Need help building a solution.

[mitchg]

I have a Alegebra problem in which I half figured out the answer (by guessing), but haven't been able to build a solution for. I went through my book and can't figure it out. Can someone help please?

x/(x-1)-(3/x=1/2

The answe I got was 2, which works. I can't show my work effectively. Please help.

Mitch

 

[Lizzie]

x/(x-1)-(3/x=1/2

The answe I got was 2, which works. I can't show my work effectively. Please help.

Mitch

Do you mean \(\displaystyle \L{}\frac{x}{(x-1)-(\frac{3}{x})} = \frac{1}{2}\)?
Or did you mean \(\displaystyle \L{}\frac{x}{(x-1)}-\frac{3}{x} = \frac{1}{2}\)?

 

[mitchg]

The second equation. I'm not sure how you posed the equation format, thus the reason for the text version.

 

[soroban]

Hello, Mitch!

\(\displaystyle \L\frac{x}{x\,-\,1}\,-\,\frac{3}{x}\:=\:\frac{1}{2}\)

When given an equation with fractions, multiply through by the LCD: \(\displaystyle 2x(x-1)\)

. . \(\displaystyle \L2x(\not{x-1})\left(\frac{x}{\not{x-1}}\right)\,-\,2\not{x}(x-1)\left(\frac{3}{\not{x}}\right)\:=\:\not{2}x(x-1)\left(\frac{1}{\not{2}}\right)\)

. . We get: \(\displaystyle \L\,2x^2\,-\,6(x\,-\,1)\:=\:x(x\,-\,1)\)

. . which simplifies: \(\displaystyle \,\L x^2\,-\,5x\,_\,6\:=\:0\)

. . and factors: \(\displaystyle \,\L(x\,-\,2)(x\,-\,3)\:=\:0\)

. . and has roots: \(\displaystyle \L\,x\,=\,2,\:3\)

 

[Lizzie]

oops, sorry. I forgot to respond to this. But, soroban, as always, did a wonderful job explaining it!

 

[Gene]

As to how to format, at the top of the post you should see some buttons. (4 of them?) One of them should have a lesson on LaTex.
------------------
Gene

 

[mitchg]

Thank you! I see where I went wrong. I also see that by guessing, I only had half the solution.

Thanks again!