Posted by Steve on Wednesday, May 20, 2009 at 1:11pm.
Sorry for number six, the bottom of the first fraction is an x^2 NOT a y^2 very sorry.
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.
3(x-2)>4(2x+11)
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