Algebra

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! :)