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Real numbers x  and y satisfy
\(\begin{align*} x + xy^2 &= 250y, \\ x - xy^2 &= -240y. \end{align*}\)
Enter all possible values of x separated by commas.
(ive came up with 35 and -35, but it isnt correct)

 Jul 1, 2021
edited by Guest  Jul 1, 2021
edited by Guest  Jul 1, 2021
 #1
avatar+120966 
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You're close.....just one more solution

 

x + xy^2   =  250y

x - xy^2  =    -240 y        add these

 

 

2x  =  10y

x  =  (10/2) y

x = 5y            sub into the  first  equation

 

5y  +  (5y)y^2   =  250y

5y^3  =  250y   - 5y

5y^3  = 245y            {   here's where you  might  have dropped a solution by dividing out y }

5y^3  - 245y =  0

y^3  - 49y  =  0

y (y^2 - 49)  =  0

y  (y + 7)(y - 7)  =  0

 

Setting all three factors to  0 and solving for y

 

y =  0  ⇒   x =  5(0) =  0

 

y = -7 , x = -35

 

y = 7 ,  x =  35

 

cool cool cool

 Jul 1, 2021
edited by CPhill  Jul 1, 2021
edited by CPhill  Jul 1, 2021
 #2
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thank you!!!

Guest Jul 1, 2021

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