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May 27th, 2015, 01:59 AM  #1 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Direct variation
I'm currently working through my algebra notes, because my calculus book told me I would be lost without an understanding of algebra. I've come to the conclusion that I was drunk when writing my notes, lol. Here is what I wrote: $\displaystyle "if" \frac{12}{3}=4 "and" \frac{a}{b}=k "then" kb = a "and"4 \cdot 3 = k "and"\frac{a}{16} = 4 "so" 4 \cdot 16 = 64$ I believe I was right up until $\displaystyle 4 \cdot 3 = k$, which should have been $\displaystyle 12 \div 3 = k$ and the rest needs correction too......... But given I am so far off, help 
May 27th, 2015, 02:31 AM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Maybe what you were trying to write was "Given that $\dfrac{a}{b} = k$, then $a = 12$ and $b = 3$ implies $k = 4$." and "If $\dfrac{a}{16} = 4$, then $a = 4\cdot 16 = 64$. If your notes are supposed to be a single thread of logic, then I don't understand it at all. Especially why $4\cdot3 = k$ is followed by $\dfrac{a}{16} = 4$. 

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