Question

a charitable organization in Lanberry is hosting a black tie benefit. Yesterday, the organization sold 55 regular tickets and 49 VIP tickets, raising $9,663.

Answer 1

Answer:

• regular: $67
• VIP: $122

Step-by-step explanation:

Let r and v represent the costs of a regular and VIP ticket, respectively. The two sales can be represented by ...

  55r +49v = 9663

  79r +84v = 15541

We can subtract 7 times the second equation from 12 times the first to eliminate the v terms.

  12(55r +49v) -7(79r +84v) = 12(9663) -7(15541)

  660r +588v -553r -588v = 115,956 -108,787 . . . . . eliminate parentheses

  107r = 7169 . . . . . . . simplify

  r = 67 . . . . . . . . . . . divide by 107

Using this value in the first equation, we have ...

  55(67) +49v = 9663

  v = 5978/49 . . . . . . . . . subtract 3685, divide by 49

  v = 122

A regular ticket costs $67; a VIP ticket costs $122.

_____

Additional comment

When using "elimination" to solve a system of equations, you're looking for coefficients of the same variable that are related by small factors. Preferably, one coefficient is the same as, or a small multiple of, the other. Here, 55 and 79 (the coefficients of x) are not related by an integer, or a couple of small integers. On the other hand, the y-coefficients 49 and 84 have a common factor of 7, and are in the ratio 7:12, a pair of small numbers. This is why we chose to eliminate the y-variable.

The x-variable could be eliminated using 55 and -79 as multipliers of the equations. This results in larger numbers, and more chance for error. (Errors tend to creep in when computing or copying large numbers.)

Of course, any of several machine methods could be used to solve these equations, including graphing and matrix-solving functions. Here, we tried to honor the requirement to solve by elimination.