[MOVED] Solve Equation 9x/3x-1 + 4/3x+1 = 3

[Jade]

Solve Equation 9x/3x-1 + 4/3x+1 = 3

I am having a lot of trouble with this one. I know that you should change to 3x^2+4x/3=3 - - but I don't know how to get there. Please help! :shock:

 

[soroban]

Re: Solve Equation 9x/3x-1 + 4/3x+1 = 3

Hello, Jade!

I am having a lot of trouble with this one . . . no wonder!

\(\displaystyle \L\;\;\;\frac{9x}{3x\,-\,1}\,+\,\frac{4}{3x\,+\,1}\:=\:3\)

I know that you should change to \(\displaystyle \,\frac{3x^2\,+\,4x}{3}\:=\:3\;\) . . . Who said that?

- - but I don't know how to get there . . . Neither do I!

Multiply through by the LCD, \(\displaystyle (3x\,-\,1)(3x\,+\,1):\)

. . \(\displaystyle (\sout{3x\,-\,1})(3x\,+\,1)\,\frac{9x}{\sout{3x\,-\,1}}\,+\,(3x\,-\,1)(\sout{3x\,+\,1})\,\frac{4}{\sout{3x\,+\,1}} \;= \;(3x\,-\,1)(3x\,+\,1)\,3\)

We have: \(\displaystyle \:9x(3x\,+\,1)\,+\,4(3x\,-\,1)\;=\;3(9x^2\,-\,1)\)

Simplify: \(\displaystyle \;27x^2\,+\,9x\,+\,12x\,-\,4\;=\;27x^2\,-\,3\;\;\Rightarrow\;\;21x\:=\:1\)

Therefore: \(\displaystyle \L\:x\,=\,\frac{1}{21}\)