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Find all ordered pairs x, y of real numbers such that x+y=10 and x^3+y^3=300.

For example, to enter the solutions (2, 4) and (-3, 9), you would enter "(2,4),(-3,9)" (without the quotation marks).

 Jun 25, 2024
 #1
avatar+129733 
+1

x + y  = 10   square both sides

x^2 + y^2 + 2xy = 100

x^2 + y^2  = 100 - 2xy

 

x^3 + y^3 =  300

(x + y) ( x^2 + y^2 - xy)  = 300

(10) ( 100 - 3xy) = 300

100 - 3xy = 30

100 - 30 =3xy

70 / 3  = xy

y = 70 / (3x)

 

x + y  =10

x + 70 / (3x) = 10     multiply through by x

x^2 + (70/3) = 10x

x^2 -10x + 70/3 = 0

3x^2 - 30x + 70  = 0

 

x =  [30 +/- sqrt [ 900 - 840 ] ] / 6  =  [ 30 +/- sqrt 60 ] / 6  =  5 + sqrt (15) / 3  or 5 - sqrt (15) / 3

 

(x,y)  = ( 5 + sqrt (15) / 3 , 5 -sqrt (15) / 3 )   or (5 - sqrt (15) / 3  , 5 + sqrt (15) / 3 )

 

cool cool cool

 Jun 25, 2024

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