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Mystery Algebra Question: What is the cost of each ticket?

  • The cost of a hockey ticket and a football ticket is $30. The cost of a movie ticket and a football ticket is $25. The cost of a hockey ticket and a movie ticket is $19. What is the cost of each ticket?

  • Answer:

    {x + y = 30, z + y = 25, x + z = 19} x = hockey = 12 y = football = 18 z = movie = 7 30 - x = y z + (30 - x) = 25 z = -5 + x x + (-5 + x) = 19 2x - 5 = 19 2x = 24 x = 12 12 + y = 30 y = 18 z + 18 = 25 z = 7

Megan Franklin at Yahoo! Answers Visit the source

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Given H+F=30; M+F=25; H+M=19 Let's use subtraction for a pair of these so we can cancel one of the variables out H+F=30 -(H+M)=-19 =F-M=11 Now we have F-M=11 & F+M=25. Add them. F-M+F+M=11+25==> 2F=36; THEREFORE F=18 From there we can determine the others H+F=30; if F=18 then H+(18)=30; H=30-18==>12 F+M=25; if F=18 then (18)+M=25; then M=25-18==>7 F=18, H=12,M=7 check, check, check :)

S.

H + F = 30 : eqn1 F + M = 25 : eqn2 H + M = 19 : eqn3 eqn4: 0 * eqn1 + eqn2 F + M = 25 : eqn4 eqn5: -eqn1 + eqn3 - F + M = -11 : eqn5 eqn6: -eqn5 - eqn4 -2 M = -14 : eqn6 M = 7 : eqn7 using eqn5: F = ( -11-7(1))/(-1) F = 18 : eqn8 using eqn1: H = ( 30-7(0)-18(1))/(1) H = 12 : eqn9

Davis P

Let f be the cost of a football ticket, h a hockey ticket, and m a movie ticket. f = $18 h = $12 m = $ 7 Have fun!

Jumpin' Jack Flash

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