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Math Help - Linear Equations

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    3

    Linear Equations

    Hello, I have been having trouble with these 3 questions that are on my upcoming test. I have tried it many times without getting an answer and I could really use some help.
    Here are the questions:
    1)
    2x - 3y = 5
    4y + 2z = -6
    5x + 7z = -15
    Solve for X Y and Z using substitution or elimination.
    EDIT :I Recently tried to use a homework help online service who gave me the answer ( I didnt understand their work though) For this they got x=4 y=1 and z=-5

    2)
    2x + 3y - z = 9
    x - 3y + z = -6
    3x + y - 4z = 31
    Solve for X Y and Z using substitution or elimination. (Same as before)

    3)
    6/x - 4/y = 1
    -3/x + 3/y = 2
    Solve for X and Y (6/x, 4/y , -3/x and 3/y and fractions) using substitution
    EDIT: Same homework help (Once again Didn't Understand what the heck their work meant but..) They Got for X=6/11 and Y=2/5
    Thank you in advance and please show your work so I can understand it.
    MOVED POST TO URGENT SECTION DUE TO TEST TIME CHANGE.
    Last edited by ChrisBarrett; November 16th 2008 at 01:57 PM. Reason: Urgency level.
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  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hi.

    Quote Originally Posted by ChrisBarrett View Post
    Hello, I have been having trouble with these 3 questions that are on my upcoming test. I have tried it many times without getting an answer and I could really use some help.
    Here are the questions:
    1)
    2x - 3y = 5
    4y + 2z = -6
    5x + 7z = -15
    Solve for X Y and Z using substitution or elimination.
    EDIT :I Recently tried to use a homework help online service who gave me the answer ( I didnt understand their work though) For this they got x=4 y=1 and z=-5
    I dont have the time to solve all 3 exercices, but the first one:
    I think the easiest way is solving by substitution, that means

    2x - 3y = 5 => 2x = 5 +3y => x = (5+3y)/2

    You have to other equation
    (2) 4y + 2z = -6 here is no x in it -> nothing to substitute

    5x + 7z = -15 but here is a x, you know x, x = (5+3y)/2

    Therefor 5((5+3y)/2) + 7z = -15

    (3) (25+15y)/2+7z = -15

    Using (2) 4y + 2z = -6 => z = (-6-4y)/2

    (3) (25+15y)/2+7z = -15 you know z, thus

    (25+15y)/2+7*((-6-4y)/2) = -15 (solve it!) => y = 1

    then solve z by z = (-6-4y)/2 (you already know y ...), same for x

    Rapha
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