Question

if 2^x-y=32 and 2^x+y =16 then the value of x^2+y^2​

Answer 1

Step-by-step explanation:

Lets try this out

2^x-y=32

2^x+y=16

To find x²+y²

By substitution method

2^x=32+y

now substituting 2^x in 2nd

So

32+y+y=16

2y=-16

y=-8

so now in 1st

if we put the value of y

2^x-(-8)=32

2^x=32-8

2^x=24

2^x=2*2*2*3

2^x=2³*3

Ok so

May be the Value of x=be 3 because of 2³=2^x

so

x²+y²=3²+(-8)²

=9+64

=73

Hope it helps

Hope it is correct

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