Question 738782
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Myself, I wouldn't go to all of the trouble to set up a system of equations for this one.  I would do it this way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x\ +\ 20(40\ -\ x)\ =\ 15\,\cdot\,40]


Then I would solve for *[tex \LARGE x] and calculate *[tex \LARGE 40\ -\ x].


However, this is just two steps in to solving the following two-variable system by substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 40]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x\ +\ 20y\ =\ 15\,\cdot\,40]


Note that I didn't need to use decimal fractions because the second equation and this one:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.12x\ +\ 0.20y\ =\ 0.15\,\cdot\,40]


are equivalent.



John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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