Question

Given that (2*(x -3) ) / 8*( 2 y - 3) = 16*( x - y )​

Answer 1

Answer:

Step-by-step explanation:

$\frac{2^{x-3}}{8^{2y-3}}=16^{x-y}\\\\\frac{2^{x-3}}{(2^{3})^{2y-3}}=(2^{4})^{x-y}\\\\\frac{2^{x-3}}{2^{3*(2y-3)}}=2^{4*(x-y)}\\\\\frac{2^{x-3}}{2^{6y-9}}=2^{4x-4y}\\\\2^{x-3 -(6y-9)}=2^{4x-4y}\\\\2^{x-3-6y+9}=2^{4x-4y}\\\\2^{x-6y+6}=2^{4x-4y}$

base is same, so equate the powers

x - 6y + 6 = 4x - 4y

Subtract 'x' from both sides

-6y + 6 = 4x -x - 4y

-6y + 6 = 3x - 4y

Add '6y' to both sides

6 = 3x - 4y + 6y

6 = 3x + 2y

Hence proved

Answer 2

Answer:

2 ^ (x - 3) ÷ 2 ^ 3(2y - 3)=2 ^ 4(x - y)

The two are common so we give it out then

x - 3 ÷ 6y - 9 =4x - 4y

x - 3 -4x - 4y = 6y -9

-3x -2y = -6

6=3x+2y