Solving a system of equations
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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

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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


Solution to the system of equations

Solution to the System of Equations

(60, 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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