how to solve the equation

I tried [over] 4 days
Good -- you have some work to show us. Please post one of your attempts. I'm assuming you want the Real solution only.

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x^2 + y = 7

x + y^2 = 11
 
Since \(\displaystyle x^2+ y= 7\), \(\displaystyle y= 7- x^2\). Then \(\displaystyle y^2+ x= (7- x^2)^2+ x= x^4- 14x^2+ 49+ x= x^4- 14x^2+ x+ 49= 11\). \(\displaystyle x^4- 14x^2+ x+ 38= 0\).

Solve that "quartic" equation". (Hint: there is a small integer solution.)
 
System of Equations

x2+y = 7
x+y2 = 11

Using substitution,

I will use the first equation and solve for y

y=-x2+7 (first equation)
x+y2=11 (second equation)


In the second equation, y gets replaced with -x2+7

x+(-x2+7)2=11 (-x2+7)2=(-x2+7)*(-x2+7) FOIL
x+x4-7x2-7x2+49=11
x4-14x2+x+49-11=0
x4-14x2+x+38=0 factor out (x-2) using long division

I can't show polynomial long division, but if you need an example:
http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/long_division/long_division.html

(x-2)(x3+2x2-10x-19)=0
Real solution x=2

plus into first equation, y=-x2+7
y=-(22)+7=-4+7=3
Real solution y=3

Answer= (2,3)
 
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