The solution of the equations `x^(2)+xy+xz=18, y^(2)+yz+yx+12=0 and z^(2)+zx+zy=30` are (3,-2,5) and (-3,2,-5)

Answer

To verify the solutions of the equations \(x^2 + xy + xz = 18\), \(y^2 + yz + yx + 12 = 0\), and \(z^2 + zx + zy = 30\) for the pairs \((3, -2, 5)\) and \((-3, 2, -5)\), we will substitute these values into each equation and check if they hold true.

### Step 1: Substitute \((3, -2, 5)\) into the equations

1. **Equation 1:**
\[ x^2 + xy + xz = 18 \]
Substituting \(x = 3\), \(y = -2\), \(z = 5\):
\[ 3^2 + (3)(-2) + (3)(5) = 9 - 6 + 15 = 18 \]
This holds true.

2. **Equation 2:**
\[ y^2 + yz + yx + 12 = 0 \]
Substituting \(y = -2\), \(z = 5\), \(x = 3\):
\[ (-2)^2 + (-2)(5) + (-2)(3) + 12 = 4 - 10 - 6 + 12 = 0 \]
This holds true.

3. **Equation 3:**
\[ z^2 + zx + zy = 30 \]
Substituting \(z = 5\), \(x = 3\), \(y = -2\):
\[ 5^2 + (5)(3) + (5)(-2) = 25 + 15 - 10 = 30 \]
This holds true.

### Step 2: Substitute \((-3, 2, -5)\) into the equations

1. **Equation 1:**
\[ x^2 + xy + xz = 18 \]
Substituting \(x = -3\), \(y = 2\), \(z = -5\):
\[ (-3)^2 + (-3)(2) + (-3)(-5) = 9 - 6 + 15 = 18 \]
This holds true.

2. **Equation 2:**
\[ y^2 + yz + yx + 12 = 0 \]
Substituting \(y = 2\), \(z = -5\), \(x = -3\):
\[ (2)^2 + (2)(-5) + (2)(-3) + 12 = 4 - 10 - 6 + 12 = 0 \]
This holds true.

3. **Equation 3:**
\[ z^2 + zx + zy = 30 \]
Substituting \(z = -5\), \(x = -3\), \(y = 2\):
\[ (-5)^2 + (-5)(-3) + (-5)(2) = 25 + 15 - 10 = 30 \]
This holds true.

### Conclusion
Both sets of values \((3, -2, 5)\) and \((-3, 2, -5)\) satisfy all three equations. Therefore, the statement is true.

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