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March 3, 2015

March 3, 2015

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

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 6:40pmOk, 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 6:57pmThank 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?

- Vectors -
**Reiny**, Tuesday, May 25, 2010 at 7:14pmThe 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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