help please with adding and subtracting rational expressions

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July 19th 2009, 07:40 PM
icecreamfrk09

help 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)
July 19th 2009, 07:43 PM
AlephZero

Quote:

Originally Posted by icecreamfrk09

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

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!
July 19th 2009, 07:46 PM
icecreamfrk09

the second one please and im sorry im new to this website
July 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:

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

$= \frac{x-4+5x+15}{(x+3)(x-4)} = \frac{6x+11}{(x+3)(x-4)}$
July 19th 2009, 08:03 PM
icecreamfrk09

thank you so much but how to you subtract the rational expressions?
July 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:

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

Now simplify, and you should have your answer.

Same method for #3.

Let us know what you get, and we'll check your answers.
July 19th 2009, 08:17 PM
AlephZero

Quote:

Originally Posted by icecreamfrk09

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)

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.
July 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??
July 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.
July 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?
July 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
July 19th 2009, 08:47 PM
icecreamfrk09

okayy so both those answers are right?
July 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:

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