finding values of x and y using Octal Base system

In finding the values of x and y, if (x567) + (2yx5) = (71yx) ( all in base 8) I proceeded as under.

I assumed x=abc and y=def and followed.

   (abc+010 def+101 110+abc 111+101)=(111 001 def abc) //adding ()+()=() and equating LHS=RHS.
    abc=111-010=101 which is 5 in base 8 and then def=001-101 which is -4
    so x=5 and y=-4

Now the Question is that the answer mentioned in my book is x=4 and y=3.

Is the above method correct.If so,then what's issue here ??

Solution

you can't compare the digits beginning with the most significant digit, because you don't know the carry from the digit below. Also a digit cannot have a negative value.

You can start with the least significant digit, because there is no carry:

7 + 5 = 14

so x = 4 with a carry of 1 at the next digit. now you can rewrite your equation to:

(4567) + (2y45) = (71y4)

now you can look at the second least significant digit (the carry in mind):

6 + 4 + 1 (carry) = 13

so y = 3, also with a carry of 1.

the whole equation is:

(4567) + (2345) = (7134)

which is true for the octal system.