Question 597144
<font face="Times New Roman" size="+2">


This process works for "two kinds of ticket" problems and "two kinds of coins" problems.


Let *[tex \Large x] represent the <i>number</i> (as yet unknown) of one of the items.  Let *[tex \Large y] represent the <i>number</i> of the other item.


The sum *[tex \Large x\ +\ y] is then equal to the total number of items (300 in this problem).


Each type of item has a value *[tex \Large \left(v_x\right)] or *[tex \Large \left(v_y\right)] per item.  For your particular ticket problem, the value of each of the *[tex \Large x] items is $5, while the value of each of the *[tex \Large y] items is $12.  In a coin problem, dimes are worth 10 cents, quarters 25 cents, and so on.


The value of all of the *[tex \Large x] items is the value of one of them times the number of them, that is:  *[tex \Large v_xx], while the value of the *[tex \Large y] items is, similarly: *[tex \Large v_yy].  The sum of these values is the total value given in the problem.  For your problem, $2410.


From this information, you can create two equations (a necessary thing to be able to do because you have two variables, namely *[tex \Large x] and *[tex \Large y]).


In general:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left{x\ +\ y\ =\ \text{Total Number of Items}\cr v_xx\ +\ v_yy\ =\ \text{Total Value of Items}]



Specifically for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left{x\ +\ y\ =\ 300\cr 5x\ +\ 12y\ =\ 2410]


Given the simplicity of the first equation in these types of problems, I recommend that you use the substitution method to solve the 2X2 system of equations.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>