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2015-04-22T11:58:01+05:30

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210 = 55 * 3 + 45
55 = 45 * 1 + 10
45 = 10 * 4 + 5
10 = 5 * 2

hence 5 is the HCF.

    5 =  45  - 10 * 4          substitute the values of 10 and 45 from above equations.
       = 210 - 55 * 3 - (55 - 45 ) 4
       =  210 - 55 * 3 - 55 * 4 + 45 * 4
       =  210 - 55 * 7 + (210 - 55 * 3) * 4
       = 210 * 5 - 55 * 19

so  x = 5 and  y = -19
============================
another way : to find an infinite number of  solutions.

       210 x + 55 y = 5
       42 x + 11 y = 1
        11 (4 x + y) = 1 + 2 x

y is an odd number as RHS is odd and so LHS has to be odd.  let y = 2 z+1.  one of x and y is negative and the other is positive.

    11 (4x + 2z+1) = 1 + 2x  
       42 x + 22 z + 10 = 0       
       21 x + 11 z + 5 = 0            --- (1)
       11 (x+z)  = - 5 (2 x + 1)

   x+z is a multiple of  -5  and perhaps (2x+1) /11
     2 x +1  is an odd multiple of 11.
so let       2x + 1 =  (2 v + 1) 11
         2 x  = 22 v + 10
           x = 11 v + 5
  then by use of (1),
        11 z = - 5 - 21 * (11 v + 5) = - 5 - 231 v - 105 = - 231 v - 110
             z = - 21 v - 10
             y =  - 42 v - 19

    5 = HCF =   210 * (11 v + 5) - 55 * (42 v + 19)
 
           x = 11 v + 5
             y = - (42 v + 19)

   v       x       y            210 * x + 55 * y
   0      5      -19              5
   -1    -6       23                5
   1     16     -61              5


1 5 1