If `x^(2) + y^(2) = 16 ` and `x + y = 6` , then find the value of `xy` .

Options

A. 10
B. 12
C. 16
D. 18

Answer

To solve the problem, we have the following equations:

1. \( x^2 + y^2 = 16 \) (Equation 1)
2. \( x + y = 6 \) (Equation 2)

We need to find the value of \( xy \).

### Step 1: Use the identity for the square of a sum
We know that:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]

### Step 2: Substitute the known values
From Equation 2, we have \( x + y = 6 \). Therefore:
\[ (x + y)^2 = 6^2 = 36 \]

### Step 3: Substitute into the identity
Now we can substitute this into the identity:
\[ 36 = x^2 + y^2 + 2xy \]

### Step 4: Substitute \( x^2 + y^2 \) from Equation 1
From Equation 1, we know \( x^2 + y^2 = 16 \). So we substitute this into the equation:
\[ 36 = 16 + 2xy \]

### Step 5: Solve for \( 2xy \)
Now we can rearrange the equation to isolate \( 2xy \):
\[ 2xy = 36 - 16 \]
\[ 2xy = 20 \]

### Step 6: Solve for \( xy \)
Now, divide both sides by 2 to find \( xy \):
\[ xy = \frac{20}{2} = 10 \]

### Final Answer
Thus, the value of \( xy \) is \( 10 \).

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