Question 135303
Solve the system of equations: 
:
9^(3x-y)=(1/81)^-1 
and
(sqrt5^(2x-y))=1
:
On the 1st equation, ^-1 is the reciprocal so we have:
 9^(3x-y)= 81
:
On the 2nd equation we can get rid of the radical by squaring both sides
5^(2x-y)) = 1
:
so we have:
{{{9^(3x-y) = 81}}}
and
{{{5^(2x-y) = 1}}}
:
In the 1st equation we ask, "What exponent of 9 = 81" Obviously its 9^2 = 81
Therefore:
3x - y = 2
:
On the second equation, What exponent of 5 = 1. We know that an exponent of 0 makes any number = 1
 5^0 = 1; Therefore 2x - y = 0
:
Solve these two equation using elimination
3x - y = 2
2x - y = 0
-----------Subtracting eliminates y
x = 2
:
3(2) - y = 2
-y = 2 - 6
-y = -4
y = 4
:
You can check this using these values in the original equation