y should be an integer and hence can take any value from the set {-12, -11, -10 ... 10, 11, 12} i.e. any one of 25 values (these are 25 values -12 to -1 (12 values), 0, 1 to 12 (another 12 values)) 13 of them are even and 12 of them are odd. 2x+y=12:Every time y is even, x will be integer. e.g. y = 12, x = 0 (because x = (12 - even)/2 will be an integer) ;Every time y is odd, x will be non-integer e.g. y = 1, x = 5.5 (because x = (12 - odd)/2 will not be an integer)
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做那条一次函数出来 然后y做两条横轴 又要在这个区间 又要在这条直线上 所以就有13个点即x（0到12）
y should be an integer and hence can take any value from the set {-12, -11, -10 ... 10, 11, 12} i.e. any one of 25 values (these are 25 values -12 to -1 (12 values), 0, 1 to 12 (another 12 values)) 13 of them are even and 12 of them are odd. 2x+y=12:Every time y is even, x will be integer. e.g. y = 12, x = 0 (because x = (12 - even)/2 will be an integer) ;Every time y is odd, x will be non-integer e.g. y = 1, x = 5.5 (because x = (12 - odd)/2 will not be an integer)
斜率是-2， X轴上对应整数为：（0， 1， 2，3，4，5，6）， X=0时， Y有正负解，但不成对。成对的有12组，共13组。这里的有序偶对应该有点关系，但具体概念不清楚。
题目是y的绝对值，需要考虑y为负的情况