If $2^{2 x - y} = 32$ and $2^{x + y} = 16$ , then $x^{2} + y^{2}$ is equal to

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Solution

The correct option is B 10
$2^{2 x - y} = 32 = 2^{5}, 2^{x + y} = 2^{4} 2 x - y = 5 a n d x + y = 4 y = 2 x - 5 x + 2 x - 5 = 4 3 x = 9, x = 3, y = 1. x^{2} + y^{2} = 3^{2} + 1^{2} = 9 + 1 = 10.$

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