Solve z=a-b/b+c

$$z = \frac { a - b } { b } + c$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{a}{-z+c-1}\text{, }&a\neq 0\text{ and }z\neq c-1\\b\neq 0\text{, }&a=0\text{ and }z=c-1\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$zb=a-b+bc$$

Add b to both sides.

$$zb+b=a+bc$$

Subtract bc from both sides.

$$zb+b-bc=a$$

Combine all terms containing b.

$$\left(z+1-c\right)b=a$$

The equation is in standard form.

$$\left(z-c+1\right)b=a$$

Divide both sides by z+1-c.

$$\frac{\left(z-c+1\right)b}{z-c+1}=\frac{a}{z-c+1}$$

Dividing by z+1-c undoes the multiplication by z+1-c.

$$b=\frac{a}{z-c+1}$$

Variable b cannot be equal to 0.

$$b=\frac{a}{z-c+1}\text{, }b\neq 0$$

Solve for a

$$a=b\left(z-c+1\right)$$

$$b\neq 0$$