LMande

07-06-2006, 10:30 PM

y/y^2-1+5y/y-y^2

I can't figure this one out! Any help would be greatly appreciated!

I can't figure this one out! Any help would be greatly appreciated!

View Full Version : Tough One: figure out y/y^2-1+5y/y-y^2

LMande

07-06-2006, 10:30 PM

y/y^2-1+5y/y-y^2

I can't figure this one out! Any help would be greatly appreciated!

I can't figure this one out! Any help would be greatly appreciated!

Denis

07-06-2006, 10:49 PM

y/y^2-1+5y/y-y^2

I can't figure this one out! Any help would be greatly appreciated!

1: clarify using brackets; like, is it y / (y^2 - 1) ?

2: you can't figure WHAT out? There is no question; WHAT are you suppose to do?

I can't figure this one out! Any help would be greatly appreciated!

1: clarify using brackets; like, is it y / (y^2 - 1) ?

2: you can't figure WHAT out? There is no question; WHAT are you suppose to do?

Mrspi

07-06-2006, 11:01 PM

y/y^2-1+5y/y-y^2

I can't figure this one out! Any help would be greatly appreciated!

I will assume that this is your problem:

y 5y

------ + -------

y^2 - 1 y - y^2

Factor each denominator:

y 5y

----------- + ------------

(y+1)(y-1) y(1 - y)

Note that (1 - y) is the OPPOSITE of y - 1.....multiply numerator and denominator of the second fraction by -1:

y -5y

--------- + ------------

(y+1)(y-1) y(y - 1)

The common denominator for the two fractions is y(y+1)(y-1)

Multiply numerator and denominator of the first fraction by y, and multiply numerator and denominator of the second fraction by (y + 1). This will make both fractions have y(y+1)(y-1) as a denominator:

y * y -5y(y+1)

------------- + ---------------

(y+1)(y-1)y y(y-1)(y+1)

Now, the fractions have the same denominator. Add the numerators, and put the result over the common denominator:

y^2 + (-5y)(y+1)

------------------------

y(y+1)(y-1)

Simplify the numerator. Reduce the fraction, if possible.

I hope this helps you.

I can't figure this one out! Any help would be greatly appreciated!

I will assume that this is your problem:

y 5y

------ + -------

y^2 - 1 y - y^2

Factor each denominator:

y 5y

----------- + ------------

(y+1)(y-1) y(1 - y)

Note that (1 - y) is the OPPOSITE of y - 1.....multiply numerator and denominator of the second fraction by -1:

y -5y

--------- + ------------

(y+1)(y-1) y(y - 1)

The common denominator for the two fractions is y(y+1)(y-1)

Multiply numerator and denominator of the first fraction by y, and multiply numerator and denominator of the second fraction by (y + 1). This will make both fractions have y(y+1)(y-1) as a denominator:

y * y -5y(y+1)

------------- + ---------------

(y+1)(y-1)y y(y-1)(y+1)

Now, the fractions have the same denominator. Add the numerators, and put the result over the common denominator:

y^2 + (-5y)(y+1)

------------------------

y(y+1)(y-1)

Simplify the numerator. Reduce the fraction, if possible.

I hope this helps you.

Denis

07-07-2006, 12:03 AM

Something I find useful:

y -5y

--------- + ------------

(y+1)(y-1) y(y - 1)

At this stage of Mrspi's equations, let a = y+1 and b = y-1; then we have:

y / ab - 5y / by

= y / ab - 5 / b

= (y - 5a) / ab

Now substitute back in:

[y - 5(y + 1)] / [(y + 1)(y - 1)]

= -(4y + 5) / (y^2 - 1)

...and you're finished: I use that method because I'm a one-finger typer,

well really because I'm lazy :shock:

y -5y

--------- + ------------

(y+1)(y-1) y(y - 1)

At this stage of Mrspi's equations, let a = y+1 and b = y-1; then we have:

y / ab - 5y / by

= y / ab - 5 / b

= (y - 5a) / ab

Now substitute back in:

[y - 5(y + 1)] / [(y + 1)(y - 1)]

= -(4y + 5) / (y^2 - 1)

...and you're finished: I use that method because I'm a one-finger typer,

well really because I'm lazy :shock:

LMande

07-07-2006, 12:12 AM

This is great that you all are here to help! I hope someday to be able to help others too! I've just started classes after being out of school for 14 years (yikes). To think, once upon a time all this made sense to me. I mean it still does, just trying to translate it onto a computer makes it more difficult for me. Sorry I wasn't so clear in my question (or lack of).

I really appreciate all your help!

I really appreciate all your help!

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