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Jnr Cert Honours student has brain freeze ! help

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  • 09-11-2011 9:07pm
    #1
    Spook_ie
    Registered Users Posts: 10,896 ✭✭✭✭


    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. :)


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    #2
    ray giraffe
    Registered Users 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.


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    #3
    Spook_ie
    Registered Users Posts: 10,896 ✭✭✭✭Spook_ie


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


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    #4
    bnt
    Registered Users Posts: 12,973 ✭✭✭✭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.

    From out there on the moon, international politics look so petty. You want to grab a politician by the scruff of the neck and drag him a quarter of a million miles out and say, ‘Look at that, you son of a bitch’.

    — Edgar Mitchell, Apollo 14 Astronaut



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    #5
    Spook_ie
    Registered Users Posts: 10,896 ✭✭✭✭Spook_ie


    Thanks BNT - I appreciate all the help


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