Question

Please answer! this is an important exam that im taking right now.

Answer 1

Answer:

Required ordered pair is (0,0) for system of equation

Step-by-step explanation:

The given system of equation is

A). $9^{12x} = 9^{3y}$

B). $y^{4} - 9y^{2} x^{2} =16y$

On simplifying the equation A

$9^{12x} = 9^{3y}$

Take log on both side,

(12x) (log9) = (3y) (log9)

4x=y

To find the solution of the system of the equation :

Replacing value of y=4x in equation B,

$y^{4} - 9y^{2} x^{2} =16y$

$(4x)^{4} - 9(4x)^{2} x^{2} =16(4x)$

$256x^{4} - 144x^{2} x^{2} =64x$

$256x^{4} - 144x^{4}=64x$

$112x^{4}=64x$

We get,

x=0 and x=0.829

Since, 0.829 is not integer number

Only solution of equations is x=0

For the value of y

Replace value x in y=4x=0

Thus, Required ordered pair is (0,0)