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let z=2^x and w=2^y
the first equation becomes z+w=10 with that substitution
now, replace 4 with 2^2, so the second equation becomes
now, 2^2x is that same as 2^x2, which is also the same as (2^x)^2...using a similar argument, for y 4^y is really (2^y)^2
but we called 2^x z and 2^y we called w, so we substitute in and get
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now using the first equation z+w=10, solve for z and then substitute into the second equation.
and the second equation is then (10-w)^2 +w^2=68
expand the (10-w)^2 with foil to get 100-20w+w^2+w^2=68 and then group like terms and move everything over
dividing out by 2, w^2-10w+16=0
w=2 and w=8
now, plug back in for w=2^y
2^y=2 and 2^y=8...so y=1 and y=3
plugging back in for y=1 gives x=3 and y=3 gives x=1
so the 2 solutions are (1,3) and (3,1)