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0
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4
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real numbers x and y satisfy

\(x+xy^2= 250y\)

\(x-xy^2= -240y\)

 

enter all possible values of x, separated by commas.

 Dec 24, 2019
 #1
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0

The solutions (x,y) are (0,0), (20,4), and (-20,-4).

 Dec 24, 2019
 #2
avatar+22096 
+1

Substituting thse values in shows that they do not all work....

   desmos graphically show     0,0    35, 7     -35, -7

 

https://www.desmos.com/calculator/tfuqvjljxr

ElectricPavlov  Dec 24, 2019
 #3
avatar
+1

Add the two equations and you get 2x = 10y, so x = 5y.

Substitute that into the top equation and you get

\(\displaystyle 5y + 5y^{3}=250y, \text{ from which,}\\5y^{3}-245y=0,\\5y(y^{2}-49)=0, \text{ so }y = 0\text{ or }y=\pm7,\\ \text{with corresponding x values}\\x=0\text{ or }x=\pm35.\)

.
 Dec 24, 2019
 #4
avatar+109563 
+1

Add the equations and we get that

 

2x  =  10y

 

x = 5y

 

So....we have that

 

5y + 5y^3  = 250y

 

y^3 + y  - 50y   =0

 

y^3 - 49y   = 0         factor

 

y ( y^2 - 49)   =0

 

y( y - 7) (y + 7)   = 0

 

Set  each factor to 0 and solve for   y  and we get that    y = 0    y = 7    and y = -7

 

So  x  = 0 , 35  and  -35

 

So

 

(x,y)  =  (0,0)    (7,35)     (-7 - 35)

 

 

cool cool cool

 Dec 24, 2019

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