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Solving A System Of Equations

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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  1. Solving A System Of Equations Elida Alvarez &Erika Frias December 19, 2013 5B

  2. 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.

  3. Define Variables A = number of Adults admissions C = number of Children admissions

  4. System Of Equations 4A + 2C = 664 A + C = 272

  5. Solution Method • We are going to solve this solution by the process of elimination.

  6. 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.

  7. Step 2 of Solution 4(60) + 2C = 664 240 + 2C = 664 -240 = 240 ___________________________________ 2C = 424 2 2 ____________________________________ C = 212

  8. Solution to the System of Equations (60, 212)

  9. Check of Solution 4(60) + 2(212) = 664 60 + 212 = 272

  10. Solution in the Problem Situation The number of Adults admissions is 60. The number of Children admissions is 212.

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