Solve g^-1(x)=8x/3x-4

$$g ^ { - 1 } ( x ) = \frac { 8 x } { 3 x - 4 }$$

Solve for g

$$\left\{\begin{matrix}g=\frac{3x}{8}-\frac{1}{2}\text{, }&x\neq \frac{4}{3}\\g\neq 0\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3x-4.

$$g^{-1}x\left(3x-4\right)=8x$$

Use the distributive property to multiply g^{-1}x by 3x-4.

$$3g^{-1}x^{2}-4g^{-1}x=8x$$

Reorder the terms.

$$3\times \frac{1}{g}x^{2}-4\times \frac{1}{g}x=8x$$

Variable g cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by g.

$$3\times 1x^{2}-4x=8xg$$

Do the multiplications.

$$3x^{2}-4x=8xg$$

Swap sides so that all variable terms are on the left hand side.

$$8xg=3x^{2}-4x$$

Divide both sides by 8x.

$$\frac{8xg}{8x}=\frac{x\left(3x-4\right)}{8x}$$

Dividing by 8x undoes the multiplication by 8x.

$$g=\frac{x\left(3x-4\right)}{8x}$$

Divide x\left(-4+3x\right) by 8x.

$$g=\frac{3x}{8}-\frac{1}{2}$$

Variable g cannot be equal to 0.

$$g=\frac{3x}{8}-\frac{1}{2}\text{, }g\neq 0$$

Solve for x

$$x=\frac{8g+4}{3}$$

$$x=0\text{, }g\neq 0$$