If `2^(2x-y)=32` and `2^(x+y)=16` then `x^2+y^2` is equal to

Options

A. 9
B. 10
C. 11
D. 13

Answer

To solve the equations \(2^{2x - y} = 32\) and \(2^{x + y} = 16\), we can follow these steps:

### Step 1: Rewrite the equations in terms of powers of 2
We know that:
- \(32 = 2^5\)
- \(16 = 2^4\)

Thus, we can rewrite our equations as:
1. \(2^{2x - y} = 2^5\)
2. \(2^{x + y} = 2^4\)

### Step 2: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
1. \(2x - y = 5\) (Equation 1)
2. \(x + y = 4\) (Equation 2)

### Step 3: Solve the system of equations
Now we have a system of linear equations:
1. \(2x - y = 5\)
2. \(x + y = 4\)

We can solve for \(y\) in terms of \(x\) using Equation 2:
\[ y = 4 - x \]

### Step 4: Substitute \(y\) into Equation 1
Now substitute \(y\) into Equation 1:
\[ 2x - (4 - x) = 5 \]
This simplifies to:
\[ 2x - 4 + x = 5 \]
Combining like terms gives:
\[ 3x - 4 = 5 \]

### Step 5: Solve for \(x\)
Add 4 to both sides:
\[ 3x = 9 \]
Now divide by 3:
\[ x = 3 \]

### Step 6: Find \(y\)
Now that we have \(x\), we can find \(y\) using \(y = 4 - x\):
\[ y = 4 - 3 = 1 \]

### Step 7: Calculate \(x^2 + y^2\)
Now we can find \(x^2 + y^2\):
\[ x^2 + y^2 = 3^2 + 1^2 = 9 + 1 = 10 \]

### Final Answer
Thus, \(x^2 + y^2 = 10\).

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