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Mathematics 17 Online
OpenStudy (anonymous):

Need to find the LCD of this equation

OpenStudy (anonymous):

OpenStudy (precal):

Use 24(2x-1)(2x+1)

OpenStudy (saifoo.khan):

just use (2x-1)(2x+1)

OpenStudy (saifoo.khan):

24 is on another side.

OpenStudy (precal):

I don't think it really matters, both should generate the same solution

OpenStudy (anonymous):

i got 0=4x^2 - 1

OpenStudy (saifoo.khan):

um... no.

OpenStudy (anonymous):

Yeah I don't think I did that right

OpenStudy (precal):

I would use the 24 because it will get rid of the 24 on the other side.

OpenStudy (anonymous):

I did use the 24

OpenStudy (anonymous):

I got 48x-24-48x+24 which = 0

OpenStudy (precal):

\[(1/(2x-1))-(1/(2x+1))=(1/24)\] multiply by the LCD 24(2x-1)(2x+1) \[24(2x+1)-24(2x-1)=(2x-1)(2x+1)\] can you finish this?

OpenStudy (anonymous):

0 = 4x^2 -1

OpenStudy (precal):

\[48=4x^{2}-1\]

OpenStudy (anonymous):

the 48's cancel. 24(2x) = 48 -24(2x) = -48 48 - 48 = 0

OpenStudy (anonymous):

It'll look like 48x + 24 - 48x - 24 which = 0 on the left side. On the right it'll be 4x^2 -1

OpenStudy (precal):

You forgot to distribute the -24 to -1 and get +24; 24+24 is 48. In the end, whatever x value you get should check out true when you plug it into the equation.

OpenStudy (precal):

Once you solve for x; x=7/2 x=-7/2

OpenStudy (anonymous):

I honestly don't understand, but it's fine. I know quite well how to get the LCD and then solve equations with it, I just think the fractions threw me off on this problem. You were right by the way.

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