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If (x + 1)(y + 1) = 40 and xy(x + y) = 380, then find all possible values of x^2 + y^2.

 Nov 11, 2019
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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

 

 

 

cool cool cool

 Nov 11, 2019

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