Thread: 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.

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

Hello, EJdive43!

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

We have: .

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x . . . . . . . . . . . . . . . . . . . .

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