SOLUTION: Solve the system of equations for x and y by the addition method 2x-4y+3z=17 x+2y-z=0 4x-4-z=6 Eliminate 1 variable at a time

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Solve the system of equations for x and y by the addition method 2x-4y+3z=17 x+2y-z=0 4x-4-z=6 Eliminate 1 variable at a time      Log On


   

Question 2831: Solve the system of equations for x and y by the addition method
2x-4y+3z=17
x+2y-z=0
4x-4-z=6
Eliminate 1 variable at a time
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
Your question was:
2x-4y+3z=17 ........... [1]
x+2y-z=0 .............. [2]
4x-4y-z=6 ............. [3]
subtracting equation 3 from 2 we get, 
1x + 2y - z = 0
4x - 4y - z = 6
-  +    +    -
---------------
-3x + 6y = -6

Taking 3 common,
- x + 2y = -2

y= (-2 + x)/2
y = (x - 2)/2 ......... [4]



x + 2y - z = 0
x + 2(x-2)/2 - z = 0
x + x - 2 - z = 0
2x - 2 - z = 0


x = ( z + 2 ) / 2 ............[5]


Substituting 5 in 4,
+y+=+%28x-2%29%2F2+
+y+=+%28%28%28z%2B2%29%2F2%29-2%29%2F2+
+y+=+%28z-2%29%2F4+


Now we have x and y in terms of z,
x = (z+2)/2
y = (z-2)/4


Putting these relative values of x and y in equation 2 we get,
x + 2y - z = 0


+%28z%2B2%29%2F2+%2B+2%28%28z-2%29%2F4%29+-z+=+0+
+%28z+%2B+2+%2B+z+-+2%29%2F2+-+z+=+0+
+2z+-+z+=+0+


Therefore, z = 0 ................[6]


Substituting this value of z in relative values of x and y we get,
x = (z+2)/2 = (0+2)/2 = 2/2 = 1
y = (z-2)/4 = (0-2)/4 = -2/4 = -1/2


Thus, x=1,y=-1/2,z=0


Cross check:
+x+%2B+2y+-+z+=+0+
+%281%29+%2B+2%28-1%2F2%29+-+%280%29+=+0+
+1+-+1+-+0+=+0+
+0+-+0+=+0+
Hence,these values of x,y and z are correct.


Hope this helps,
Best of luck.