Question 597144
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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}]

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