Solve z=9/2(bm/bx/frac{b+x{bx}+3})

$$z = \frac { 9 } { 2 } ( \frac { \frac { b m } { b x } } { \frac { b + x } { b x } + 3 } )$$

Solve for m

$$m=\frac{2z\left(3bx+x+b\right)}{9b}$$

$$b\neq 0\text{ and }x\neq 0\text{ and }\left(x=-\frac{1}{3}\text{ or }b\neq -\frac{x}{3x+1}\right)$$

Steps for Solving Linear Equation

Cancel out b in both numerator and denominator.

$$z=\frac{9}{2}\times \frac{\frac{m}{x}}{\frac{b+x}{bx}+3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{bx}{bx}.

$$z=\frac{9}{2}\times \frac{\frac{m}{x}}{\frac{b+x}{bx}+\frac{3bx}{bx}}$$

Since \frac{b+x}{bx} and \frac{3bx}{bx} have the same denominator, add them by adding their numerators.

$$z=\frac{9}{2}\times \frac{\frac{m}{x}}{\frac{b+x+3bx}{bx}}$$

Combine like terms in b+x+3bx.

$$z=\frac{9}{2}\times \frac{\frac{m}{x}}{\frac{3bx+x+b}{bx}}$$

Divide \frac{m}{x} by \frac{3bx+x+b}{bx} by multiplying \frac{m}{x} by the reciprocal of \frac{3bx+x+b}{bx}.

$$z=\frac{9}{2}\times \frac{mbx}{x\left(3bx+x+b\right)}$$

Cancel out x in both numerator and denominator.

$$z=\frac{9}{2}\times \frac{bm}{3bx+x+b}$$

Multiply \frac{9}{2} times \frac{bm}{3bx+x+b} by multiplying numerator times numerator and denominator times denominator.

$$z=\frac{9bm}{2\left(3bx+x+b\right)}$$

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

$$z=\frac{9bm}{6xb+2x+2b}$$

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

$$\frac{9bm}{6xb+2x+2b}=z$$

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

$$9bm=z\times 2\left(3bx+x+b\right)$$

Reorder the terms.

$$9bm=2z\left(3bx+x+b\right)$$

Use the distributive property to multiply 2z by 3bx+x+b.

$$9bm=6xzb+2zx+2zb$$

The equation is in standard form.

$$9bm=6bxz+2xz+2bz$$

Divide both sides by 9b.

$$\frac{9bm}{9b}=\frac{2z\left(3bx+x+b\right)}{9b}$$

Dividing by 9b undoes the multiplication by 9b.

$$m=\frac{2z\left(3bx+x+b\right)}{9b}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{2xz}{9m-2z-6xz}\text{, }&\left(z=\frac{9m}{2}\text{ and }m\neq 0\text{ and }x\neq 0\text{ and }x\neq -\frac{1}{3}\right)\text{ or }\left(m\neq 0\text{ and }x\neq 0\text{ and }x\neq -\frac{1}{3}\text{ and }z\neq \frac{9m}{2\left(3x+1\right)}\text{ and }z\neq 0\text{ and }m\neq \frac{2xz}{3}+\frac{2z}{9}\right)\text{ or }\left(m\neq 0\text{ and }x=-\frac{1}{3}\text{ and }z\neq 0\right)\\b\in \mathrm{R}\setminus -\frac{x}{3x+1}\text{, }x\neq 0\text{ and }x\neq -\frac{1}{3},0\text{, }&x\neq 0\text{ and }z=0\text{ and }m=0\end{matrix}\right.$$