Question

I ordered pair (x,y) satisfying the system of equations 4^x÷y+y÷x=32 and log3 (x - y) = 1 - log3 (x + y).
(C) x-y-1
(D) x + 2y - 4
(B) xy - 2
(Ax+y-3​

Answer 1

Given :   4^(x÷y + y÷x) =32   , log₃ (x - y) = 1 - log₃(x + y).

To find : Ordered pair Satisfying Given Equation

Solution:

log₃ (x - y) = 1 - log₃(x + y).

=>  log₃ (x - y) + log₃(x + y). = 1

=>  log₃  ( (x-y)(x + y)) = 1

=> 3¹ = (x-y)(x + y)

=> (x-y)(x + y)   = 3

=> x² - y ²  = 3

32 =  2⁵

4^(x÷y + y÷x)  = (2²)^(x÷y + y÷x)  = 2^(2(x/y + y/x))

=> 2 ( x/y  + y/x ) = 5

=> x/y  + y/x  = 5/2

=> x² + y ²  = 5xy/2

(x + y)²  - 2xy =  5xy/2

=> (x + y)² = 9xy/2

(x - y)²  + 2xy =  5xy/2

=> (x - y)² =  xy/2  

=> (x + y)² (x - y)²  = (9xy/2)(xy/2)

=> (x² - y²)²  = 9x²y²/2²

=> x² - y² = 3xy/2

=> 2x² = 4xy  => x = 2y

x² - y ²  = 3  

= (2y)² - y² = 3

=> 3y² = 3

=> y² = 1

=> y = 1

& x = 2

x = 2  ,  y = 1  

x+y-3​   => 2 + 1 - 3 = 0   Satisfied

xy - 2  => 2*1 - 2 =  0  Satisfied

x-y-1 = 2 - 1 - 1 = 0 Satisfied

x + 2y - 4 = 2 + 2 - 4 = 0   Satisfied

x = 2  ,  y = 1    

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Answer 2

Answer:

Step-by-step explanation: