`x^(2)-xy+y^(2)=7, x^(4)+x^(2)y^(2)+y^(4)=133` then (x,y)=

Options

A. `(2,3)`
B. `(-2,-3)`
C. `(3,2)`
D. `(-3,-2)`

Answer

To solve the equations \( x^2 - xy + y^2 = 7 \) and \( x^4 + x^2y^2 + y^4 = 133 \), we can follow these steps:

### Step 1: Factor the second equation
We start with the second equation:
\[ x^4 + x^2y^2 + y^4 = 133 \]
This can be factored as:
\[ (x^2 - xy + y^2)(x^2 + xy + y^2) = 133 \]

### Step 2: Substitute the first equation into the factorization
From the first equation, we know:
\[ x^2 - xy + y^2 = 7 \]
Substituting this into the factorization gives:
\[ 7(x^2 + xy + y^2) = 133 \]

### Step 3: Solve for \( x^2 + xy + y^2 \)
Dividing both sides by 7:
\[ x^2 + xy + y^2 = \frac{133}{7} = 19 \]

### Step 4: Set up a system of equations
Now we have two equations:
1. \( x^2 - xy + y^2 = 7 \)
2. \( x^2 + xy + y^2 = 19 \)

### Step 5: Add the two equations
Adding both equations:
\[ (x^2 - xy + y^2) + (x^2 + xy + y^2) = 7 + 19 \]
This simplifies to:
\[ 2x^2 + 2y^2 = 26 \]
Dividing by 2:
\[ x^2 + y^2 = 13 \]

### Step 6: Subtract the first equation from the second
Now, subtract the first equation from the second:
\[ (x^2 + xy + y^2) - (x^2 - xy + y^2) = 19 - 7 \]
This simplifies to:
\[ 2xy = 12 \]
Dividing by 2:
\[ xy = 6 \]

### Step 7: Solve for \( x^2 \) and \( y^2 \)
Now we have:
1. \( x^2 + y^2 = 13 \)
2. \( xy = 6 \)

Let \( x^2 = a \) and \( y^2 = b \). Then we have:
\[ a + b = 13 \quad \text{(1)} \]
\[ ab = 36 \quad \text{(since } (xy)^2 = 6^2 = 36\text{)} \]

### Step 8: Form a quadratic equation
Using the equations \( a + b = 13 \) and \( ab = 36 \), we can form the quadratic:
\[ t^2 - (a+b)t + ab = 0 \]
Substituting the values:
\[ t^2 - 13t + 36 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{13 \pm \sqrt{13^2 - 4 \cdot 36}}{2} \]
Calculating the discriminant:
\[ 13^2 - 144 = 169 - 144 = 25 \]
Thus:
\[ t = \frac{13 \pm 5}{2} \]
Calculating the roots:
\[ t_1 = \frac{18}{2} = 9, \quad t_2 = \frac{8}{2} = 4 \]

### Step 10: Find \( x \) and \( y \)
So, we have \( x^2 = 9 \) and \( y^2 = 4 \) (or vice versa):
\[ x = 3, y = 2 \quad \text{or} \quad x = -3, y = -2 \]
\[ x = 2, y = 3 \quad \text{or} \quad x = -2, y = -3 \]

### Final Solutions
Thus, the pairs \( (x, y) \) are:
1. \( (3, 2) \)
2. \( (-3, -2) \)
3. \( (2, 3) \)
4. \( (-2, -3) \)