1.)rewrite in order to solve

3/2x-6 + 5/x-3 = 1/6

2.)rewrite in order to complete the square. (i hate these)

y^2 +3y-1=0

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- Aug 20th 2008, 04:31 PMPeyton SawyerMath Help plz!
1.)rewrite in order to solve

3/2x-6 + 5/x-3 = 1/6

2.)rewrite in order to complete the square. (i hate these)

y^2 +3y-1=0 - Aug 20th 2008, 05:50 PMskeeter
in future, please use grouping symbols for the sake of clarity ...

3/(2x-6) + 5/(x-3) = 1/6

3/[2(x-3)] + 5/(x-3) = 1/(2*3) note that x cannot equal 3.

get a common denominator ...

3*3/[3*2(x-3)] + 5*2*3/[2*3(x-3)] = 1(x-3)/[2*3(x-3)]

now that you have a common denominator, the numerators form the equation ...

9 + 30 = x - 3

now can you find x ?

y^2 + 3y - 1 = 0

y^2 + 3y = 1

half the coefficient of the linear term (that's the 3), square, and add result to both sides ...

y^2 + 3y + (3/2)^2 = 1 + (3/2)^2

[y + (3/2)]^2 = 13/4

y + (3/2) = +/- sqrt(13/4) = +/- sqrt(13)/2

y = [3 +/- sqrt(13)]/2

get over your hatred of these ... they are not going to go away. - Aug 21st 2008, 05:41 PMsaranya
- Aug 21st 2008, 10:09 PMProve It

Do you see that the denominator of the second term is half of the first term's denominator? So multiply the top and bottom of the second term by 2 to get the same denominator for both...

Cross multiplication gets rid of the denominators

So yes you are right. - Aug 22nd 2008, 12:58 PMskeeterQuote:

Cross multiplication gets rid of the denominators

better check going from this step to the next, x = 36 is not correct

So yes you are right.