A consistent system can be dependent with infinitely many solutions or independent with one solution. An inconsistent system has no solution. What is the case here? {1+y=x ; x+y=1
To determine the case for the system of equations {1+y=x ; x+y=1}, we can solve it using elimination or substitution methods.
Using the elimination method, we can add the two equations together:
(1+y) + (x+y) = x+(x+y) + 1
1 + 2y + x = 2x + y + 1
x - 2x = 2y - y
-x = y
From this equation, we can see that x and y are dependent on each other since y = -x. Therefore, this system of equations is a dependent system with infinitely many solutions.
Alternatively, using the substitution method, we can solve for y in terms of x from the first equation:
1 + y = x
y = x - 1
Then, we substitute this expression for y into the second equation:
x + (x-1) = 1
2x - 1 = 1
2x = 2
x = 1
Substituting this value of x back into the first equation, we find:
1 + y = 1
y = 0
So, the solution to this system of equations is x = 1 and y = 0.
In conclusion, the system of equations {1+y=x ; x+y=1} is a dependent system with infinitely many solutions.