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Jnr Cert Honours student has brain freeze ! help

  • 09-11-2011 09:07PM
    #1
    Registered Users, Registered Users 2 Posts: 10,892 ✭✭✭✭


    Ok, I'm a bit stuck on a question for my maths HomeWork. It is a form of simultaneous equation.
    The question is as follows:

    The sum of four times one number and three times the second number is 61.
    If twice the first number less the second number is 13, find the numbers.

    The answers in the back of my book are 7 and 10 however I keep getting the incorrect answer. :confused:

    Any help would be much appreciated. :)


Comments

  • Registered Users, Registered Users 2 Posts: 338 ✭✭ray giraffe


    Spook_ie wrote: »
    The sum of four times one number and three times the second number is 61.
    If twice the first number less the second number is 13, find the numbers.

    The answers in the back of my book are 7 and 10

    4x+3y=61
    2x-y=13

    Maybe you could multiply the second equation by 3, so you have 3y in the first and -3y in the second.


  • Registered Users, Registered Users 2 Posts: 10,892 ✭✭✭✭Spook_ie


    Thanks a million for the quick reply! I understand completely now! :D


  • Registered Users, Registered Users 2 Posts: 13,228 ✭✭✭✭bnt


    Another way I like would involve rearranging the second equation to read y=2x-13. Then, in the first equation, replace y with (2x-13) and solve for x.

    In its pure form, fascism is the sum total of all irrational reactions of the average human character.

    ― Wilhelm Reich



  • Registered Users, Registered Users 2 Posts: 10,892 ✭✭✭✭Spook_ie


    Thanks BNT - I appreciate all the help


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