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Solving A System Of Equations. Elida Alvarez &Erika Frias December 19, 2013 5B. Problem Situation. The zoo charges $4 for every adult and $2 for every child . Today , the zoo had 272 people in attendance and collected a total of $664.

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solving a system of equations

Solving A System Of Equations

Elida Alvarez &Erika Frias

December 19, 2013

5B

problem situation
Problem Situation

The zoo charges $4 for every adult and $2 for every child. Today, the zoo had 272 people in attendance and collected a total of $664.

Find the number of $4 and $2 admission fees collected by the zoo today.

define variables
Define Variables

A = number of Adults admissions

C = number of Children admissions

system of equations
System Of Equations

4A + 2C = 664

A + C = 272

solution method
Solution Method
  • We are going to solve this solution by the process of elimination.
step 1 of solution
Step 1 of Solution

4A + 2C = 664

  • 2A + 2C = 544

____________________________________

2A = 120

_______ __________

2 2

A = 60

Multiply the second equation by 2 to eliminate the variable C.

step 2 of solution
Step 2 of Solution

4(60) + 2C = 664

240 + 2C = 664

-240 = 240

___________________________________

2C = 424

2 2

____________________________________

C = 212

check of solution
Check of Solution

4(60) + 2(212) = 664

60 + 212 = 272

solution in the problem situation
Solution in the Problem Situation

The number of Adults admissions is 60.

The number of Children admissions is 212.

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