how to solve the equation

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

Please familiarize yourself with the forum guidelines. We look forward to seeing your efforts. :cool:


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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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