please help me i dont remember how to do these anymore

1. add (1/x+3) + (5/x-4)

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

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

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- Jul 19th 2009, 07:40 PMicecreamfrk09help please with adding and subtracting rational expressions
please help me i dont remember how to do these anymore

1. add (1/x+3) + (5/x-4)

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

3. subract (3/x-3) - (4/x+5) - Jul 19th 2009, 07:43 PMAlephZero
- Jul 19th 2009, 07:46 PMicecreamfrk09
the second one please and im sorry im new to this website

- Jul 19th 2009, 07:55 PMAlephZero
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:

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

$\displaystyle = \frac{x-4+5x+15}{(x+3)(x-4)} = \frac{6x+11}{(x+3)(x-4)}$ - Jul 19th 2009, 08:03 PMicecreamfrk09
thank you so much but how to you subtract the rational expressions?

- Jul 19th 2009, 08:11 PMAlephZero
Well, by the same method. You find common denominators, and just subtract instead of adding. Here's how to start the second one:

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

Now simplify, and you should have your answer.

Same method for #3.

Let us know what you get, and we'll check your answers. - Jul 19th 2009, 08:17 PMAlephZero
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 PMicecreamfrk09
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 PMAlephZero
Unfortunately, no. For the second one, I have shown you how to get to

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

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 PMicecreamfrk09
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 PMyeongil
No, you can leave the denominator in factored form. It's considered to in "simplified form" that way.'

01 - Jul 19th 2009, 08:47 PMicecreamfrk09
okayy so both those answers are right?

- Jul 19th 2009, 08:55 PMAlephZero