The sum of two numbers is 8 and the difference of their squares is 32. Find the numbers.

Answer

To solve the problem step by step, we will use the information given in the question to form equations and then solve for the two numbers.

### Step 1: Define the Variables
Let the first number be \( x \) and the second number be \( y \).

### Step 2: Set Up the Equations
According to the problem, we have two pieces of information:
1. The sum of the two numbers is 8:
\[ x + y = 8 \quad \text{(Equation 1)} \]
2. The difference of their squares is 32:
\[ x^2 - y^2 = 32 \quad \text{(Equation 2)} \]

### Step 3: Use the Difference of Squares Formula
We can use the identity for the difference of squares:
\[ x^2 - y^2 = (x + y)(x - y) \]
Substituting Equation 1 into this identity gives:
\[ x^2 - y^2 = (8)(x - y) \]
Setting this equal to Equation 2:
\[ 8(x - y) = 32 \]

### Step 4: Solve for \( x - y \)
Now, we can solve for \( x - y \):
\[ x - y = \frac{32}{8} = 4 \quad \text{(Equation 3)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( x + y = 8 \) (Equation 1)
2. \( x - y = 4 \) (Equation 3)

We can add these two equations to eliminate \( y \):
\[ (x + y) + (x - y) = 8 + 4 \]
This simplifies to:
\[ 2x = 12 \]
Dividing both sides by 2 gives:
\[ x = 6 \]

### Step 6: Find the Value of \( y \)
Now, substitute \( x = 6 \) back into Equation 1 to find \( y \):
\[ 6 + y = 8 \]
Subtracting 6 from both sides gives:
\[ y = 2 \]

### Final Answer
Thus, the two numbers are:
\[ x = 6 \quad \text{and} \quad y = 2 \]