Answer:
To find the value of zw, we need to solve for x, y, and z first using the given equations.
From the equation xy = zy, we can divide both sides by y (assuming y ≠ 0) to get x = z.
Substituting x = z into the equation xz = 10x + 16, we have z^2 = 10z + 16.
Rearranging this quadratic equation, we get z^2 - 10z - 16 = 0.
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
z = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -10, and c = -16. Plugging these values into the formula, we have:
z = (-(-10) ± √((-10)^2 - 41(-16))) / (2*1)
= (10 ± √(100 + 64)) / 2
= (10 ± √164) / 2
= (10 ± 2√41) / 2
= 5 ± √41
So, z can be either 5 + √41 or 5 - √41.
Since x = z, we have two possible values for x as well:
x = 5 + √41 or x = 5 - √41.
Now, let's solve for y using the equation xy = zy.
For x = 5 + √41:
(5 + √41)y = (5 + √41)(5 + √41)
= 25 + 10√41 + 10√41 + 41
= 66 + 20√41
Dividing both sides by (5 + √41), we get:
y = (66 + 20√41) / (5 + √41)
For x = 5 - √41:
(5 - √41)y = (5 - √41)(5 - √41)
= 25 - 10√41 - 10√41 + 41
= 66 - 20√41
Dividing both sides by (5 - √41), we get:
y = (66 - 20√41) / (5 - √41)
Now that we have the values of x and y, we can find zw using the equation wx = 6x + 1.
For x = 5 + √41:
w(5 + √41) = 6(5 + √41) + 1
= 30 + 6√41 + 1
= 31 + 6√41
Therefore, zw = (31 + 6√41)(5 + √41).
For x = 5 - √41:
w(5 - √41) = 6(5 - √41) + 1
= 30 - 6√41 + 1
= 31 - 6√41
Therefore, zw = (31 - 6√41)(5 - √41).
So, depending on the values of x and y, the value of zw can be either (31 + 6√41)(5 + √41) or (31 - 6√41)(5 - √41).
Explanation: