• Jul 19th 2009, 07:40 PM
icecreamfrk09

2. subtract (3/x) - (7/x-2)

3. subract (3/x-3) - (4/x+5)
• Jul 19th 2009, 07:43 PM
AlephZero
Quote:

Originally Posted by icecreamfrk09

2. subtract 3/x - 7/x-2

3. subract 3/x-3 - 4/x+5

Do you want $\frac{1}{x} + 3 + \frac{5}{x} -4,$ or do you want

$\frac{1}{x+3} + \frac{5}{x-4}$?

If you want the second one, you need to use parentheses!
• Jul 19th 2009, 07:46 PM
icecreamfrk09
the second one please and im sorry im new to this website
• Jul 19th 2009, 07:55 PM
AlephZero
Quote:

Originally Posted by icecreamfrk09
the second one please and im sorry im new to this website

Fair enough!

What you need to do for these is to find a common denominator, which is most easily done by multiplying the two denominators. Here goes:

$\frac{1}{x+3}\cdot \frac{x-4}{x-4} + \frac{5}{x-4} \cdot \frac{x+3}{x+3} = \frac{x-4}{(x+3)(x-4)} + \frac{5(x+3)}{(x+3)(x-4)}$

$= \frac{x-4+5x+15}{(x+3)(x-4)} = \frac{6x+11}{(x+3)(x-4)}$
• Jul 19th 2009, 08:03 PM
icecreamfrk09
thank you so much but how to you subtract the rational expressions?
• Jul 19th 2009, 08:11 PM
AlephZero
Quote:

Originally Posted by icecreamfrk09
thank you so much but how to you subtract the rational expressions?

Well, by the same method. You find common denominators, and just subtract instead of adding. Here's how to start the second one:

$\frac{3}{x}\cdot\frac{x-2}{x-2} - \frac{7}{x-2}\cdot\frac{x}{x} = \frac{3(x-2)-7x}{x(x-2)}.$

Same method for #3.

• Jul 19th 2009, 08:17 PM
AlephZero
Quote:

Originally Posted by icecreamfrk09

2. subtract (3/x) - (7/x-2)

3. subract (3/x-3) - (4/x+5)

By the way, and not to be a stick-in-the-mud, but you're still not expressing these in a mathematically correct way, which is important because it's hard for us to tell what you mean. The correct way to use the parentheses would be

1. 1/(x+3) + 5/(x-4)
2. 3/x - 7/(x-2)
3. 3/(x-3) - 4/(x+5)

In other words, you need to group all the numerators and denominators together, so we know what is being divided into what. I hope that makes sense.
• Jul 19th 2009, 08:19 PM
icecreamfrk09
okayy not sure if i did these right but i got (3-7x)/x for the second one and -1 for the third one is that right??
• Jul 19th 2009, 08:26 PM
AlephZero
Quote:

Originally Posted by icecreamfrk09
okayy not sure if i did these right but i got (3-7x/x) for the second one and -1 for the third one is that right??

Unfortunately, no. For the second one, I have shown you how to get to

$\frac{3(x-2)-7x}{x(x-2)}.$

What you need to do now is carry out the multiplication in the numerator and combine like terms. You won't have to do anything to the denominator.

Give it a shot.
• Jul 19th 2009, 08:41 PM
icecreamfrk09
okayy i got (-4x-6)/x(x-2) for the second and (-x+27)/(x+5)(x-3) for the third but do you multiply the denominator for third one?
• Jul 19th 2009, 08:43 PM
yeongil
No, you can leave the denominator in factored form. It's considered to in "simplified form" that way.'

01
• Jul 19th 2009, 08:47 PM
icecreamfrk09
okayy so both those answers are right?
• Jul 19th 2009, 08:55 PM
AlephZero
Quote:

Originally Posted by icecreamfrk09
okayy so both those answers are right?

Yes, very good job!

However the second one can be simplified further, depending upon what form your instructor wants:

$\frac{-4x-6}{x(x-2)} = \frac{-2(2x+3)}{x(x-2)}.$