Solve for m

harpazo

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

Solve for m.

y = mx + b
 

TheJason

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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

\(\displaystyle m = \dfrac{1}{x}[y - b]\)
 
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harpazo

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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

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I would subtract through by \(b\) first:

\(\displaystyle y-b=mx\)

Now divide through by \(x\) where \(x\ne0\):

\(\displaystyle m=\frac{y-b}{x}\)
 
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harpazo

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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.