ISBN: 978-0-07-338315-6 section 5.3 question 14(c) How do you solve a Diophantin

Question

ISBN: 978-0-07-338315-6 section 5.3 question 14(c)

How do you solve a Diophantine equation 221x+35y =11? Need a step-by-step approach. Please review what I have written.

221x+35y = 11; gcd(221, 35) =1

221 =35(6)+ 11

35 = 11(3) +2

11 = 2(5) +1

2 = 2(1) + 0, thus we found that the gcd is actually 1

1 = 11 - 5(2) --> 1 = 221-35(6) - 5(35-11(3)---> 1 = 221 -35(6) +35(-5) - 11(15)--->

1 = 221 -35(6) +35(-5) -[15(221-35(6))]--->1 = 221(1)+35(-6)+35(-5) +221(-15) +35(-6*-15)--->

1 = 221(-14)+35(-11)+35(90)---->1=221(-14) + 35(79), which the entire quanity must now be

multiplied by 11......11(1=221(-14) + 35(79))---> 11 = 221(-154) + 35(869), which is 100% wrong!!!! should equal the left side which is 11. Please show what's wrong with what I'm doing. I must find all solutions to 221x+35y =11, but I can't even get one!

Explanation / Answer

You have got 221 = 35*6 + 11

so (35*6 + 11)x + 35y = 11,

35*6*x+11(x - 1) + 35y = 0

35(6x + y) + 11(x - 1) = 0

Since gcd (11, 35) = 1, so 6x + y = 11t, x - 1 = -35t

answer: x = 1 - 35t, y = -6 + 221t, t is any real number

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