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Step-by-Step Solution
Step 1: Write down the system of linear equations
The given system is:
x - 2y + 5z = 0 \quad\quad (1)
-2x + 4y + z = 0 \quad (2)
-7x + 14y + 9z = 0 \quad (3)
Step 2: Combine equations to eliminate variables
We first combine equation (1) and a multiple of (1) with (2) to find a simpler relation:
Multiply equation (1) by 2: 2x - 4y + 10z = 0 .
Add this to equation (2): (-2x + 4y + z) + (2x - 4y + 10z) = 0 + 0 .
On simplifying, we get:
0x + 0y + 11z = 0 \;\Rightarrow\; z = 0.
Step 3: Substitute z = 0 back into the equations
With z = 0 , equation (1) becomes:
x - 2y + 5(0) = 0 \;\Rightarrow\; x - 2y = 0 \;\Rightarrow\; x = 2y.
Thus the solutions have the form:
(x, y, z) = (2y, y, 0).
Step 4: Apply the condition on x^2 + y^2 + z^2
We are given:
15 \le x^2 + y^2 + z^2 \le 150.
Substituting x = 2y and z = 0 , we get:
x^2 + y^2 + z^2 = (2y)^2 + y^2 + 0^2 = 4y^2 + y^2 = 5y^2.
Hence the condition becomes:
15 \le 5y^2 \le 150 \;\Rightarrow\; 3 \le y^2 \le 30.
Step 5: Identify integer values of y satisfying 3 \le y^2 \le 30
The perfect squares in the range [3, 30] are 4, 9, 16, 25. Thus
y^2 \in \{4,\;9,\;16,\;25\} \quad \Longrightarrow \quad y \in \{\pm 2,\;\pm 3,\;\pm 4,\;\pm 5\}.
This gives 8 possible integer values for y .
Step 6: Determine corresponding x and z
For each integer y , x = 2y and z = 0. Hence each y gives a unique triplet (x,y,z) . Thus we have:
If y = 2 , then x = 4, z = 0
If y = -2 , then x = -4, z = 0
If y = 3 , then x = 6, z = 0
If y = -3 , then x = -6, z = 0
If y = 4 , then x = 8, z = 0
If y = -4 , then x = -8, z = 0
If y = 5 , then x = 10, z = 0
If y = -5 , then x = -10, z = 0
There are total 8 such solutions.
Step 7: Conclude the number of elements in set S
The number of integer triplets (x,y,z) satisfying the given system and the condition on x^2 + y^2 + z^2 is
8.
