Question 4179: Need another soloution to the question previously asked
Solve the following system of equations
3x-2y+4z=1
x+y+2z=3
2x-3y+6z=6
I previously did not understand your answer and per your advisement is seeking another solution (in other words had not idea what you were "talking about") :-)
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! lol no worries...Right then, the tedious way:
eqn2 can be re-written as x=3-y-2z. We will now sub this into eqn1 and 3, to give:
eqn1 --> 3(3-y-2z)-2y+4z=1 --> 9-3y-6z-2y+4z=1 --> -5y-2z=-8
eqn3 --> 2(3-y-2z)-3y+6z=6 --> 6-2y-4z-3y+6z=6 --> -5y+2z=0
Add these 2 eqns to give: -10y=-8 --> y=0.8
Subtract the 2 eqns to give: -4z = -8 --> z=2
But these into eqn2 to find x: x=3-0.8-2(2), so x=-1.8
Is this what i gave as the answers te first time? I cannot remember!
jon
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