Question

Find the number ordered pairs (x, y) if $x,y\in \\{0,1,2,3,......,10\\}$ and if $|x-y| > 5$.

Answer 1

Hint: We will take two cases because modulus is there that is $|x-y| > 5$. So the first case will be when x is greater than y and this implies $x-y > 5$. And the second case will be when x is smaller than y and this implies $x-y < -5$. Then we will find the ordered pairs which will satisfy both these equations and add the answers from both the cases to get the total ordered pairs.

Complete step-by-step answer:
Also it has been given in the question that $x,y\in \\{0,1,2,3,......,10\\}$.
So assuming that x is greater than y than $x-y > 5$ is the first case.
$\Rightarrow x-y > 5.......(1)$
Rearranging and isolating x in equation (1) we get,
$\Rightarrow x > 5+y.......(2)$
Now from equation (2) we can see that the minimum value of y is 0 and this means that x will start from 6 if we have to satisfy equation (2). Now we will make a table to find the values of y which will satisfy equation (2) for values of x till 10.

x	y	Ordered pairs
6	0	1
7	0, 1	2
8	0, 1, 2	3
9	0, 1 , 2, 3	4
10	0, 1, 2, 3, 4	5

So from this table we can see that the total number of ordered pairs of (x, y) is $1+2+3+4+5=15$.
Now assuming that x is less than y than $x-y < -5$ is the second case.

& \Rightarrow -(x-y) > 5 \\\ & \Rightarrow x-y < -5.......(3) \\\ \end{aligned}$$ Rearranging and isolating x in equation (3) we get, $$\Rightarrow y > 5+x.......(4)$$ Now from equation (4) we can see that the minimum value of x is 0 and this means that y will start from 6 if we have to satisfy equation (4). Now we will make a table to find the values of x which will satisfy equation (2) for values of y till 10. y| x| Ordered pairs ---|---|--- 6| 0| 1 7| 0, 1| 2 8| 0, 1, 2| 3 9| 0, 1 , 2, 3| 4 10| 0, 1, 2, 3, 4| 5 So from this table we can see that the total number of ordered pairs of (x, y) is $$1+2+3+4+5=15$$. So the total number of ordered pairs from both the cases is $$15+15=30$$. Note: We should know the concept of modulus and its properties to solve this question. Also we should know how to solve inequalities. We in a hurry can make a mistake in solving equation (3) if we fail to change the sign when multiplying by negative sign on both sides.