Adding and Subtracting Rational Expressions

[PreAlgebraDummy15]

Ok I'm adding and subtracting rational expressions:

For this one:
y^2-3y 3y^2+4y y^2+3y
-------- - ----------- - ---------- (y^2= y squared, etc.)
2y+1 2y+1 2y+1




I get the general idea of it...I've tried it numerous times and checked the answer in the back of my book and I'm not getting it right. If someone could help me, that would be so great, thanks!!!! [/tex]

 

[nelly vega]

are you sure you dont need to use parenthesis?

 

[stapel]

Ok I'm adding and subtracting rational expressions:

For this one:
y^2-3y 3y^2+4y y^2+3y
-------- - ----------- - ---------- (y^2= y squared, etc.)
2y+1 2y+1 2y+1

I will guess that the above means the following:

. . . . .(y² - 3y)/(2y + 1) - (3y² + 4y)/(2y + 1) - (y² + 3y)/(2y + 1)

I've tried it numerous times and...I'm not getting it right. If someone could help me, that would be so great, thanks!

We'll be glad to help you find your error, but you will need to show your work in order for us to do that.

Thank you.

Eliz.

 

[soroban]

Hello, PreAlgebraDummy15!

\(\displaystyle \L\frac{y^2\,-\,3y}{2y\,+\,1}\,-\,\frac{3y^2\,+\,4y}{2y\,+\,1}\,-\,\frac{y^2\,+\,3y}{2y\,+\,1}\)

I get the general idea of it.

Really? If you "get" the general idea, this one is easy!

I've tried it numerous times and checked the answer in the back of my book and I'm not getting it right.
If someone could help me, that would be so great.

It's hard to see your work from here
\(\displaystyle \;\;\)but it looks like you played the \(\displaystyle Q\club\) instead of the \(\displaystyle 5\spade\)
\(\displaystyle \;\;\)and you hit \(\displaystyle C#\) instead of \(\displaystyle B\flat.\)


First, we need a common denominator . . .

\(\displaystyle \L\;\;\frac{y^2\,-\,3y}{\underbrace{2y\,+\,1}}\,-\,\frac{3y^2\,+\,4y}{\underbrace{2y\,+\,1}}\,-\,\frac{y^2\,+\,3y}{\underbrace{2y\,+\,1}}\)

It already has a common denominator!


Next, combine the numerators ... write this over the common denominator.

\(\displaystyle \L\;\;\frac{(y^2\,-\,3y)\,-\,(3y^2\,+\,4y)\,-\,(y^2\,+\,3y)}{2y\,+\,1}\)


The rest is Algebra I . . .

\(\displaystyle \L\;\;\frac{y^2\,-\,3y \,-\,3y^2\,-\,4y\,-\,y^2\,-\,3y}{2y\,+\,1} \;=\;\frac{-3y^2\,-\,10y}{2y\,+\,1} \;=\;\frac{-y(3y\,+\,10)}{{2y\,+\,1}\)