Solve f(x)=3x+2quadg(x)=x+3/2x+1

$$f ( x ) = 3 x + 2 \quad g ( x ) = \frac { x + 3 } { 2 x + 1 }$$

Solve for g

$$g=-\frac{6x^{2}+2x-3}{2x\left(2x+1\right)}$$

$$x\neq -\frac{1}{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x+1.

$$3x\left(2x+1\right)+2gx\left(2x+1\right)=x+3$$

Use the distributive property to multiply 3x by 2x+1.

$$6x^{2}+3x+2gx\left(2x+1\right)=x+3$$

Use the distributive property to multiply 2gx by 2x+1.

$$6x^{2}+3x+4x^{2}g+2gx=x+3$$

Subtract 6x^{2} from both sides.

$$3x+4x^{2}g+2gx=x+3-6x^{2}$$

Subtract 3x from both sides.

$$4x^{2}g+2gx=x+3-6x^{2}-3x$$

Combine x and -3x to get -2x.

$$4x^{2}g+2gx=-2x+3-6x^{2}$$

Combine all terms containing g.

$$\left(4x^{2}+2x\right)g=-2x+3-6x^{2}$$

The equation is in standard form.

$$\left(4x^{2}+2x\right)g=3-2x-6x^{2}$$

Divide both sides by 4x^{2}+2x.

$$\frac{\left(4x^{2}+2x\right)g}{4x^{2}+2x}=\frac{3-2x-6x^{2}}{4x^{2}+2x}$$

Dividing by 4x^{2}+2x undoes the multiplication by 4x^{2}+2x.

$$g=\frac{3-2x-6x^{2}}{4x^{2}+2x}$$

Divide -2x+3-6x^{2} by 4x^{2}+2x.

$$g=\frac{3-2x-6x^{2}}{2x\left(2x+1\right)}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{g^{2}+14g+19}-g-1}{2\left(2g+3\right)}\text{; }x=-\frac{\sqrt{g^{2}+14g+19}+g+1}{2\left(2g+3\right)}\text{, }&g\neq -\frac{3}{2}\\x=-3\text{, }&g=-\frac{3}{2}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{g^{2}+14g+19}-g-1}{2\left(2g+3\right)}\text{; }x=-\frac{\sqrt{g^{2}+14g+19}+g+1}{2\left(2g+3\right)}\text{, }&g\leq -\sqrt{30}-7\text{ or }\left(g\neq -\frac{3}{2}\text{ and }g\geq \sqrt{30}-7\right)\\x=-3\text{, }&g=-\frac{3}{2}\end{matrix}\right.$$