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
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
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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 :)
First convert each base 2 numbers to base 10 and solve either substitution method or elimination method
Solve with elimination method
Not what I am looking for
To solve the given simultaneous equations in base two, we need to convert all the numbers (coefficients and constants) to base ten, perform the calculations, and then convert the final result back to base two.
Let's convert the equations to base ten:
Equation 1: 11x + 10y = 10001
Equation 2: 10x - y = 10
Now we can solve these equations using any appropriate method. Let's use the substitution method:
From Equation 2, we can isolate x:
10x = y + 10
x = (y + 10) / 10
Substituting this value of x into Equation 1:
11((y + 10) / 10) + 10y = 10001
11y + 110 + 10y = 10001
21y = 9891
y = 469
Now substitute the value of y back into Equation 2 to find x:
10x - 469 = 10
10x = 10 + 469
10x = 479
x = 47.9
Since x is not an integer, the given system of equations does not have an exact solution.
However, if we round x to the nearest integer, we get x ≈ 48.
Now let's convert the decimal solution back to base two:
x = 48 in base two is 110000
y = 469 in base two is 111010101
Therefore, the solution to the simultaneous equations in base two is:
x = 110000
y = 111010101