Two variables... One equation?

We have $(4x^{2}-12x+25)(y^{2}-4y+8) = 64$ Factoring out the 4 we get, $4(x^{2}-2 * \frac{3}{2} + \frac{9}{4} + 4)((y-2)^{2}+4) = 64$ $((x-3/2)^{2}+4)((y-2)^{2}+4) = 16$

But we have $(x-3/2)^{2}+4 \ge 4$ and $(y-2)^{2}+4 \ge 4$ .

So their product is $\ge 16$ .

So the only possibility is that they are equal to 4 that is $x = 3/2$ and $y = 2$ . Hence the sum is $1.5+2 = 3.5$

So if we take box the value is $\boxed{3}$ !

$(4{ x }^{ 2 }-12x+25)({ y }^{ 2 }-4y+8)=64\\ We\quad have\\ 4{ x }^{ 2 }-12x+25=(4{ x }^{ 2 }-12x+9)+16={ (2x-3) }^{ 2 }+{ 4 }^{ 2 }\\ { y }^{ 2 }-4y+8=({ y }^{ 2 }-4y+4)+4={ (y-2) }^{ 2 }+{ 2 }^{ 2 }\\ Let\quad be\quad a=(2x-3)\quad and\quad b=(y-2)\\ Then\quad we\quad have\\ ({ a }^{ 2 }+{ 4 }^{ 2 })({ b }^{ 2 }+{ 2 }^{ 2 })={ 8 }^{ 2 }\\ { (ab) }^{ 2 }+{ (2a) }^{ 2 }+{ (4b) }^{ 2 }+{ 8 }^{ 2 }={ 8 }^{ 2 }\\ { (ab) }^{ 2 }+{ (2a) }^{ 2 }+{ (4b) }^{ 2 }=0\quad \\ ab=0,\quad 2a=0\quad and\quad 4b=0\quad because\quad the\quad 3\quad terms\quad of\quad the\quad \\ LHS\quad are\quad squares\quad and\quad hence\quad can´t\quad be\quad negatives,\quad this\quad \\ means\quad a=0\quad and\quad b=0\\ 2x-3=0\Rightarrow x=3/2\\ y-2=0\Rightarrow y=2\\ These\quad are\quad the\quad unique\quad solutions\quad \\ Hence\quad (\sum _{ i=1 }^{ n }{ { x }_{ i }+{ y }_{ i } } )-0.5=[(3/2)+2]-(1/2)=3$

First we use: (4x^2-12x+25)=((2x-3)^2+16) and (y^2-4y+8)=((y-2)^2+4) we then get: ((2x-3)^2+16)((y-2)^2+4) = 64 since 16 * 4 = 64 and (2x-3)^2 and (y-2)^2 is only greater or equal to 0 where equality holds for x= 1,5 and y = 2

We then have 1,5 + 2 - 0,5 = 3