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Math Help - Help, please

  1. #1
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    Help, please

    I have one question left on my assignment, but it's confusing me, as algebra is not my strongest subject!!

    If somebody could figure this out for me, it would be greatly appreciated!!

    "Solve the following system algebraically. Show your work"

    3/4x-2/5y=2

    1/2x-3/5y=-2
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by star4822 View Post
    I have one question left on my assignment, but it's confusing me, as algebra is not my strongest subject!!

    If somebody could figure this out for me, it would be greatly appreciated!!

    "Solve the following system algebraically. Show your work"

    3/4x-2/5y=2

    1/2x-3/5y=-2
    please clarify, is it:

    (3/4)x-(2/5)y=2

    (1/2)x-(3/5)y=-2

    or

    3/(4x)-2/(5y)=2

    1/(2x) - 3/(5y)=-2
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  3. #3
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    Sorry, it's meant to be:

    (3/4)x-(2/5)y=2

    (1/2)x-(3/5)y=-2
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by star4822 View Post
    Sorry, it's meant to be:

    (3/4)x-(2/5)y=2

    (1/2)x-(3/5)y=-2
    ok, here goes. we will do the elimination method, rather than the substitution method

    (3/4)x-(2/5)y=2 ..................................(1)

    (1/2)x-(3/5)y=-2 .................................(2)

    x - (8/15)y = 8/3 .................................(3) = (1) * (4/3)
    x - (6/5)y = -4 ....................................(4) = (2) * 2

    now subtract the (4) equation from the (3) equation, we get:

    (2/3)y = 20/3 .....................................(5) = (3) - (4)
    => y = 10 ..........................................(6) = (5) * 3/2

    but from equation (4), x - (6/5)y = -4
    => x - (6/5)(10) = -4
    => x = 60/5 - 4 = 8

    so we get x = 8, y = 10
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