Simplifying by combining terms

[headbang]

Ok.. What about this one?

\(\displaystyle \frac{1}{y+1}+\frac{2}{x+2}+\frac{\frac{x+2}{y+1}-2}{x+2}\)

this is where I am at:

\left(\frac{x+2}{y+1}-2\right)

 

[topsquark]

Ok.. What about this one?

\(\displaystyle \frac{1}{y+1}+\frac{2}{x+2}+\frac{\frac{x+2}{y+1}-2}{x+2}\)

this is where I am at:

Let's clear that mess in the 3rd term:
\(\displaystyle \frac{1}{y+1}+\frac{2}{x+2}+\frac{\frac{x+2}{y+1}-2}{x+2} \cdot \frac{(y + 1)}{(y + 1)}\)

\(\displaystyle \frac{1}{y+1}+\frac{2}{x+2}+\frac{(x+2) - 2(y + 1)}{(x+2)(y + 1)}\)

\(\displaystyle \frac{1}{y+1}+\frac{2}{x+2}+\frac{x - 2y}{(x+2)(y + 1)}\)
Now just add the fractions. Give it a try and tell us how it goes from here.

-Dan

 

[headbang]

Can the ansver be:

\(\displaystyle \frac{2+x}{4y}\) And i stil don't understand how you solved the messy 3rd term..?

what i did from where you left me..

\(\displaystyle \frac{x+2+2y+2+x-y}{(x+2)(y+1)}\)

\(\displaystyle \frac{4+2x+y}{(8x+2)(y+1)}\)

\(\displaystyle \frac{2*2+x*x+y}{x*y+2y+y+y+2}\)

 

[MarkFL]

Ok.. What about this one?

\(\displaystyle \frac{1}{y+1}+\frac{2}{x+2}+\frac{\frac{x+2}{y+1}-2}{x+2}\)

this is where I am at:

\left(\frac{x+2}{y+1}-2\right)

I moved the posts associated with your new problem into a separate thread. We ask that new questions not be tagged onto an existing thread...this way the original thread does not potentially become convoluted and hard to follow. :D

 

[topsquark]

Can the ansver be:

\(\displaystyle \frac{2+x}{4y}\) And i stil don't understand how you solved the messy 3rd term..?

what i did from where you left me..

\(\displaystyle \frac{x+2+2y+2+x-y}{(x+2)(y+1)}\)

First things first: The equation above is almost correct. You missed a 2 in the last term in the numerator. It should be
\(\displaystyle \frac{x+2+2y+2+x-2y}{(x+2)(y+1)}\)

Okay, the third term in your problem is what is known as a "compound fraction." Compound fractions have fractions in the numerator, denominator, or both. Let's take this one step by step.
\(\displaystyle \frac{\frac{x + 2}{y + 1} - 2}{x + 2}\)

The goal here is to remove the fraction in the numerator. Clearly if we multiply the numerator by y + 1 the fraction goes away. And what we do to the numerator we need to do to the denominator. So:
\(\displaystyle \frac{\frac{x + 2}{y + 1} - 2}{x + 2} \cdot \frac{y + 1}{y + 1}\)

Now simplify:
\(\displaystyle \frac{\frac{x + 2}{y + 1} - 2}{x + 2} \cdot \frac{y + 1}{y + 1} = \frac{\left ( \frac{x + 2}{y + 1} \right ) (y + 1) - 2(y + 1)}{(x + 2)(y + 1)}\)

\(\displaystyle \frac{(x + 2) - 2(y +1)} {(x + 2)(y + 1)}\)

And now go from there.

-Dan

 

[headbang]

\(\displaystyle \frac{x+2+2y+2+x-2(y+1)}{(x+2)(y+1)}\)

\(\displaystyle 2y+2+x-2=x+2y\)

Ok?