SOLUTION: Can I please have your assistance? Solve the system : x + y + z = 6, 2x - y + z = 3, x + y - z = 0

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Question 1042299: Can I please have your assistance? Solve the system : x + y + z = 6, 2x - y + z = 3, x + y - z = 0
Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!




Keep row 1 (R1) as is;
R2 ----- R2-2*R1;
R3 -----R3-R1.

Revision to


R2 -------(-1)R2;
R3 --------(1/2)R3


Revision to


further....
Revision to

-



Not fully finished, but

Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!
Can I please have your assistance? Solve the system : x + y + z = 6, 2x - y + z = 3, x + y - z = 0
x + y + z = 6 -------- eq (i)
2x - y - z = 3 ------- eq (ii)
x + y - z = 0 -------- eq (iii)
3x = 9 ------- Adding eqs (i) & (ii)
Instead of 2x - y - z = 3, equation (ii) should have been: 2x - y + z = 3 , and so:
Answer by ikleyn(52908)   (Show Source): You can put this solution on YOUR website!
.
Solve the system : x + y + z = 6, 2x - y + z = 3, x + y - z = 0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 x + y + z = 6,    (1)
2x - y + z = 3,    (2)
 x + y - z = 0     (3)

Add equations (2) and (3) (both sides). You will get

3x = 3.   Hence, x = 1.

Substitute x = 1 into equations ((1) and (3). You will get

1 + y + z = 6,
1 + y - z = 0,

or, which is the same,

y + z =  5,    (3)
y - z = -1.    (4)

Now add the equations (3) and (4) (both sides). You will get

2y = 4.   Hence, y = 2.

Now from (3) z = 5 - y = 5 - 2 = 3.

Answer.  x = 1,  y = 2,  z = 3.


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