Question

Coefficient of 16 x² y ² in - 16 x² y^4 z​

Answer 1

Answer:

16^(x²+y) + 16^(y²+x) = 1 …….(1)

Equation (1) is only and only possible when 16^(x²+y) = 1/2 and 16^(y²+x) = 1/2

Then 1/2 + 1/2 = 1

Take 1st part…

16^(x²+y) = 1/2

[{2}⁴]^(x²+y) = 2^(-1)

2^{4(x²+y)} = 2^(-1)

Comparing power

Therefore, 4(x²+y) = -1

x²+y = -1/4

x² + y + 1/4 = 0 ………… (2)

Take 2nd part

16^(y²+x) = 1/2

[{2}⁴]^(y²+x) = 2^(-1)

2^{4(y²+x)} = 2^(-1)

Comparing power

Therefore, 4(y²+x) = -1

y²+x = -1/4

y² + x + 1/4 = 0 …………(3)

Adding equation (2) and (3)

We have,

x² + y + 1/4 + y² + x + 1/4 = 0

(x² + x + 1/4) + (y² + y + 1/4) = 0

(x + 1/2 )² + (y + 1/2)² = 0

This equation is possible if only if both terms are equal to zero.

therefore,

x + 1/2 = 0 => x = -1/2

y + 1/2 = 0 => y = -1/2

Hence, the solution of given equation is x = -1/2 & y = - 1/2.

Answer 2

Answer:

solution of given equation is x = -1/2 & y = - 1/2.