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

• Jan 2nd 2013, 11:46 AM
EJdive43
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√x2+1 -1/(√x2+1 +1)
I need to know how would you go about solving for the G(x) because i'm so stuck!
Note: The "x2+1" is under the square root. It jus didn't look right so I had to make a note.
• Jan 2nd 2013, 12:19 PM
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))??
• Jan 2nd 2013, 12:24 PM
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)])
• Jan 2nd 2013, 04:42 PM
Soroban
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:

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

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

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

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

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