Question 70907
Let x be units digit then by condition of problem tens digit is x-3
The origonal number is 10(x-3) + x

 The sum of the digits is x-3 + x

 The origonal number is six more than four times the sum of the digits 

 10(x-3)+x = 4(x-3 + x) + 6

 Simplify the left and right sides

  11x - 30 = 8x - 6

   From this 

   3x  = 24

    x = 8
   
 The number is 58