Posted by Darren on Tuesday, May 25, 2010 at 6:24pm.
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.

Vectors  Reiny, Tuesday, May 25, 2010 at 6:40pm
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 ? 
Vectors  Darren, Tuesday, May 25, 2010 at 6:57pm
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 (2x2y+5z+10=0) and (2x+y4z+7=0).
So far, I eliminated x by subtracting the two equations. Leaving me with 3y+9z+3=0
What would I do next? 
Vectors  Reiny, Tuesday, May 25, 2010 at 7:14pm
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 = (y1)/6 = z/2