SOLUTION: I do not even know were to start for this problem because i only know how to solve problems of two equations with two variables. I would like to know how to solve it as well as a a

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I do not even know were to start for this problem because i only know how to solve problems of two equations with two variables. I would like to know how to solve it as well as a a      Log On


   



Question 269048: I do not even know were to start for this problem because i only know how to solve problems of two equations with two variables. I would like to know how to solve it as well as a answer for x,y, and z

x + y + z = 1
5x + 4y + 4z = 6
4x + 5y + 6z = −3

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x + y + z = 1
5x + 4y + 4z = 6
4x + 5y + 6z = −3
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One method is elimination. Eliminate one of the variables, make it 2 eqns with 2 variables.
Pick a variable to eliminate. Makes no difference, any one. I'll do z.
Multiply eqn 1 by 4 and subtract it from eqn 2.
5x + 4y + 4z = 6
4x + 4y + 4z = 4
------------------- subtract
x = 2
That's even better, usually it's not that easy.
Sub 2 for x in 2 of the eqns, I'll do 1 and 2
y + z = -1
4y+4z = -4
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Now there's a problem, the 2nd eqn is -4 times the first.
--> dependent, no unique solution.
An infinite # of solutions.