SOLUTION: 2x+y-2z=10;3x+2y+2z=1; 5x+4y+3z=4 solve .

Question 1132289: 2x+y-2z=10;3x+2y+2z=1; 5x+4y+3z=4
solve .
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
2x + y - 2z = 10
3x +2y + 2z = 1
5x +4y + 3z = 4
:
Adding the 1st and 2nd equation
3x +2y + 2z = 1
2x + y - 2z = 10
------------------Addition eliminates z
5x + 3y = 11
:
Multiply the 1st equation by 3, multiply the 3rd equation and add
6x + 3y - 6z = 30
10x +8y + 6z = 8
---------------------addition eliminates z
16x + 11y = 38
:
Multipl 5x +3y = 11 by 16, multiply 16x + 11y = 38 by 5
80x + 48y = 176
80x + 55y = 190
----------------------Subtraction eliminates x, find y
0 - 7y = -14
y = -14/-7
y = + 2
:
Find x using 5x + 3y = 11
5x + 3(2) = 11
5x = 11 - 6
5x = 5
x = 1
:
Find z using 2x + y - 2z = 10
2(1) + 2 - 2z = 10
-2z = 10 - 4
z = 6/-2
z = -3
;
;
Check solution in 3x + 2y + 2z = 1
3(1) + 2(2) + 2(-3) = 1
3 + 4 - 6 = 1

Answer by MathTherapy(10556) (Show Source): You can put this solution on YOUR website!

2x+y-2z=10;3x+2y+2z=1; 5x+4y+3z=4
solve .

2x + y - 2z = 10 -------- eq (i)
3x + 2y + 2z = 1 -------- eq (ii)
5x + 4y + 3z = 4 -------- eq (iii)
6x + 4y + 4z = 2 -------- Multiplying eq (ii) by 2 ----- eq (iv)
x + z = - 2 ------ Subtracting eq (iii) from (iv) ------ eq (v)
- 4x - 2y + 4z = - 20 --- Multiplying eq (i) by - 2 ---- eq (vi)
- x + 6z = - 19 --------- Adding eqs (vi) & (ii) ------- eq (vii)
7z = - 21 ---------- Adding eqs (vii) & (v)

x - 3 = - 2 -------- Substituting - 3 for z in eq (v)

2(1) + y - 2(- 3) = 10 ------- Substituting 1 for x, and - 3 for z in eq (i)
2 + y + 6 = 10

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