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Question 1111208: solve by using elimination method
3x+2y+z=8
5x-3y+4z=3
2x+y+3z=7
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 3x+2y+z=8
5x-3y+4z=3
2x+y+3z=7
:
multiply the 2nd equation by -1
3x + 2y + z = 8
-5x + 3y -4z =-3
2x+ y + 3z = 7
------------------addition eliminates x and z
0 + 6y + 0 = 12
y = 12/6
y = 2
substituting 2 for y in the 1st equation
3x + 2(2) + z = 8
3x + z = 8 - 4
3x + z = 4
Do the same on the 3rd equation
2x + 2 + 3z = 7
2x + 3z = 7 - 2
2x + 3z = 5
:
Multiply the 1st 2 unknown equation by 3, subtract the above equation
9x + 3z = 12
2x + 3z = 5
----------------subtraction eliminates z, find x
7x + 0 = 7
x = 1
Find z using the 2nd original equation
5x - 3y + 4z = 3
Replace x and y
5(1) - 3(2) + 4z = 3
5 - 6 + 4z = 3
-1 + 4z = 3
4z = 3 + 1
4z = 4
z = 1
We have x=1; y=2; z=1
:
:
Check in the 2nd original equation 5x - 3y + 4z = 3,
5(1) - 3(2) + 4(1) = 3
5 - 6 + 4 = 3
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