x,y,z are positive real numbers such that x+y+z=9 and xy+9xz+25yz=9xyz. Find the sum of all possible values of xyz.
Try all x,y,z combinations that work. There will not be very many.
(x,y,z) 9xyz xy+9xz+25yz
(1,7,1) 63 >63
(1,1,7) 63 >63
(2,1,6) 108 >108
(2,6,1) 108 >108
(2,5,2) 180 >180
(2,2,5) 180 >180
(2,4,3) 216 >216
(2,3,4) 216 >216
(3,3,3) 243 225 + 81 + 9 > 243
(3,4,2) 216 200 + 54 + 12 > 216
(3,2,4) 216 200 + 108 + 6 > 216
(3,1,5) 235 125 + 135 + 9 > 235
(3,5,1) 235 125 + 27 + 15 < 235
(4,3,2) 216 150 + 72 + 12 > 216
(4,2,3) 216 150 + 108 + 8 > 216
(4,1,4) 144 100 + 144 + 4 > 144
(4,4,1) 144 100 + 36 + 16 > 144
(5,3,1) 135 75 + 45 + 15 = 135
(5,1,3) 135 75 + 135 + 6 > 135
(5,2,2) 180 100 + 90 + 10 > 180
(6,1,2) 108 50 + 108 + 6 > 108
(6,2,1) 108 50 + 54 + 12 > 108
(7,1,1) 63 25 + 63 + 7 > 63
The only solution is (x,y,z) = (5,3,1), for which xyz = 15
that is such a mission though lol, is that the only way?
Though the answer may be correct, the solution isn't ,because x,y,z are real numbers not integers
To find the sum of all possible values of xyz, we need to first solve the given system of equations.
Given:
x + y + z = 9 ...(Equation 1)
xy + 9xz + 25yz = 9xyz ...(Equation 2)
Let's try to simplify Equation 2 before proceeding further.
Expanding the terms, we have:
xy + 9xz + 25yz = 9xyz
=> xy + 9xz + 25yz - 9xyz = 0
=> xy + (9z - 9yz)x + 25yz = 0
=> x(y + (9z - 9yz)x) + 25yz = 0
=> xy + (9z - 9yz)x = -25yz
Now, we can isolate x from the equation above by dividing both sides by (y + (9z - 9yz)x):
x = -25yz / (y + (9z - 9yz)x) ...(Equation 3)
Now, let's substitute the value of x from Equation 3 into Equation 1:
-25yz / (y + (9z - 9yz)x) + y + z = 9
Multiplying the entire equation by (y + (9z - 9yz)x) to eliminate the denominator, we get:
-25yz + y(y + (9z - 9yz)x) + z(y + (9z - 9yz)x) = 9(y + (9z - 9yz)x)
Simplifying further:
-25yz + y^2 + y(9z - 9yz)x + z^2 + z(9z - 9yz)x = 9y + 81z - 81yz + 9z^2 - 9yz^2)x
Combining like terms:
y^2 + z^2 - 34yz + (9z - 9yz)x(y + z) = 9y + 81z + 9z^2 - 9yz^2)x ...(Equation 4)
Now, since x, y, and z are positive real numbers, the term (y + z) must be greater than zero. So, for Equation 4 to hold true, the coefficients of x and y on both sides must be equal. This leads us to the following equations:
9z - 9yz = 0 ...(Equation 5)
-34yz = 9y ...(Equation 6)
Solving Equation 5 for y:
9z - 9yz = 0
=> 9z = 9yz
=> 1 = y
Substituting y = 1 into Equation 6:
-34z = 9
=> z = -9/34
Therefore, for the given system of equations, there is only one unique solution:
x = 1, y = 1, z = -9/34.
Now, let's find the value of xyz using this solution:
xyz = 1 * 1 * (-9/34) = -9/34
So, the sum of all possible values of xyz is:
-9/34.