The solution of the equations x+y+xy=11, `x^(2)y+xy^(2)=30` are ….

Answer

To solve the equations \( x + y + xy = 11 \) and \( x^2y + xy^2 = 30 \), we can follow these steps:

### Step 1: Rewrite the equations
The first equation can be rewritten as:
\[ xy + x + y = 11 \]
The second equation can be factored as:
\[ xy(x + y) = 30 \]

### Step 2: Introduce new variables
Let:
\[ u = x + y \quad \text{and} \quad v = xy \]
Then, we can rewrite the equations in terms of \( u \) and \( v \):
1. \( u + v = 11 \) (Equation 1)
2. \( uv = 30 \) (Equation 2)

### Step 3: Solve for \( v \) in terms of \( u \)
From Equation 1, we can express \( v \):
\[ v = 11 - u \]

### Step 4: Substitute \( v \) into Equation 2
Substituting \( v \) in Equation 2 gives:
\[ u(11 - u) = 30 \]
Expanding this:
\[ 11u - u^2 = 30 \]
Rearranging gives:
\[ u^2 - 11u + 30 = 0 \]

### Step 5: Factor the quadratic equation
To factor the quadratic equation, we look for two numbers that multiply to \( 30 \) and add to \( 11 \). The numbers \( 5 \) and \( 6 \) work:
\[ (u - 5)(u - 6) = 0 \]

### Step 6: Find the values of \( u \)
Setting each factor to zero gives:
\[ u - 5 = 0 \quad \Rightarrow \quad u = 5 \]
\[ u - 6 = 0 \quad \Rightarrow \quad u = 6 \]

### Step 7: Find corresponding values of \( v \)
Using \( u + v = 11 \):
1. If \( u = 5 \):
\[ v = 11 - 5 = 6 \]
2. If \( u = 6 \):
\[ v = 11 - 6 = 5 \]

### Step 8: Solve for \( x \) and \( y \)
Now we have two pairs of \( (u, v) \):
1. \( (5, 6) \)
2. \( (6, 5) \)

Using the relationship \( x + y = u \) and \( xy = v \), we can find \( x \) and \( y \) for each pair.

#### For \( (u, v) = (5, 6) \):
The equations are:
\[ x + y = 5 \]
\[ xy = 6 \]
This leads to the quadratic:
\[ t^2 - 5t + 6 = 0 \]
Factoring gives:
\[ (t - 2)(t - 3) = 0 \]
Thus, \( t = 2 \) or \( t = 3 \), so \( (x, y) = (2, 3) \) or \( (3, 2) \).

#### For \( (u, v) = (6, 5) \):
The equations are:
\[ x + y = 6 \]
\[ xy = 5 \]
This leads to the quadratic:
\[ t^2 - 6t + 5 = 0 \]
Factoring gives:
\[ (t - 1)(t - 5) = 0 \]
Thus, \( t = 1 \) or \( t = 5 \), so \( (x, y) = (1, 5) \) or \( (5, 1) \).

### Final Solutions
The solutions for the equations are:
1. \( (x, y) = (2, 3) \)
2. \( (x, y) = (3, 2) \)
3. \( (x, y) = (1, 5) \)
4. \( (x, y) = (5, 1) \)