# Algebraic fractions with 3 or more fractions

• Nov 18th 2008, 12:54 AM
silverbird
Algebraic fractions with 3 or more fractions
Can someone tell me all the different methods you know of to simplifye this
$2/(y+2)- 1/(y+3) + 5/(y-1)$

http://img530.imageshack.us/img530/8...uation1eg7.jpg

and does anyone know if this type of qustion have a specific name
• Nov 18th 2008, 02:53 AM
HallsofIvy
Quote:

Originally Posted by silverbird
Can someone tell me all the different methods you know of to simplifye this
$2/(y+2)- 1/(y+3) + 5/(y-1)$

http://img530.imageshack.us/img530/8...uation1eg7.jpg

and does anyone know if this type of qustion have a specific name

No special name, just "combining fractions".

"All the different ways"? That could take a while!
Here's one:
Using the "associative law" combine the first two fractions: to subract
$\frac{2}{y+2}- \frac{1}{y+3}$
get a common denominator. Multiply both numerator and denominator of the first fraction by y+3 and of the second fraction by y+2:
$\frac{2}{y+2}\frac{y+3}{y+3}- \frac{1}{y+3}\frac{y+2}{y+ 2}= \frac{2y+ 2}{(y+2)(y+3)}- \frac{y+2}{(y+2)(y+3)}$
$= \frac{2y+ 2- y- 2}{(y+2)(y+3)}= \frac{y}{(y+2)(y+3)}$

Now the problem has become $\frac{y}{(y+2)(y+3)}+ \frac{5}{y-1}$.

Again, get common denominators: multiply numerator and denominator of the first fraction by y-1 and of the second fraction by $(y+2)(y+3)= y^2+ 5y+ 6$
$\frac{y}{(y+2)(y+3)}\frac{y-1}{y-1}+ \frac{5}{y-1}\frac{y^2+ 5y+ 6}{(y+2)(y+3)}= \frac{y^2- y+ 5y^2+ 25y+ 30}{(y+2)(y+3)(y-1)}$
$= \frac{6y^2+ 24y+ 30}{(y+2)(y+3)(y-1)}$
You could, of course, multiply that denominator but I recommend leaving it factored.

By the associative law we could also combine the last two fractions, the add the first. Be careful to include the "-" sign with the second fraction only. This is NOT
$\frac{2}{y+2}- \left[\frac{1}{y+2}+ \frac{5}{y-1}$

Or we could use the "commutative law swap the orders and combine in any order we please. Because of the associative law and commutative law, the order in which we add and subtract doesn't matter and most people would combine them "all at once", getting the common denominator by multiplying the numerator and denominator of the first fraction by (y+3)(y- 1), the second fraction by (y+2)(y-1), and the third fraction by (y+2)(y+3):
$\frac{2}{y+2}\frac{(y+3)(y-1)}{(y+3)(y-1)}- \frac{1}{y+3}\frac{(y+2)(y-1)}{(y+2)(y-1)+ \frac{5}{y-1}\frac{(y+2)(y+3)}{(y+2)(y+3)}$
$= \frac{2(y+3)(y-1)}{(y+2)(y+3)(y-1)}- \frac{(y+2)(y-1)}{(y+2)(y+3)(y-1)}+ \frac{5(y+2)(y+3)}{(y+2)(y+3)(y-1)}$
$= \frac{2y^2+ 4y- 6- y^2- y+ 2+ 5y^2+ 25y+ 30}{(y+2)(y+3)(y-1)}$
$= \frac{6y^2+ 24y+ 30}{(y+2)(y+3)(y-1)}$
• Nov 19th 2008, 02:52 AM
silverbird
genius.
Thx heaps you dont know how much that helps (Clapping)