Algebra 2:Can someone show me how you would go about solving for G(x)?

The equation goes: 2G(x)-1/(G(x)+1=2√x^{2}+1 -1/(√x^{2}+1 +1)

I need to know how would you go about solving for the G(x) because i'm so stuck!

Note: The "x^{2}+1" is under the square root. It jus didn't look right so I had to make a note.

Re: Algebra 2:Can someone show me how you would go about solving for G(x)?

is it (2G(x)-1)/(G(x)) or 2G(x)-(1/G(x))??

Re: Algebra 2:Can someone show me how you would go about solving for G(x)?

Because if it's (2G(x)-1)/(G(x)) then just split up the numerator and you'll get 2-1/G(x)= 2√x 2+1 -1/(√x2+1 +1) and that's the equivalent of 2-[2√x2+1 -1/(√x2+1 +1) ]=1/G(x)

So that'll give you G(x)=1/(2-[2√x2+1 -1/(√x2+1 +1)])

Re: Algebra 2:Can someone show me how you would go about solving for G(x)?

Hello, EJdive43!

How is this any different from any other "solve for G" problem?

Quote:

$\displaystyle \text{Solve for G: }\:\frac{2G-1}{G+1} \:=\:\frac{2\sqrt{x^2+1} - 1}{\sqrt{x^2+1} + 1}$

We have: .$\displaystyle (2G-1)(\sqrt{x^2+1} + 1) \:=\:(G+1)(2\sqrt{x^2+1} - 1)$

. . $\displaystyle 2G\sqrt{x^2+1} + 2G - \sqrt{x^2+1} - 1 \:=\:2G\sqrt{x^2+1} - G + 2\sqrt{x^2+1} - 1$

x . . . . . . . . . . . . . . . . . . . . $\displaystyle 3G \:=\:3\sqrt{x^2+1}$

. . . . . . . . . . . . . . . . . . . . . . $\displaystyle G \:=\:\sqrt{x^2+1}$