Question 184241
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First thing, your answers are correct.


Second thing, we are going to express the dollar amounts in 1000s, so Compacts cost 12, Intermediates cost 18, and Full-size cost 24, and the total amount spent is 1500.


Let <i>x</i> be the number of Intermediates and


Let <i>y</i> be the number of Full-Size and then


2<i>x</i> is the number of Compacts.


We know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 2x + x + y = 100 \ \ \Rightarrow\ \ 3x + y = 100]


And, since there are 2<i>x</i> Compacts each costing 12, the cost of the Compacts is 24<i>x</i>.  Likewise the cost of the Intermediates is 18<i>x</i> and the cost of the Full-Size is 24<i>y</i>.  Finally, the sum of the costs is 1500, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 24x + 18x + 24y = 1500\ \ \Rightarrow\ \ 42x + 24y = 1500]


Multiply the first equation by -14:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ -42x -14y = -1400]


Add this new equation, term-by-term, to the second equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 0x + 10y = 100 \ \ \Rightarrow\ \ 10y = 100 \ \ \Rightarrow\ \ y = 10]


Hence, 10 Full-Size cars.  Substitute this value into the original first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 3x + 10 = 100 \ \ \Rightarrow\ \ 3x = 90 \ \ \Rightarrow\ \ x = 30]


Hence, 30 Intermediates and 60 Compacts.


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