<PRE><B>Question 1203611</B>
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The solution from tutor @mathtutor_2020 shows a solution using what is commonly called Gauss-Jordan elimination.  That is one good method for solving a system of 3 linear equations.&lt;br&gt;
Tutor MathLover uses a purely algebraic approach; but her propensity to use substitution instead of elimination leads down an unnecessarily difficult path to the solution.&lt;br&gt;
And her solution only shows a path to the solution -- it doesn&#39;t do anything to teach YOU how to get the solution.&lt;br&gt;
Here is an algebraic solution using only elimination.&lt;br&gt;
In a system of 3 linear equations, the objective is to first reduce the system to 2 equations by combining the given equations in a way that one of the variables is eliminated.  Then that system of 2 equations can be solved by again eliminating one of the variables by combining the two equations in an appropriate way.&lt;br&gt;
(1) x+4y-6z=-1
(2) 2x-y+2z=-7
(3) -x+2y-4z=5&lt;br&gt;
Seeing the &quot;x&quot; in (1) and the &quot;-x&quot; in (3), add those two equations to get an equation in only y and z:&lt;br&gt;
(4) 6y-10z=4
(4) 3y-5z=2&lt;br&gt;
Now find another way to eliminate x using a different pair of the original equations.  One way to do that is to double (3) and add to (2):&lt;br&gt;
2x-y+2z=-7
-2x+4y-8z=10
(5) 3y-6z=3&lt;br&gt;
Equations (4) and (5) are now a system of 2 equations in y and z.  We could simplify (5); however, in their current forms (4) and (5) both contain the term 3y, so subtracting (4) from (5) will give us an equation in only z:&lt;br&gt;
3y-6z=3
-3y+5z=-2
-z=1
(6) z=-1&lt;br&gt;
Now substitute z=-1 in either (4) or (5) and solve for y:&lt;br&gt;
3y+6=3
3y=-3
(7) y=-1&lt;br&gt;
And last substitute y=-1 and z=-1 in any of the original equations to solve for x:&lt;br&gt;
x-4+6=-1
(8) x=-3&lt;br&gt;
ANSWER: (x,y,z) = (-3,-1,-1)&lt;br&gt;
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