How do I manipulate algebraic expressions?

[Schniz2]

Homework Statement

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

Ive scribbled all the things i can thinkof on paper and still can't seem to get from one to the other, its driving me nuts!

Any help would be much appreciated :D.

Cheers.

 

[cristo]

They appear to add 8 and subtract 8 from the numerator. That is,

$$\frac{2y^2}{4+y^2}=\frac{2y^2+8-8}{4+y^2}=\frac{2(4+y^2)-8}{4+y^2}=2-\frac{8}{4+y^2}$$

 

[Schniz2]

They appear to add 8 and subtract 8 from the numerator. That is,

$$\frac{2y^2}{4+y^2}=\frac{2y^2+8-8}{4+y^2}=\frac{2(4+y^2)-8}{4+y^2}=2-\frac{8}{4+y^2}$$

Ahhh, i feel like such a fool for not seeing that.

Thanks ;)

 

[HallsofIvy]

Another way to see that is simply "long division". $y^2+ 0y+ 4$ divides into $2y^2+ 0y+ 0$ 2 times with a remainder of [itex]2y^2- 2(y^2+ 4)= 2y^2-2y^2- 8= -8[/tex] so<br /> $$\frac{2y^2}{y^2+ 4}= 2- \frac{8}{y^2+ 4}$$[/itex]

 

[Schniz2]

Another way to see that is simply "long division". $y^2+ 0y+ 4$ divides into $2y^2+ 0y+ 0$ 2 times with a remainder of [itex]2y^2- 2(y^2+ 4)= 2y^2-2y^2- 8= -8[/tex] so<br /> $$\frac{2y^2}{y^2+ 4}= 2- \frac{8}{y^2+ 4}$$[/itex]

$<br /> Thanks ;). Very clear to me now.$