# Solve for m

#### harpazo

##### Pure Mathematics
Math practice for you on the Fourth of July 2018. Enjoy.

Solve for m.

y = mx + b

#### Jason

Staff member
Moderator
Solve for $$\displaystyle m$$.

$$\displaystyle y = mx + b$$

$$\displaystyle \dfrac{y}{x} = \dfrac{mx}{x} + \dfrac{b}{x}$$

$$\displaystyle \dfrac{y}{x} = m + \dfrac{b}{x}$$

$$\displaystyle -m = -\dfrac{y}{x} + \dfrac{b}{x}$$

$$\displaystyle \dfrac{-m}{-1} = \dfrac{-\dfrac{y}{x}}{-1} + \dfrac{\dfrac{b}{x}}{-1}$$

$$\displaystyle m = \dfrac{y}{x} - \dfrac{b}{x}$$

or

$$\displaystyle m = \dfrac{1}{x}[y - b]$$

harpazo

#### harpazo

##### Pure Mathematics
Solve for $$\displaystyle m$$.

$$\displaystyle y = mx + b$$

$$\displaystyle \dfrac{y}{x} = \dfrac{mx}{x} + \dfrac{b}{x}$$

$$\displaystyle \dfrac{y}{x} = m + \dfrac{b}{x}$$

$$\displaystyle -m = -\dfrac{y}{x} + \dfrac{b}{x}$$

$$\displaystyle \dfrac{-m}{-1} = \dfrac{-\dfrac{y}{x}}{-1} + \dfrac{\dfrac{b}{x}}{-1}$$

$$\displaystyle m = \dfrac{y}{x} - \dfrac{b}{x}$$

or

$$\displaystyle m = \dfrac{1}{x}[y - b]$$
Very good.

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
I would subtract through by $$b$$ first:

$$\displaystyle y-b=mx$$

Now divide through by $$x$$ where $$x\ne0$$:

$$\displaystyle m=\frac{y-b}{x}$$

anemone and harpazo

#### harpazo

##### Pure Mathematics
I would subtract through by $$b$$ first:

$$\displaystyle y-b=mx$$

Now divide through by $$x$$ where $$x\ne0$$:

$$\displaystyle m=\frac{y-b}{x}$$
This would be my choice method, too.