SOLUTION: Could you please take the time to help me solve this question: Solve the following system of three linear equations: 2x + y + z = 3, -x + 2y + 2z = 1 and x - y - 3z = -6. Noting th

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Could you please take the time to help me solve this question: Solve the following system of three linear equations: 2x + y + z = 3, -x + 2y + 2z = 1 and x - y - 3z = -6. Noting th      Log On


   



Question 101338: Could you please take the time to help me solve this question: Solve the following system of three linear equations: 2x + y + z = 3, -x + 2y + 2z = 1 and x - y - 3z = -6. Noting that each equation represents a plane, interpret your results.
Thank you, very much for taking the time to help me, most appreciated

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
first take these two equations:
2x+%2B+y+%2B+z+=+3
-x+%2B+2y+%2B+2z+=+1 (multiply both sides by 2 in this one)
then we will have
2x+%2B+y+%2B+z+=+3
-2x+%2B+4y+%2B+4z+=+2
now we will eliminate x by adding them and it will be:
%28y%2B4y%29+%2B%28z%2B4z%29=3%2B2
5y+%2B5z+=5 now divide both sides by 5,and we will have:
y%2Bz=1
then we have
y=-z%2B1 which we can put in third equation
x+-+y+-+3z+=+-6
and we will have

x+%2Bz+-1+-+3z+=+-6
x+%2Bz+-1+-3z+=-6
x+-2z+=-6+%2B1
x+-2z+=-5
from this equation
x=2z-5
since we have x+=+2z+-+5 and y+=+-z+%2B+1, we can use them to calculate z:
let use the second equation
-x+%2B+2y+%2B+2z+=+1
substitute x and y+
-%282z-5%29+%2B+2%28-z+%2B+1%29%2B+2z+=+1
-2z+%2B+5+-2z+%2B+2+%2B+2z+=+1
-2z=+-6
z=-6%2F-2
z=3+
now is easy to calculate x and y
x=+2z+%96+5
x+=+2%2A3+%96+5
x+=+1
y+=+-z+%2B+1
y+=+-3+%2B+1
y+=+-2