Combine the 2nd and 3rd equations. 3x will disappear:
2y + 5z = 7 This is true for any y and z, and is independent of x.
7-5z
Solving for y, 2y = 7 - 5z, so that y = ---------
2
We have to eliminate x again by using the 1st and 2nd equations:
-2x - 6y - 2z = -2. We want the coeff. of x in the first eqn to be -3.
Therefore, mult. all terms of -2x - 6y - 2z = -2 by 3/2:
(3/2)(-2x - 6y - 2z = -2) = -3x -9y -3z =-3
Now add this version of the 1st row to the 2nd row:
-3x -9y -3z =-3
3x +2y +5z = 7
----------------------
-7y + 2z = 4
7 - 5z
We found earlier that y = ----------, and can elim. y by subst. this fraction into
2
-7y + 2z = 4. Then -7y + 2z = 4 becomes -7( (7-5z)/2 ) + 2Z = 4.
Elim. the fraction by mult. all terms by 2:
-7(7-5z) + 4z = 8 which becomes -49 + 35z + 4z = 8
so that 39z = 57, and so z = 57/39.
Subst. this back into -7y + 2z = 4 to calculate y. Make a final subst. to calculate x.
Given the choice, I would solve this system using matrices.
Turns out that the solution, using matrices, does not exist.