SOLUTION: y/4+z/4=z/3+x/3=x/2+y/2 X+y+z=27

Question 1054915: y/4+z/4=z/3+x/3=x/2+y/2
X+y+z=27
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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y/4+z/4=z/3+x/3=x/2+y/2
X+y+z=27
~~~~~~~~~~~~~~~~~

You are given these THREE equations

y/4+z/4 = z/3+x/3,   (1)
z/3+x/3 = x/2+y/2,   (2)
x   + y  + z = 27.   (3)

Multiply (1) by 12 (both sides) and (2) by 6 (both sides). You will get

3y + 3z = 4z + 4x,   (1')
2z + 2x = 3x + 3y,   (2')
x +   y +  z = 27    (3')

Or, which is the same (simplifying)

4x - 3y +  z = 0,    (1'')
x  + 3y - 2z = 0,    (2'')
x  +  y +  z = 27.   (3'')

Add (1'') and (2'') (both sides). You will get

5x - z = 0,   or   z = 5x.

Now substitute z = 5x into (1'') and (3''). You will get

4x - 3y + 5x =  0,    (4)  instead of (1'') and
x  +  y + 5x = 27     (5)  instead of (3'').

Simplify:

9x - 3y =  0,         (4')
6x +  y = 27,         (5')

or

3x - y =  0,          (4'')
6x + y = 27.          (5'')

Add (4'') and (5''). You will get

9x = 27  --->  x =  = 3.

Then from (4'')  y = 3x = 3*3 = 9.

Finally,  z = 5x = 5*3 = 15.

Answer.  x = 3,  y = 9,  z = 15.

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