+0

# help meeee!

0
128
4

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
0

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

Dec 24, 2019
#2
+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
+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
+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)

Dec 24, 2019