SOLUTION: x=by+cz y=cz+ax z=ax+by prove a/1+a + b/1+b + c/1+c = 1 pls help with solution...thanks

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Question 472703: x=by+cz
y=cz+ax
z=ax+by
prove a/1+a + b/1+b + c/1+c = 1
pls help with solution...thanks
Found 2 solutions by richard1234, robertb:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Adding all three of our given statements we obtain x+y+z = 2ax + 2by + 2cz. Matching the x,y,z coefficients, we get 2a = 1, 2b = 1, 2c = 1, so a,b,c = 1/2. Replacing 1/2 into a/(1+a) + b/(1+b) + c/(1+c) yields a sum of 1.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Subtract the 2nd equation from the 3rd: z - y = by - cz, whence z(1+c) = y(1+b).
Similarly, x(1+a) = y(1+b).
==> x(1+a) = y(1+b) = z(1+c).
==> y+=+%28%281%2Ba%29%2F%281%2Bb%29%29x+ and z+=+%28%281%2Ba%29%2F%281%2Bc%29%29x+
Subsitute into the 1st equation: x+=+b%2A%28%281%2Ba%29%2F%281%2Bb%29%29x++%2B+c%2A%28%281%2Ba%29%2F%281%2Bc%29%29x+.
Since x is not identically 0, we can cancel x throughout.
==> 1+=+b%2A%28%281%2Ba%29%2F%281%2Bb%29%29++%2B+c%2A%28%281%2Ba%29%2F%281%2Bc%29%29.
==> 1%2F%281%2Ba%29+=+b%2F%281%2Bb%29+%2B+c%2F%281%2Bc%29
==> 1+-+a%2F%281%2Ba%29+=+b%2F%281%2Bb%29+%2B+c%2F%281%2Bc%29
The conclusion follows after one more step.