document.write( "Question 472703: x=by+cz
\n" ); document.write( "y=cz+ax
\n" ); document.write( "z=ax+by
\n" ); document.write( "prove a/1+a + b/1+b + c/1+c = 1\r
\n" ); document.write( "\n" ); document.write( "pls help with solution...thanks \n" ); document.write( "
Algebra.Com's Answer #324040 by robertb(5830)\"\" \"About 
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).\r
\n" ); document.write( "\n" ); document.write( "Similarly, x(1+a) = y(1+b).\r
\n" ); document.write( "\n" ); document.write( "==> x(1+a) = y(1+b) = z(1+c).
\n" ); document.write( "==> \"y+=+%28%281%2Ba%29%2F%281%2Bb%29%29x+\" and \"z+=+%28%281%2Ba%29%2F%281%2Bc%29%29x+\"\r
\n" ); document.write( "\n" ); document.write( "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+\".
\n" ); document.write( "Since x is not identically 0, we can cancel x throughout.\r
\n" ); document.write( "\n" ); document.write( "==> \"1+=+b%2A%28%281%2Ba%29%2F%281%2Bb%29%29++%2B+c%2A%28%281%2Ba%29%2F%281%2Bc%29%29\".\r
\n" ); document.write( "\n" ); document.write( "==> \"1%2F%281%2Ba%29+=+b%2F%281%2Bb%29+%2B+c%2F%281%2Bc%29\"\r
\n" ); document.write( "\n" ); document.write( "==> \"1+-+a%2F%281%2Ba%29+=+b%2F%281%2Bb%29+%2B+c%2F%281%2Bc%29\"\r
\n" ); document.write( "\n" ); document.write( "The conclusion follows after one more step.\r
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