SOLUTION: x/6+y/2+z/6=1/2 x+y+z=5 x/3+y/2+z/6=1

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Question 1097642: x/6+y/2+z/6=1/2
x+y+z=5
x/3+y/2+z/6=1

Found 2 solutions by Boreal, MathTherapy:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x/6+y/2+z/6=1/2
x+y+z=5
x/3+y/2+z/6=1
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x+3y+z=3
x+y+z=5
2x+3y+z=6
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-x-3y-z=-3
2x+3y+z=6
x=3
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6+3y+z=6; 3y+z=0,z=-3y
3+y-3y=5
3-2y=5
y=-1
(3, -1, 3)
Check
(1/2)-(1/2)+3/6=1/2 TRUE
3-1+3=5, TRUE
1+(-1/2)+(1/2)=1 TRUE

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

x/6+y/2+z/6=1/2
x+y+z=5
x/3+y/2+z/6=1
x%2F6+%2B+y%2F2+%2B+z%2F6+=+1%2F2 ------- eq (i)
x+%2B+y+%2B+z+=+5 ------- eq (ii)
x%2F3+%2B+y%2F2+%2B+z%2F6+=+1 ------- eq (iii)
 x + 3y + z = 3 ----- Multiplying eq (i) by LCD, 6 ------- eq (iv)
 x +  y + z = 5 ----- eq (ii)
2x + 3y + z = 6 ----- Multiplying eq (iii) by LCD, 6 ----- eq (v)

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2y = - 2 ----- Subtracting eq (ii) from eq (iv)
highlight_green%28matrix%281%2C5%2C+y%2C+%22=%22%2C+%28-+2%29%2F2%2C+%22=%22%2C+-+1%29%29

===============
highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+3%29%29 ----- Subtracting eq (v) from eq (iv)

===============
3 + - 1 + z = 5 ----- Substituting 3 for x, and - 1 for y in eq (ii) 
highlight_green%28matrix%281%2C5%2C+z%2C+%22=%22%2C+5+-+2%2C+%22=%22%2C+3%29%29