Solve b^2(x-1)-b(x+5)-6(x+1)=0

$$b ^ { 2 } ( x - 1 ) - b ( x + 5 ) - 6 ( x + 1 ) = 0$$

Solve for x (complex solution)

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

Steps for Solving Linear Equation

Use the distributive property to multiply b^{2} by x-1.

$$b^{2}x-b^{2}-b\left(x+5\right)-6\left(x+1\right)=0$$

Use the distributive property to multiply b by x+5.

$$b^{2}x-b^{2}-\left(bx+5b\right)-6\left(x+1\right)=0$$

To find the opposite of bx+5b, find the opposite of each term.

$$b^{2}x-b^{2}-bx-5b-6\left(x+1\right)=0$$

Use the distributive property to multiply -6 by x+1.

$$b^{2}x-b^{2}-bx-5b-6x-6=0$$

Add b^{2} to both sides. Anything plus zero gives itself.

$$b^{2}x-bx-5b-6x-6=b^{2}$$

Add 5b to both sides.

$$b^{2}x-bx-6x-6=b^{2}+5b$$

Add 6 to both sides.

$$b^{2}x-bx-6x=b^{2}+5b+6$$

Combine all terms containing x.

$$\left(b^{2}-b-6\right)x=b^{2}+5b+6$$

Divide both sides by b^{2}-b-6.

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

Dividing by b^{2}-b-6 undoes the multiplication by b^{2}-b-6.

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

Divide \left(2+b\right)\left(3+b\right) by b^{2}-b-6.

$$x=\frac{b+3}{b-3}$$

Solve for x

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

Steps for Solving Linear Equation

Use the distributive property to multiply b^{2} by x-1.

$$b^{2}x-b^{2}-b\left(x+5\right)-6\left(x+1\right)=0$$

Use the distributive property to multiply b by x+5.

$$b^{2}x-b^{2}-\left(bx+5b\right)-6\left(x+1\right)=0$$

To find the opposite of bx+5b, find the opposite of each term.

$$b^{2}x-b^{2}-bx-5b-6\left(x+1\right)=0$$

Use the distributive property to multiply -6 by x+1.

$$b^{2}x-b^{2}-bx-5b-6x-6=0$$

Add b^{2} to both sides. Anything plus zero gives itself.

$$b^{2}x-bx-5b-6x-6=b^{2}$$

Add 5b to both sides.

$$b^{2}x-bx-6x-6=b^{2}+5b$$

Add 6 to both sides.

$$b^{2}x-bx-6x=b^{2}+5b+6$$

Combine all terms containing x.

$$\left(b^{2}-b-6\right)x=b^{2}+5b+6$$

Divide both sides by b^{2}-b-6.

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

Dividing by b^{2}-b-6 undoes the multiplication by b^{2}-b-6.

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

Divide \left(2+b\right)\left(3+b\right) by b^{2}-b-6.

$$x=\frac{b+3}{b-3}$$

Solve for b

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