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

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- Jul 23rd 2013, 12:46 PMMathnood768rational 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 PMShadowKnight8702Re: 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 PMMathnood768Re: 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 PMShadowKnight8702Re: rational expressions #1
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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 PMMathnood768Re: rational expressions #1
i tried to do 7 but got a different wrong answer again...

- Jul 23rd 2013, 03:02 PMHallsofIvyRe: rational expressions #1
Then

**show**us what you did! - Jul 23rd 2013, 03:17 PMMathnood768Re: rational expressions #1
- Jul 23rd 2013, 03:23 PMPlatoRe: rational expressions #1
- Jul 23rd 2013, 03:29 PMMathnood768Re: 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 PMHallsofIvyRe: 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.