



 
Dear Anonymous, What happens is that each equation 1, 2, ..., 6 is solved for its remainder term and then the result is substituted into the one among b, c, ... that you're working on. These equations are used in t he order 5, 4, 3, 2, 1 The first equation you have is equation 5 rewritten. To get from it to a b, you rewrite equation 4 as 28 = 57  1x29 and substitute that for 28. Then you simplify in order to get c. Next you rewrite 3 as 29 = 861x57 and substitute that for 29 in what you have. Simplify, then repeat by solving equation 2 for 57, then simplify again and do the same using equation 1 and 86. When you're done, everything is in terms of 1232 and 573. Hope this helps.  


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