Graphing $|x+y|+|x-y|=4$ I'm trying to graph $|x+y|+|x-y|=4$. I rewrote the expression as follows to get a function that resembles the direction of unit vectors at $\pi/4$ to the horizontal axis (take it to be $x$)$$\biggl|\dfrac{x+y}{\sqrt{2}}\biggr|+\biggl|\dfrac{x-y}{\sqrt{2}}\biggr|=2\sqrt{2}$$
However, I'm not able to proceed further. Any hints are appreciated. Notice this is an exam problem, so time-efficient methods are key. Please provide any hints accordingly.
 A: The lines $x+y=0$ and $x-y=0$ define a coordinate system rotated $45^\circ$ from the canonical axes and scaled by a factor of $\sqrt2$. Accordingly, $|x+y|+|x-y|=4$ defines a diamond (rotated square) in this new coordinate system, which corresponds to a (axis-aligned, centred) square in the canonical one.
To determine the size of this square, note that $(x,y)=(2,0)$ satisfies the equation. Thus the graph of $|x+y|+|x-y|=4$ is a square of side $4$ centred on the origin.
A: The given graph forms a square with side 4.Simple logic is that when (X+y)+(x-y)=c
When (X+y) approaches c,(x-y) approaches 0.
From the above thing the upper boundary will be 2x=4.
You get X=2,y=2.
Because there is mod if you can solve one boundary it gets mirrored about both X and y axis
A: It is a square with sides as $x=\pm 2$ and $y=\pm 2$,and diagonals as $y=\pm x$. for the explanation see my answer in MSE below.
How to draw graph of $|x+y-1| + |2x + y + 1|=1$?
A: I think you dismiss the solution by cases too quickly.
Actually you only need to do one case, and the rest is developed by symmetry.
The easiest case is when $x+y\geq 0$ and $x - y \geq 0,$
equivalently $x\geq -y$ and $x \geq y,$
or in other words when $(x,y)$ is to the right of both of the lines
$x = y$ and $x= -y.$
In that case the formula simplifies to 
$$ (x+y) + (x - y) = 4, $$
that is, $2x = 4,$ so $x = 2.$ We have a vertical line segment from $(2,-2)$ to $(2,2).$
Now observe that the result of $\lvert x+y\rvert +\lvert x-y\rvert$ does not change if we swap $x$ and $y$.
So we also have the mirror image through the line $y=x,$ which says we also have the horizontal line segment from $(-2,2)$ to $(2,2).$
Notice how this is two adjacent sides of the square with vertices
$(-2,2),$ $(2,2),$ $(2,-2),$ and $(-2,-2).$
Also observe that the result of $\lvert x+y\rvert +\lvert x-y\rvert$ does not change if we replace $x$ and $y$ with $-x$ and $-y$.
So we also have symmetry via a reflection through the origin, which is also a $180$-degree rotation around the origin.
That gives us the other two sides of the square.
A: I would take cases. When $x > y$
$$|x+y| + |x-y| = 4 \implies x = 2$$
Try to figure out the rest of the cases on your own then graph everything together.
A: Once $x$ is fixed you can look for solutions in three distinct regions: $y>x$, $-x<y<x$, $y<-x$. By doing this you can simplify the absolute values and find solutions.
A: Looks like you have said
$u = \frac {\sqrt {2}}{2} (x + y)\\
v = \frac {\sqrt {2}}{2} (x - y)$
Which rotates coordinate system 45 degrees.
And your expression becomes
$|\sqrt 2 u| + |\sqrt 2 v| = 4$ or $|u| + |v| = \sqrt 2$
So what does this graph look like in an unrotated frame?
Do you know about the taxicab metric?  
https://en.wikipedia.org/wiki/Taxicab_geometry
What we have here is the set of points that are $\sqrt 2$ units from the origin by the taxicab metric.

And rotate coordinates back to the original system.
