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Question 19511: Hi, I am stuck somewhere in here, can anybody point me back to the proper path? Thanks,
I am solving for a linear system with 3 variables
2x+3y-4z=4
x-6y+z=-16
-x+3z=8
I multiplied the first equation by 2 and got 4x+6y-8z=8
I then added this to the second and got a answer of 5x-7z=-8
I then multiplied the third equation by 5 and got z=4 but then I just petered out from here.... Any help would be great, Thanks sloozy
Answer by stanbon(48558)   (Show Source):
You can put this solution on YOUR website!Correct: z=4.
Now substitute that value into the original 3rd equation, as follows:
-x + 3z =8
-x + 3(4)= 8
=x = -4
x = 4
Now substitute those values into the 1st equation to find "y".
2x + 3y -4z =4
2(4) + 3y -4(4) = 4
3y -8 = 4
3y = 12
y = 4
Now you have x=y=z=4
Cheers,
stan H.
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