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# help

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Solve the system x + y = 9/2, 2x^2 y^2 - 13xy + 18 = 0.

Dec 6, 2019

#1
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x + y = 9/2, 2x^2 y^2 - 13xy + 18 = 0

2x^2y^2  -  13xy  +  18   =    0    factor as

(2xy  -  9 ) (xy - 2)   =  0

Settting each factor to 0  and solving we have

2xy - 9  = 0                xy  - 2  = 0

xy  = 9/2                    xy  =  2

y = / [ 2x]                y  = 2/x

And using  x + y  = 9/2   we can solve this

x +  / [2x]  = 9/2       multiply through by 2x

2x^2  + 9 =  (9/2)(2x)    rearrange

2x^2  - 9x + 9   = 0

(2x - 3) ( x - 3)  = 0

So

2x - 3  =0        x - 3  = 0

x = 3/2              x   =  3

And y  = 9 ] [ 2(3/2)]  = 3      And    y  = 9/ {2*3] = 3/2

So....two solutions are    ( 3/2, 3)   and  ( 3 , 3/2)

And for the other solutions, we have that

x + 2/x  = 9/2       multiply through by x

x^2 + 2 = (9/2)x

2x^2 - 9x  + 4  = 0

(2x - 1) ( x - 4)  = 0

2x - 1  = 0              x  - 4  =0

x =1/2                   x  = 4

And y  = 2/ [1/2]   =  4        And  y = 2 / 4   = 1/2

So....the other two solutions are   (1/2, 4)   and ( 4, 1/2)  ​ Dec 6, 2019