Solve 1-2bx/b-3=2(4x-1)/3b-b^2

$$\frac { 1 - 2 b x } { b - 3 } = \frac { 2 ( 4 x - 1 ) } { 3 b - b ^ { 2 } }$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{1}{2\left(b+2\right)}\text{, }&b\neq -2\text{ and }b\neq 3\text{ and }b\neq 0\\x\in \mathrm{C}\text{, }&b=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by b\left(b-3\right), the least common multiple of b-3,3b-b^{2}.

$$b\left(1-2bx\right)=-2\left(4x-1\right)$$

Use the distributive property to multiply b by 1-2bx.

$$b-2xb^{2}=-2\left(4x-1\right)$$

Use the distributive property to multiply -2 by 4x-1.

$$b-2xb^{2}=-8x+2$$

Add 8x to both sides.

$$b-2xb^{2}+8x=2$$

Subtract b from both sides.

$$-2xb^{2}+8x=2-b$$

Combine all terms containing x.

$$\left(-2b^{2}+8\right)x=2-b$$

The equation is in standard form.

$$\left(8-2b^{2}\right)x=2-b$$

Divide both sides by -2b^{2}+8.

$$\frac{\left(8-2b^{2}\right)x}{8-2b^{2}}=\frac{2-b}{8-2b^{2}}$$

Dividing by -2b^{2}+8 undoes the multiplication by -2b^{2}+8.

$$x=\frac{2-b}{8-2b^{2}}$$

Divide 2-b by -2b^{2}+8.

$$x=\frac{1}{2\left(b+2\right)}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{1}{2\left(b+2\right)}\text{, }&b\neq -2\text{ and }b\neq 3\text{ and }b\neq 0\\x\in \mathrm{R}\text{, }&b=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by b\left(b-3\right), the least common multiple of b-3,3b-b^{2}.

$$b\left(1-2bx\right)=-2\left(4x-1\right)$$

Use the distributive property to multiply b by 1-2bx.

$$b-2xb^{2}=-2\left(4x-1\right)$$

Use the distributive property to multiply -2 by 4x-1.

$$b-2xb^{2}=-8x+2$$

Add 8x to both sides.

$$b-2xb^{2}+8x=2$$

Subtract b from both sides.

$$-2xb^{2}+8x=2-b$$

Combine all terms containing x.

$$\left(-2b^{2}+8\right)x=2-b$$

The equation is in standard form.

$$\left(8-2b^{2}\right)x=2-b$$

Divide both sides by -2b^{2}+8.

$$\frac{\left(8-2b^{2}\right)x}{8-2b^{2}}=\frac{2-b}{8-2b^{2}}$$

Dividing by -2b^{2}+8 undoes the multiplication by -2b^{2}+8.

$$x=\frac{2-b}{8-2b^{2}}$$

Divide 2-b by -2b^{2}+8.

$$x=\frac{1}{2\left(b+2\right)}$$

Solve for b

$$\left\{\begin{matrix}\\b=2\text{, }&\text{unconditionally}\\b=-2+\frac{1}{2x}\text{, }&x\neq \frac{1}{10}\text{ and }x\neq 0\text{ and }x\neq \frac{1}{4}\end{matrix}\right.$$