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Could you please help me solve this question:
Solve this system of equations using elementary operations.

1. x/3 + y/4 + z/5 =14
2. x/4 + y/5 + z/3 =-21
3. x/5 + y/3 + z/4 =7

So far, I have multiplied all three equations by 60, to turn them into whole numbers. Then everytime I try elmination, it wouldn't work. Could someone please help.

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Darren

  1. Ok, I will do the same, new equations are

    1. 20x + 15y + 12z = 840
    2. 15x + 12y + 20z = -1260
    3. 12x + 20y + 15z = 420

    #1x4 --> 80x + 60y + 48z = 4200
    #2x5 --> 75x + 60y + 100z = -6300
    subtract --> 5x - 52z = 10500 --- #4

    #1x4 --> 80x + 60y + 48z = 4200
    #3x3 --> 36x + 60y + 45z = 1260
    subtract --> 44x + 3z = 2940 --- #5

    #4x44 --> 220x - 2288z = 462000
    #5x5 ---> 220x + 15z = 14700
    subtr --> 2273z = 447300
    z = 447300/2273 ????? (expecting a nicer number)

    Did I make an arithmetic error somewhere?
    How do my numbers compare to yours ?

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    Reiny

  2. Thank You Reiny, that's exactly what I got. But I was told the number isn't a nice one, I believe that it's correct.

    Also, I have another question to ask.

    Find the intersection of the 2 planes (2x-2y+5z+10=0) and (2x+y-4z+7=0).

    So far, I eliminated x by subtracting the two equations. Leaving me with -3y+9z+3=0

    What would I do next?

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    Darren

  4. The intersection of two planes is a straight line, unless the two planes are parallel

    Ok so far --->
    -3y + 9z + 3 = 0
    divide by -3
    y - 3z = 1
    y = 1 + 3z

    now we pick an arbitrary value for z
    let z=0, then y = 1, back in first equation
    2x -2 + 0 = -10
    x = -4
    so a point on our line is (-4,1,0)

    let z = 1
    y = 1+3 = 4
    in first
    2x -8 + 5 = -10
    x = -7/2
    and another point is (-7/2, 4,1)

    so we can find a direction vector for our line
    = [1/2 , 3,1) or [1,6,2]

    we can now write our line in parametric form as
    x = -4 + k
    y = 1 + 6k
    z = 0 + 2k

    or in symmetric form

    (x+4)/1 = (y-1)/6 = z/2

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    Reiny

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