Question 1163638
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There are always countless ways to solve a system of three linear equations in three unknowns.<br>
If you are using a purely algebraic technique, the objective is to first eliminate one of the equations to get two equations in two unknowns.  Then of course when you get to that point there are different ways of solving.<br>
Here is what I see when I look at the three given equations: adding the first two equations eliminates x.  So....<br>
3x+3y=3; -3x-2z = -11 --> 3y-2z = -8
y+z = -1<br>
For two equations in two unknowns, with the equations in this form, I prefer a solution by elimination.<br>
(1) 3y-2z = -8
(2) y+z = -1<br>
Multiply (2) by 2 and add to (1):<br>
3y-2z = -8
2y+2z = -2
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5y = -10
y = -2<br>
Substitute y=-2 in (2) to find z = 1.<br>
Substitute y=-2 in the first original equation to find x=3.<br>
ANSWER: (x,y,z) = (3,-2,1)<br>