Basic question at algebra

$x+y=3\rightarrow\;x=3-y$ Replacing this value in the second equation,we get: $(3-y)(y)=2$ $3y-y^2=2$ $y^2-3y=2$ $y^2-3y-2=0$ $(y-2)(y-1)=0\rightarrow\;y=2\;or\;y=1$ If we put the possible values of $y$ in $x+y=3$ we get $x=1\;or\;x=2$ It is obvious that $x^5+y^5$ will give the same answer irrespective if the values of $x,y$ So $x^5+y^5=1^5+2^5=2^5+1^5=\boxed{33}$

please pay attention x+y=3 xy=2 x=2 or 1 and y=1 or 2 assuming x to be 2 and y to be 1 x to power 5=32 and y to power 5=1 therefore the answer is 33

The equation which will have roots x and y is a^2 - 3a + 2 = 0

therefore x and y will gain values 2 and 1.

2^5 + 1^5 = 33

Let x =1 and y=2. x+y=3 . xy=2. x= 1 and y= 32 the answer is 33

x+y=3,thus consider x and y are two integers,3/2=1.5 means that 1<1.5<2 mean 1~1.5~2,x and y possibly be 2 or 1, just consider that x as 2 then and y as 1 so x^5+y^5=2^5+1^5=32+1=33