Posted by **Steve** on Wednesday, May 20, 2009 at 1:11pm.

Hi, I have a test tomorrow and our teacher gave us a huge review worksheet. I am still not sure how to do some of these so I would appreciate if you could do the following problems out in all steps. I tried to pick some from each section.

- Thanks :)

1. 7x+2/6x + 2x-11/6x

2. x^2/x+5 - 25/x+5

3. 5/6x-2y - 3/9x-3y

4. x/2x-4 - 2/x^2-2x

5. x/x+3 - 5/x+2

6. x-4/y^2-x-2 + x-7/x+1

- Algebra 1 lvl 1 -
**Steve**, Wednesday, May 20, 2009 at 1:13pm
Sorry for number six, the bottom of the first fraction is an x^2 NOT a y^2 very sorry.

- Algebra 1 lvl 1 -
**Reiny**, Wednesday, May 20, 2009 at 2:25pm
In each case, you should have brackets around what I assume is the numerator and the denominator.

e.g. in #1 (7x+2)/(6x) + (2x-11)/(6x)

since you have a common denominator already, just add the tops

= (9x-9)/(6x)

= (3x-3)/(2x) after dividing top and bottom by 3

#2 becomes (x^2-25)/(x+5) since we again have a common denominator

= (x+5)(x-5)/(x+5) , the top was a difference of squares

= x-5

in #3 factor the two denominators to

2(3x-y) and 3(3x-y)

in the second fraction you can cancel the 3's

and then your common denominator is 2(3x-y)

etc

in #4

x/(2x-4) - 2/(x^2-2x)

= x/[2(x-2)] - 2/[x(x-2)]

common denominator will be 2x(x-2)

so we have

x^2/[2x(x-2)] - 4/[2x(x-2)]

= (x^2 - 4)/[2x(x-2)]

= (x+2)(x-2)/[2x(x-2)]

= (x+2)/(2x)

do #5 and #6 the same way

in #6 the first denominator factors to

(x-2)(x+1), which then becomes your common denominator.

Let me know what you get.

- Algebra 1 -
**robert**, Friday, July 17, 2009 at 9:52am
3(x-2)>4(2x+11)

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