Solve the system of equations for positive real numbers:
1/xy = x/z + 1, 1/yz = y/x + 1, 1/zx = z/y + 1.
By symmetry, we conclude at least one solution is x=y=z.
Assume p=x=y=z, then
1/p^2=p/p+1=2
Can you solve for p?
how does 2 came
1/p^2=p/p+1=2
p/p=1
To solve this system of equations, we can start by isolating variables in each equation.
Equation 1:
1/xy = x/z + 1
We can rearrange this equation by multiplying both sides by (xy)z to eliminate the denominators:
z = x^2y + xyz
Equation 2:
1/yz = y/x + 1
By multiplying both sides by (yz)x, we eliminate the denominators:
x = y^2z + xyz
Equation 3:
1/zx = z/y + 1
By multiplying both sides by (xz)y, we eliminate the denominators:
y = z^2x + xyz
Now that we have rearranged all three equations, we can set them equal to each other in pairs to eliminate one variable at a time.
Equations 1 and 2:
z = x^2y + xyz
x = y^2z + xyz
Since both equations have xyz on the right-hand side, we can set them equal to each other:
x^2y + xyz = y^2z + xyz
Canceling out the common term xyz on both sides:
x^2y = y^2z
Dividing both sides by yz, we get:
x^2/yz = y/z
Now, we can use this result and equation 3 to eliminate another variable.
x^2/yz = y/z
y = z^2x + xyz
Substituting y/z with x^2/yz:
x^2/yz = z^2x + xyz
Multiplying both sides by yz to eliminate the denominators:
x^2 = z^2xyz + xyz^2
Rearranging this equation, we get:
x^2 - z^2xyz = xyz^2
Factoring out xyz on the right-hand side:
x^2 = xyz(z + z^2)
Finally, dividing both sides by z + z^2:
x^2/(z + z^2) = xyz
We have now eliminated the variable y. To solve for the remaining variables x and z, we need one more equation. Let's go back to equations 1 and 2:
z = x^2y + xyz
x = y^2z + xyz
Substituting equations 1 and 2 into each other:
z = (y^2z + xyz)^2y + xyz
x = ((x^2y + xyz)^2)z + xyz
At this point, we have two equations involving the remaining variables x and z. We can use numerical methods such as iterations or numerical solvers to approximate the values of x and z.