Solving a Proportion: Advanced

Hello, ondverg!

Sorry, I don't see how this is "advanced"./


\(\displaystyle \text{Solve the proportion: }\:\frac{\text{-}12}{y-2} \:=\:\frac{\text{-}8}{y}\)

\(\displaystyle \text{Note that: }\:y \neq 0,\,2\)

\(\displaystyle \text{Multiply both sides by }y(y-2)\!:\quad y(y-2)\,\cdot\frac{\text{-}12}{y-2} \:=\:y(y-2)\,\cdot\frac{\text{-}8}{y}\)

. . \(\displaystyle -12y \:=\:-8(y-2) \quad\Rightarrow\quad -12y \:=\:-8y + 16 \quad\Rightarrow\quad -4y \:=\:16\)

\(\displaystyle \text{Therefore: }\;y \:=\:-4\)

 
Thank you!

*It's not advanced - that's just the title listed at the top of my ALEKS problem set guide. I had to do what they call a 'basic' problem first. It's just considered advanced for people like me who don't know what they're doing. :)
 
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