# Triangular inequality - number of ordered pairs

Question:

Find the number of integral order pairs (x,y) that satisfy the system of equations: $$|x+y-4| = |x-3| + |y-1| = 5$$

I've managed to simplify it to: $$(x-3)(y-1) \geq 0$$

However, I'm unable to solve it further. How would I count the number of ordered pairs from here? Won't there be infinite number of ordered pairs satisfying this?

• By "simplify it" you mean the first equation? Mar 27, 2016 at 7:17
• I think algebra would have been a better tag. Mar 27, 2016 at 8:17

What you have is good. You can also create further inequalities based on the initial equations.

$|x-3|+|y-1|=5$ leads to $|x-3|\leq5$ and $|y-1|\leq5$ so you only need to consider $-2\leq x\leq8$ and $-4\leq y\leq6$.

This will lead to the answers of $(-2,1),(-1,0),(0,-1),(1,-2),(2,-3),(3,-4),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)$.



An alternate approach to your problem could be a graphical solution.

The equation $|x+y-4|=5$ becomes the lines $y=-1-x$ and $y=9-x$.

The equation $|x-3|+|y-1|=5$ becomes the functions: $y=-|x-3|+6$ and $y=|x-3|-4$.

Graphing these shows the area of overlap from which you can get the points.

$$|a+b|=|a|+|b| \iff ab\geq0$$
So both of $x-3$ and $y-1$ should be positive or negative
So there is a solutions:
$(m,n)\quad m>2$ and $n>0$ and $m+n=9$
$(m',n')\quad m'<3$ and $n'<1$ and $m'+n'=-1$