adding rational expressions

ash1017

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Apr 7, 2008
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My second question is also with adding rational expressions. the question is (x+1) divided by (x+3)(x-3) + 4(x-3) divided by (x-3)(x+3) + (x-1)(x-3) divided by (3-x)(x+3). I tried to take the last part times negative one. which still makes half of the denominator unequal. SO then i tried to make the denominator (x-3)(x-3)(x+3). i did this and i got 5x-11(x-1) divided by (x+3)(x-3). the back of the book gives a different answer and i am not sure where i am going wrong.
 

stapel

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ash1017 said:
My second question is also with adding rational expressions. the question is (x+1) divided by (x+3)(x-3) + 4(x-3) divided by (x-3)(x+3) + (x-1)(x-3) divided by (3-x)(x+3).
I will guess you mean the following:

. . . . .(x + 1)/[(x + 3)(x - 3)] + [4(x - 3)]/[(x - 3)(x + 3)] + [(x - 3)(x - 3)]/[(3 - x)(x + 3)]

I will guess that the instructions were somethig like "simplify the expression". I'm afraid I don't follow your description of what you tried...? Why are you adding another copy of x - 3 to the denominator? Where did this come from? How did you arrive at your answers? (You mention what your result was but, since you did not show your steps, there is no way to know how you got to that result, or what errors might have been made along the way.)

Please reply with clarification, starting from flipping the one subtraction in the third denominator:

. . . . .(x + 1)/[(x + 3)(x - 3)] + [4(x - 3)]/[(x - 3)(x + 3)] + [-1(x - 3)(x - 3)]/[(x - 3)(x + 3)]

...multiplying out the numerators:

. . . . .(x + 1)/[(x + 3)(x - 3)] + (4x - 12)/[(x + 3)(x - 3)] + (-x[sup:32b7s5mq]2[/sup:32b7s5mq] + 6x - 9)/[(x + 3)(x - 3)]

...and adding the fractions (that is, adding their numerators). You then combined "like" terms and simplified the numerator, and factored to see if anything would cancel with something in the denominator, and... then what?

Please be complete. Thank you! :D

Eliz.
 

Loren

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Aug 28, 2007
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> (x+1) divided by (x+3)(x-3) + 4(x-3) divided by (x-3)(x+3) + (x-1)(x-3) divided by (3-x)(x+3)
\(\displaystyle \frac{x+1}{(x+3)(x-3)}+\frac{4(x-3)}{(x-3)(x+3)}+\frac{(x-1)(x-3)}{(3-x)(x+3)}\)

I would change the last fraction to become...
\(\displaystyle -\frac{(x-1)(x-3)}{(x-3)(x+3)}\)

Now all the denominators are the same. So, combine the numerators to get the numerator of the sum and the common denominator is the denominator of the sum. Finally, simplify.
 
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