# rational expressions #1

• Jul 23rd 2013, 12:46 PM
Mathnood768
rational expressions #1
http://puu.sh/3K0Og.jpg

i cannot finish question 7 and i got the wrong answer for 6,

6. answer is 3/y y cannot equal -1, 0
7. answer is 2(x^2 - 3x + 5)/(x-5)(x-5) x cannot equal -3, 0, 1, 2

thank u
• Jul 23rd 2013, 01:03 PM
Re: rational expressions #1
For 6, I have no idea how you got the denominator y + 1 + y^2. The common denominator would be y^2 + y. Notice that (y+1) * y = y^2 + y. You should be able to simplify with that info in mind. Just make sure that before you start simplifying, you take the separate denominators and set them equal to zero so you know what values of y are impossible.

For 7, the factorization of 2x^2 + 9x -5 is (2x - 1)(x + 5). You should be able to finish the simplification with that info.
• Jul 23rd 2013, 01:20 PM
Mathnood768
Re: rational expressions #1
http://puu.sh/3K2B6.jpg

i tried to do it again but still go them both wrong, what happened?
• Jul 23rd 2013, 01:38 PM
Re: rational expressions #1
Attachment 28891

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That is the process for 6. Do you understand it? You should be able to solve 7 similarly. You need to work on your factorization, 3x+15 is 3(x+5). Which is just the reverse of distribution
• Jul 23rd 2013, 01:53 PM
Mathnood768
Re: rational expressions #1
i tried to do 7 but got a different wrong answer again...
• Jul 23rd 2013, 03:02 PM
HallsofIvy
Re: rational expressions #1
Then show us what you did!
• Jul 23rd 2013, 03:17 PM
Mathnood768
Re: rational expressions #1
• Jul 23rd 2013, 03:23 PM
Plato
Re: rational expressions #1
Quote:

Originally Posted by Mathnood768

Your operations are all wrong. It should be:
$\displaystyle \frac{3(x+5)+(x-5)(2x+1)}{x^2-25}$

Now finish!
• Jul 23rd 2013, 03:29 PM
Mathnood768
Re: rational expressions #1
http://puu.sh/3K8wW.jpg

i tried doing it like you said, but i dont think its correct; i wrote down what the answer should be though i just get get it
• Jul 23rd 2013, 05:39 PM
HallsofIvy
Re: rational expressions #1
Apparently you just don't understand how to "cancel". In the above problem, you tried to cancel an "x+ 5" in the denominator with an "x+ 5" which is in the first term but not the second, and cancel an "x- 5" in the denominator with an "x- 5" which is in the second term but not the first. You cannot do that! For example, if you have $\displaystyle \frac{2+ 3}{2(3)}= \frac{5}{6}$, you cannot cancel the "2" in the denominator with the 2 in the first term and the "3" in the denominator with the 3 in the second term. 5/6 is NOT equal to 1+ 1= 2!

You have to cancel factors of the numerator and denominator. If we have 6/10= 2(3)/2(5) then we can cancel the factors of "2" leaving 3/5. If we have (2+3)/2(5) we cannot cancel the "2"s. We can add the 2+ 3= 5 getting 5/2(5) and then cancel the factors of "5" leaving 1/2.