# [SOLVED] Verifying Inverse of this Function

#### unstopabl3

$$\displaystyle g(x)=\frac{2x-1}{x+3}$$

My Attempt At The Inverse

Let $$\displaystyle g(x)=z$$

$$\displaystyle (x+3)(z)=2x-1$$

$$\displaystyle xz+3z=2x-1$$

$$\displaystyle 1+3z=2x-xz$$

$$\displaystyle 1+3z=x(2-z)$$

$$\displaystyle \frac{1+3z}{2-z}=x$$

$$\displaystyle g^-1(x)=\frac{1+3x}{2-x}$$

Where did I make a mistake?

Thanks!

#### mr fantastic

MHF Hall of Fame
$$\displaystyle g(x)=\frac{2x-1}{x+3}$$

My Attempt At The Inverse

Let $$\displaystyle g(x)=z$$

$$\displaystyle (x+3)(z)=2x-1$$

$$\displaystyle xz+3z=2x-1$$

$$\displaystyle 1+3z=2x-xz$$

$$\displaystyle 1+3z=x(2-z)$$

$$\displaystyle \frac{1+3z}{2-z}=x$$

$$\displaystyle g^-1(x)=\frac{1+3x}{2-x}$$

Where did I make a mistake?

Thanks!
Why do you think there's a mistake. There is no mistake.

unstopabl3

#### unstopabl3

Okay, great then!

Thanks for the verification!