Solve =g(e^-frac{x{3}})=2+3e^-frac{x{3}}*2x+1

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

Solve for g

$$g=3\left(2x+e^{\frac{x}{3}}\right)$$

Steps for Solving Linear Equation

Multiply 3 and 2 to get 6.

$$ge^{-\frac{x}{3}}=2+6e^{-\frac{x}{3}}x+1$$

Add 2 and 1 to get 3.

$$ge^{-\frac{x}{3}}=3+6e^{-\frac{x}{3}}x$$

The equation is in standard form.

$$e^{-\frac{x}{3}}g=6xe^{-\frac{x}{3}}+3$$

Divide both sides by e^{-\frac{1}{3}x}.

$$\frac{e^{-\frac{x}{3}}g}{e^{-\frac{x}{3}}}=\frac{6xe^{-\frac{x}{3}}+3}{e^{-\frac{x}{3}}}$$

Dividing by e^{-\frac{1}{3}x} undoes the multiplication by e^{-\frac{1}{3}x}.

$$g=\frac{6xe^{-\frac{x}{3}}+3}{e^{-\frac{x}{3}}}$$

Divide 3+6e^{-\frac{x}{3}}x by e^{-\frac{1}{3}x}.

$$g=6x+3e^{\frac{x}{3}}$$