Let $x,y \in R$ such that $|x+y| + |x-y| = 2$ 
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.

How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.

Well, in the worst case when we are stuck we can do cases:
Case 1: $x+y\ge 0$ and $x-y \ge 0$ then
$x\ge -y$ and $x \ge y$ so $x \ge \max(y,-y) = |y|$.
And $|x+y| + |x-y| = (x+y)+(x-y) = 2x = 2$ so $x = 1$ and $|y| \le x=1$ so $-1 \le y \le 1$.
$x^2 - 6x +y^2 = -5+y^2$ and the max that $y^2$ can be is $1$ so the max of $x^2 -6x +y^2$ is $-4$ when $y = \pm 1$ if was assume $x+y\ge 0$ and $x-y \ge 0$.
(Likewise the max of $x^2 + y^2 + 10y = 1+y^2 + 10y$ is $12$ if $y=1$.)
Case 2:
$x +y < 0$ and $x-y \ge 0$ then
$x < -y$ and $x \ge y$ so $y \le x < -y$ so $y < 0$.
And $|x+y| + |x-y| = -x -y + x-y = -2y = 2$ so $y = -1$ and $-1\le x < 1$.
$x^2 -6x + y^2 = x^2 -6x +1$ and $x^2 \le 1$ and $x \ge -1$ so $-6x \le 6$ so the very max that $x^2-6x +y^2$ is $1+6 +1 = 8$ if $x=-1$.
(Likewise $x^2 +y^2 + 10y = x^2 + 11$ an the max that can be is $12$ if $x = -1$.)
Case 3: $x +y \ge 0$ and $x -y < 0$ then
$x \ge -y$ and $x < y$ so $-y \le x < y$ so $y > 0$.
$|x+y| + |x-y| = x+y+y-x = 2y =2$ so $y =1$ and $-1 \le x < 1$ and
the max value of $x^2-6x +y^2 = x^2-6x+1$ and $x^2 \le 1$ and $-6x \le 6$ so $x^2 -6x+1 \le 8$ and the max value is $8$ when $x=-1$.
case 4: $x+y < 0$ and $x-y < 0$ so $x< -y$ and $x < y$ so $x< \min (y,-y) = -|y|$
And $|x+y| + |x-y| = -x -y +y -2 =-2x = 2$ so $x = -1$ and $-|y| > x=-1$ so $|y|< 1$.
so $x^2 -6x +y^2 = 7 + y^2< 8$.  but $x^2-6x +y^2 = 8$ never occurs; it's an unacheived least upper bound.
Of those four cases the max value of $x^2 -6x +y^2 = 8$.
....
that's probably not an efficient or clever way but it is a complete hit the ground running systematic way.
