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x+y+z = 7
2x-y+z = 4
11y-10x = ??
By trial and error
We get x=1 y=2 z=4
So 11*2 - 10*1 = 12
12 is the answer or 3z
x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be
1/x + 1/y = 1/3
1/x + 1/z = 1/5
1/y + 1/z = 1/7
x/y = ??
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