Question

pls answer this question fast
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Answer 1

Answer:

solving first

3^x-y=9

3^x-y= 3^2

equal base

x-y=2......

x=2+y.....eqn i

solving second

3^x+y=81

3^x+y=3^4

equal base

x+y=4...eqn ii

then, put the value of x in eqn ii

2+y+y=4

2y=2

y=1

put the value of y in eqn i

x=2+1

x=3

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Answer 2

Answer:

x= 3, y=1

Step-by-step explanation:

$3^x^-^y = 9\\3^x^-^y= 3^2$

$3^x^+^y= 81\\3^x^+^y= 3^4$

Now both the bases are same so

$x-y= 2$---------------1

$x+y= 4$---------------2

Subtracting 1 from 2, we get

$x+y-x+y= 2\\2y= 2\\y= 1$

Adding 1 and 2, we get

$x+y+x-y= 6\\2x= 6\\x= 3$

HOPE IT HELPS

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