The following simultaneous equations are written in base two

11x+10y=10001
10x-y=10
solve and live your anser in base two?? plz help tnx

6 answers

from second
y = [(1 0) x - (1 0)]
substitute
(1 1) x + (1 0) [(1 0) x - (1 0)] =(1 0 0 0 1)

now what is (1 0) (1 0) ?
(1^2^1 +0*2^0)(1^2^1 +0*2^0)
= 2^1*2^1 = 2^2 = (1 0 0)

so we have
(1 1) x + ((1 0 0) x - (1 0 0)] =(1 0 0 0 1)
or
(1 1 1)x = (1 0 1 0 1)

x = [ 2^4 + 2^2 + 2^0] / (2^2 + 2^1 + 2^0)

hard to show long division, I will try later maybe but I get
2^2 - 2^1 + 2^0
or
( 1 1) which in base ten would be 3
====================================
CHECK in base ten
3 x + 2 y = 17
2 x - y = 2

3 x + 2 y = 17
4 x - 2 y = 4
-------------------
7 x = 21
x = 3 Caramba !
now do y :)

Solve with elimination method

Not what I am looking for

First convert each base 2 numbers to base 10 and solve either substitution method or elimination method