Question

(√32)^x÷2^y+1=1 and 8^y-16^4-x/2=0

Answer 1

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Answer 2

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$\frac{ {( \sqrt{32}) }^{x} }{ {2}^{y + 1} } = 1 \\ or. \: \frac{ {(2}^{ \frac{5}{2} } ) ^{x} }{ {2}^{y + 1} } = 1 \\ or. \: \frac{ {2}^{ \frac{5x}{2} } }{ {2}^{y + 1} } = 1 \\ or. \frac{5x}{2} = y + 1 \\ or. \: 5x = 2y + 2.....(1)$
${8}^{y} - {16}^{(4 - \frac{x}{2}) } = 0 \\ or. \: { {2}^{3} }^{y} = { {2}^{4} }^{(4 - \frac{x}{2}) } \\ or. \: {2}^{3y} = {2}^{(16 - 2x)} \\ or. \: 3y = 16 - 2x \\ or. \: y = \frac{16 - 2x}{3}$

Substituting the value of y in equation (1)

$5x = 2( \frac{16 - 2x}{3} ) + 2 \\ or. \: 15x = 32 - 4x + 6 \\ or. \: 19x = 38 \\ or. \: x = \frac{38}{19} = 2$
Putting the value of x we get..
$y = \frac{16 - 2 x}{3} \\ = \frac{16 - 2 \times 2}{3} = \frac{16 - 4}{3 } \\ = \frac{12}{3} = 4$
Thus x =2 and y = 4

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