+1467 Find one ordered pair $(x,y)$ of real numbers such that $x + y = 10$ and $x^3 + y^3 = 162 + x^2 + y^2.$
+1790 First of all, let's focus on the second equation.
We can factor the right side of the equation to get \((x+y)(x^2+xy+y^2)=162+x^2+y^2\)
Since x+y is equal to 10 from the first equation, the equation becomes
\(10(x^2+xy+y^2)=162+x^2+y^2\)
\(10x^2+10xy+10y^2= 162+x^2+y^2\)
\(9x^2+9y^2+10xy=162\)
Now, isolating x in the first equation, we get
\(x=10-y\)
Now plugging this value of x into the second equation, we get
\(9(10-y)^2+9y^2+10(10-y)y=162\\ 9(100-20y+y^2)+9y^2+100y-10y^2=162\\900-180y+9y^2+9y^2+100y-10y^2=162\\ 8y^2-80y+900=162\\8y^2-80y+738=0\)
However, plugging this into the quadratic formula, notice that unofruntately, the descriminant is less than 0, menaing the eventual solutions are NONREAL numbers.
Thanks! :)
