Question 310674
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Actually, you could multiply the Quantity equation by 2, but then you would have to multiply the Active Ingredient equation by 10 instead of 100.


As to your other question, let me illustrate it this way.  In the first place, I would have used -20 as a multiplier for the Q equation and 100 for the AI equation, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -20a\ -\ 20b\ =\ -20\,\cdot\,550]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 20a\ +\ 65b\ =\ 30.9\,\cdot\,550]


Now, you could multiply out the constant terms in the right-hand sides, and that would actually answer your question about "why the 550 doesn't cancel".  But let's proceed without that step.  Since I used a -20 multiplier for the Q equation, we can actually add the two equations rather than subtracting them.  This technique gives you much less chance for a sign error.  But rather than just showing the sum of the two, let's use the idea that if *[tex \Large \alpha\ =\ \beta] and *[tex \Large \gamma\ =\ \delta], then *[tex \Large \alpha\ +\ \gamma\ =\ \beta\ +\ \delta] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 20a\ -\ 20a\ +\ 65b\ -\ 20b\ =\ (30.9\,\cdot\,550)\ -\ (20\,\cdot\,550)]


Now, *[tex \Large 20a\ -\ 20a\ =\ 0] and that is why the *[tex \Large 20a] goes away.  *[tex \Large 65b\ -\ 20b\ =\ 45b], just like you wrote.  For the constant term, let's do a little factoring.  Factor out the 550:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ (30.9\,\cdot\,550)\ -\ (20\,\cdot\,550)\ =\  550(30.9-20)\ =\ 550\,\cdot\,10.9]


Putting it all back together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 45b\ =\ 550\,\cdot\,10.9]


And finally, as you stated,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ b\ =\ 133.\overline{2}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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