If (x + 1)(y + 1) = 40 and xy(x + y) = 380, then find all possible values of x^2 + y^2.
If (x + 1)(y + 1) = 40 and xy(x + y) = 380, then find all possible values of x^2 + y^2.
I'm only assuming integer values for x , y
(x + 1) ( y + 1) = 40 xy ( x + y) = 380 (2)
xy + x + y + 1 = 40
xy + x + y = 39
x + y = 39 - xy (1)
Sub (1) into (2) and we have that
xy ( 39 - xy) = 380
-(xy)^2 + 39(xy) = 380
-(xy)^2 + 39(xy) - 380 = 0 multiply through by -1
(xy)^2 - 39xy + 380 = 0 factor
(xy - 20) ( xy - 19) = 0
Set both factors to 0 and we have that
xy = 20 xy = 19
The possible values for x, y are
x y
1 19
19 1
1 20
2 10
4 5
5 4
10 2
20 1
Because ( x + 1) ( y + 1) = 40
Then only
(1, 19) and (19, 1) will work
So
x^2 + y^2 = 362