Solve f^-1(x)=3x+7/2x-6

$$f ^ { - 1 } ( x ) = \frac { 3 x + 7 } { 2 x - 6 }$$

Solve for f

$$f=\frac{2x\left(x-3\right)}{3x+7}$$

$$x\neq 3\text{ and }x\neq 0\text{ and }x\neq -\frac{7}{3}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2\left(x-3\right).

$$f^{-1}x\times 2\left(x-3\right)=3x+7$$

Use the distributive property to multiply f^{-1}x\times 2 by x-3.

$$2f^{-1}x^{2}-3f^{-1}x\times 2=3x+7$$

Multiply -3 and 2 to get -6.

$$2f^{-1}x^{2}-6f^{-1}x=3x+7$$

Reorder the terms.

$$2\times \frac{1}{f}x^{2}-6\times \frac{1}{f}x=3x+7$$

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

$$2\times 1x^{2}-6x=3xf+f\times 7$$

Do the multiplications.

$$2x^{2}-6x=3xf+f\times 7$$

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

$$3xf+f\times 7=2x^{2}-6x$$

Combine all terms containing f.

$$\left(3x+7\right)f=2x^{2}-6x$$

Divide both sides by 3x+7.

$$\frac{\left(3x+7\right)f}{3x+7}=\frac{2x\left(x-3\right)}{3x+7}$$

Dividing by 3x+7 undoes the multiplication by 3x+7.

$$f=\frac{2x\left(x-3\right)}{3x+7}$$

Variable f cannot be equal to 0.

$$f=\frac{2x\left(x-3\right)}{3x+7}\text{, }f\neq 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{9f^{2}+92f+36}}{4}+\frac{3f}{4}+\frac{3}{2}$$

$$x=-\frac{\sqrt{9f^{2}+92f+36}}{4}+\frac{3f}{4}+\frac{3}{2}\text{, }f\neq 0$$

Solve for x

$$x=\frac{\sqrt{9f^{2}+92f+36}}{4}+\frac{3f}{4}+\frac{3}{2}$$

$$x=-\frac{\sqrt{9f^{2}+92f+36}}{4}+\frac{3f}{4}+\frac{3}{2}\text{, }f\leq \frac{-16\sqrt{7}-46}{9}\text{ or }\left(f\neq 0\text{ and }f\geq \frac{16\sqrt{7}-46}{9}\right)$$