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10-21-2007

Mr. Hui, you didn't follow one the problems parts.

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Let x be the number of $1.50 souvenirs sold, and y be the number of $2.50. So x+y=200, this means that y=200-x. Now solving the equation:
$ \begin{eqnarray*} 1.50x+2.50(200-x)&=&380\\ 1.50x-2.50x+500&=&380\\ -x&=&-120\\ x&=&120 \end{eqnarray*} $
Then, you can easily find y which is 80. So the Solution to set is that there are 120 $1.50 souvenirs and 80 $2.50 souvenirs.

Edit:You have to fix the Hide feature so it can cover latex.

Edit by Temperal: You can use [more] instead.

10-21-2007	#3
hunter34 Guest Posts: n/a	Mr. Hui, you didn't follow one the problems parts. Show more: Let x be the number of $1.50 souvenirs sold, and y be the number of $2.50. So x+y=200, this means that y=200-x. Now solving the equation: $<br /> \begin{eqnarray}<br /> 1.50x+2.50(200-x)&=&380\\<br /> 1.50x-2.50x+500&=&380\\<br /> -x&=&-120\\<br /> x&=&120<br /> \end{eqnarray}<br />$ Then, you can easily find y which is 80. So the Solution to set is that there are 120 $1.50 souvenirs and 80 $2.50 souvenirs. Edit:You have to fix the Hide feature so it can cover latex. Edit by Temperal: You can use [more] instead. Last edited by Temperal; 10-21-2007 at 10:32 PM..