SOLUTION: Please help me solve this 3 variable system: x+y+z=8 x-y/3+(4/3)z=7 3z-3y-2x=7

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Question 1058261: Please help me solve this 3 variable system:
x+y+z=8
x-y/3+(4/3)z=7
3z-3y-2x=7

Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
(1)
---------------------
(2)
(2)
----------------------------
(3)
----------------------------
Multiply both sides of (1) by
and add (1) and (3)
(1)
(3)
--------------------------

Subtract (2) from this result
---------------------------

(2)
--------------------------

------------------
Now I can say:

and

----------------------
Plug these results into (1)
(1)
(1)
(1)
(1)
(1)
(1)
---------------------

-------------------------

---------------
The solutions are:
x = 1
y = 2
z = 5
-----------------
check:
(1)
(1)
{1)
OK
You can check (2) and (3)

Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!

Please help me solve this 3 variable system:
x+y+z=8
x-y/3+(4/3)z=7
3z-3y-2x=7

x + y + z = 8 -------- eq (i)
------ eq (ii)
3z - 3y - 2x = 7_____- 2x - 3y + 3z = 7 ----- eq (iii)
3x + 3y + 3z = 24 ---- Multiplying eq (i) by 3 ------- eq (iv)
3x - y + 4z = 21 ----- Multiplying eq (ii) by 3 ------ eq (v)
x + 6z = 31 ------ Adding eqs (iii) & (iv) ------- eq (vi)
4x + 5z = 29 ------ Adding eqs (i) & (v) ---------- eq (vii)
- 4x - 24z = - 124 --- Multiplying eq (vi) by - 4 ---- eq (viii)
- 19z = - 95 --------- Adding eqs (viii) & (vii)

x + 6(5) = 31 ----- Substituting 5 for z in eq (vi)
x + 30 = 31

1 + y + 5 = 8 ------ Substituting 1 for x, and 5 for z in eq (i)
6 + y = 8