How many ordered pairs of integers (x,y) are there that satisfy |x|+|y| ≤ 10?
160
from (0,10) to (0,-10) -- 21
from (1,9) to (1,-9) --- 19
from (2,8) to (2,-8) --- 17
..
from (9,1) to (9,-1) -- 3
and (10,0) ----- 1
same thing on the left side, but we can't count the centre line twice, so
sum = 2(1+3+5+...19) + 21
there are 10 terms in the series
sum = 2(5)(first + last) + 21
= 10(20) + 21 = 221
or 2(1+3+5+..+21) - 21
= 2(11/2)(1 + 21) - 21 = 221
There are 221 such ordered pairs.
To solve this problem, we first need to understand the condition |x| + |y| ≤ 10.
The expression |x| represents the absolute value of x. The absolute value of a number is its distance from zero on a number line, always resulting in a non-negative value.
Given that |x| + |y| ≤ 10, we need to find all possible combinations of values for (x, y) that satisfy this inequality.
To begin, we can break down the problem into four separate cases based on the value of x and y:
Case 1: x ≥ 0 and y ≥ 0
In this case, both x and y are positive. Since the sum of two positive numbers can never be greater than 10, we can solve this case by finding the number of ordered pairs (x, y) where 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10.
Case 2: x ≥ 0 and y < 0
In this case, x is positive while y is negative. With |x| + |y| ≤ 10, the inequality can be simplified to x - y ≤ 10. We can solve this case by finding the number of ordered pairs (x, y) where 0 ≤ x ≤ 10 and -10 ≤ y < 0.
Case 3: x < 0 and y ≥ 0
In this case, x is negative while y is positive. With |x| + |y| ≤ 10, the inequality can be simplified to -x + y ≤ 10. We can solve this case by finding the number of ordered pairs (x, y) where -10 < x ≤ 0 and 0 ≤ y ≤ 10.
Case 4: x < 0 and y < 0
In this case, both x and y are negative. Since the sum of two negative numbers can never be greater than 10, there are no solutions in this case.
By solving each of the cases and counting the number of solutions in each, we can find the total number of ordered pairs (x, y) that satisfy |x| + |y| ≤ 10.