If for x = -1 we find the same value of y in both equations, then x = -1 will be a part of the solution of the system.
Let's suppose that x = -1 is the x-coordinate of the solution of the given system
x + y = 4
x - y = 6.
x + y = 4
-1 + y = 4
-1 + 1 + y = 4 + 1
y = 5
x - y = 6
-1 - y = 6
-1 - 6 + y - y = 6 - 6 + y
-7 = y
Since we have two different values for y, we say that x = -1 is not a part of the solution for the given system.
x+xy=8 xy=-x+8 y=-1+8/x
The expression (-1-1) simplifies to -2. To determine if (-2) is a solution to the equation xy, we need to know what the equation specifically is. If the equation is of the form xy = -2, then (-1, -1) is indeed a solution since (-1)(-1) equals 1, not -2. However, without a specific equation, we cannot definitively say if (-1-1) is a solution.
If ( x = 0 ) and ( y = 1 ), then ( xy = 0 \times 1 = 0 ). Therefore, the value of ( xy ) is 0.
The solution is (x, y, z) = (2.5, 1, 2.5).
solution: y = 0 x = -1
It equals y+1
x+xy=8 xy=-x+8 y=-1+8/x
The expression (-1-1) simplifies to -2. To determine if (-2) is a solution to the equation xy, we need to know what the equation specifically is. If the equation is of the form xy = -2, then (-1, -1) is indeed a solution since (-1)(-1) equals 1, not -2. However, without a specific equation, we cannot definitively say if (-1-1) is a solution.
If ( x = 0 ) and ( y = 1 ), then ( xy = 0 \times 1 = 0 ). Therefore, the value of ( xy ) is 0.
The solution is: x = 1 and y = -1
xy + x + y + 1 = (x + 1)(y + 1).
That is the commutative property of equality.
xy = x ÷x y = 1
No.
In the equations Y=X-1 and Y=-X+1, the solution is (1,0)
The solution is (x, y, z) = (2.5, 1, 2.5).
It is (-1, 3).