4+y^2}[/tex]
Ahhh, i feel like such a fool for not seeing that.

Thanks ;)
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,792
919
Another way to see that is simply "long division". [itex]y^2+ 0y+ 4[/itex] divides into [itex]2y^2+ 0y+ 0[/itex] 2 times with a remainder of [itex]2y^2- 2(y^2+ 4)= 2y^2-2y^2- 8= -8[/tex] so
[tex]\frac{2y^2}{y^2+ 4}= 2- \frac{8}{y^2+ 4}[/tex]
 
  • #5
19
0
Another way to see that is simply "long division". [itex]y^2+ 0y+ 4[/itex] divides into [itex]2y^2+ 0y+ 0[/itex] 2 times with a remainder of [itex]2y^2- 2(y^2+ 4)= 2y^2-2y^2- 8= -8[/tex] so
[tex]\frac{2y^2}{y^2+ 4}= 2- \frac{8}{y^2+ 4}[/tex]
Thanks ;). Very clear to me now.
 

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