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Math Help - Linear Equations

  1. #1
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    Exclamation Linear Equations

    It turns out that I need to have these questions done by 6:30 (EST) so Im moving it to the Urgent Thread, ( I have to go to my school for a class event and my teacher wants me to come in for the test due to her being absent on monday). Sorry for the thread switch, Thanks in advance!
    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.
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  2. #2
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    1) We have the following equations:

    2x - 3y = 5

    4y + 2z = -6

    5x + 7z = -15

    Notice that the final two equations allow for x and y to be given in terms of z, all we do is solve the 2nd equation for y and the 3rd one for x:

    4y + 2z = -6

    4y = -6 - 2z

    y = -\frac{3}{2} - \frac{1}{2}z

    5x + 7z = -15

    5x = -15 - 7z

    x = -3 - \frac{7}{5}z

    Now, we take the 1st equation and plug in what we know for x and y:

    2x - 3y = 5

    2(-3 - \frac{7}{5}z) - 3(-\frac{3}{2} - \frac{1}{2}z) = 5

    -6 - \frac{14}{5}z + \frac{9}{2} + \frac{3}{2}z = 5

    \frac{9}{2} - \frac{12}{2} - \frac{14}{5}z + \frac{3}{2}z = 5

    -\frac{3}{2} - \frac{28}{10}z + \frac{15}{10}z = 5

    -\frac{3}{2} - \frac{13}{10}z = 5

    -\frac{13}{10}z = \frac{10}{2} + \frac{3}{2}

    -\frac{13}{10}z = \frac{13}{2}

    z = \frac{\frac{13}{2}}{-\frac{13}{10}}

    z = \frac{13}{2} *-\frac{10}{13}

    z = -\frac{10}{2} = -5

    Now we sub in the z-value:

    4y + 2z = -6

    4y - 10 = -6

    4y = 4

    y = 1

    5x + 7z = -15

    5x - 35 = -15

    5x = 20

    x = 4

    And there you go.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by ChrisBarrett View Post
    It turns out that I need to have these questions done by 6:30 (EST) so Im moving it to the Urgent Thread, ( I have to go to my school for a class event and my teacher wants me to come in for the test due to her being absent on monday). Sorry for the thread switch, Thanks in advance!
    Here are the questions:

    (1) 2x - 3y = 5
    (2) 4y + 2z = -6
    (3) 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
    Well, it's almost 6:30 somewhere, but it may be too late for you.
    Multiply (1) by 5 and multiply (3) by -2. Add the equations to eliminate x.

    10x-15y=25 [1]

    -10x-14z=30 [3]
    --------------------------------------
    15y-14z=55 [4]

    28y+14z=-42 [5] Multiply (2) by 7 and add to [4].
    -------------------------------------
    13y=13
    \boxed{y=1}

    Use (1): 2x-3(1)=5
    2x=8
    \boxed{x=5}

    Use (2): 4(1)+2z=-6
    2z=-10
    \boxed{x=-5}

    I'll hold up solving the others to see if this was any help to you or if you still have questions. Good luck.
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