# Math Help plz!

• Aug 20th 2008, 04:31 PM
Peyton Sawyer
Math 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 PM
skeeter
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 PM
saranya
Quote:

Originally Posted by Peyton Sawyer
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

for the first one i am trying ,it maybe .. i am not sure
3/2x-6 + 5/x-3 = 1/6
3/x(x-3)+5/x-3 = 1/6
3+5(2)/2(x-3) = 1/6
13/2(x-3) = 1/6
2x-6 = 78
2x = 72
x = 36
• Aug 21st 2008, 10:09 PM
Prove It
Quote:

Originally Posted by Peyton Sawyer
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

$\frac{3}{2x-6}+\frac{5}{x-3}=\frac{1}{6}$
$\frac{3}{2(x-3)}+\frac{5}{x-3}=\frac{1}{6}$

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...

$\frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}$
$\frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}$
$\frac{3}{2x-6}+\frac{10}{2x-6}=\frac{1}{6}$
$\frac{3+10}{2x-6}=\frac{1}{6}$
$\frac{13}{2x-6}=\frac{1}{6}$

Cross multiplication gets rid of the denominators

$13\times 6 = 2x-6$
$78 = 2x-6$
$72 = 2x$
$36 = x$

So yes you are right.
• Aug 22nd 2008, 12:58 PM
skeeter
Quote:

$\frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}$
$\frac{3}{2(x-3)}+\frac{2(5)}{2(x-3)}=\frac{1}{6}$
$\frac{3}{2x-6}+\frac{10}{2x-6}=\frac{1}{6}$
$\frac{3+10}{2x-6}=\frac{1}{6}$
$\frac{13}{2x-6}=\frac{1}{6}$

Cross multiplication gets rid of the denominators

$13\times 6 = 2x-6$
$78 = 2x-6$ better check going from this step to the next, x = 36 is not correct
$72 = 2x$
$36 = x$

So yes you are right.
x = 42 is the correct solution.