If `x + y = 10 and x ^(2) + y ^(2) = 68,` then find xy

Options

A. A)21
B. B)24
C. C)25
D. D)16

Answer

To solve the problem, we start with the given equations:

1. \( x + y = 10 \)
2. \( x^2 + y^2 = 68 \)

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

### Step 1: Use the identity for the square of a sum
We know that:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]
Substituting the values we have:
\[ 10^2 = x^2 + y^2 + 2xy \]
Calculating \( 10^2 \):
\[ 100 = x^2 + y^2 + 2xy \]

### Step 2: Substitute the value of \( x^2 + y^2 \)
From the second equation, we know that \( x^2 + y^2 = 68 \). Now, we substitute this into the equation:
\[ 100 = 68 + 2xy \]

### Step 3: Solve for \( 2xy \)
Rearranging the equation gives:
\[ 2xy = 100 - 68 \]
Calculating the right side:
\[ 2xy = 32 \]

### Step 4: Solve for \( xy \)
Now, divide both sides by 2 to find \( xy \):
\[ xy = \frac{32}{2} = 16 \]

Thus, the value of \( xy \) is \( 16 \).

### Final Answer:
\[ \text{The value of } xy \text{ is } 16. \]

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