Algebraic Expressions

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abigal09

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Sep 22, 2009
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Directions- Simplify each algebraic expression:

14/a[sup:13rimu8l]2[/sup:13rimu8l] - 11/b[sup:13rimu8l]2[/sup:13rimu8l]

and

3/x-4/y+5/xy

For the second problem I was starting to solve it like this:

3/xy-4/xy+5/xy

Do I multiply the numerator by xy as well?

I'm so confused, please help!
 
abigal09 said:
This is the problem:

3/x - 4/y + 5/xy
Click to expand...

Do you realise that 5/xy means 5/x times y ?
If the division ix by xy, then you need to show this way: 5/(xy)

So which one is it?
 
abigal09 said:
3/x - 4/y + 5/(xy)
Click to expand...
YAHOO!! We got it :wink:

So the LCD is xy, right?

[3xy/x - 4xy/y + 5xy/(xy)] / (xy) = (3y - 4x + 5) / (xy) ; got that?
 
I don't understand how you got to that point.
I have never seen a problem solved in that way, so I
am having trouble following you.

Could you break down your steps a little more?
 
Look, it's not easy to "teach" by typing back and forth...classroom is required...

Take the first 2 terms: 3/x - 4/y
Can you simplify that?
 
No. I am not sure how to work with the variables.

This is what I get though:

3/x - 4/y
3y/(xy) - 4x/(xy)
 
abigal09 said:
No. I am not sure how to work with the variables.

This is what I get though:

3/x - 4/y
3y/(xy) - 4x/(xy)
Click to expand...
That's pretty good;
So now you have (3y - 4x) / (xy)
OK?
 
Don't the fractions have to have the common denominator in order to subtract though?
 
abigal09 said:
Don't the fractions have to have the common denominator in order to subtract though?
Click to expand...
Yes...I omitted the division..edited showing the division...
 
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