kakz wrote:
x be the number of both maths and english
y be the number of both english and history
z be the number of both history and maths
we need to fin the value of x+y+z
68=25+25+34-x-y-z+3
68=84-(x+y+z)+3
x+y+z=19
Why it can't be solved this way?
The method you've used is only valid if x,y, and z are also counting the people who are taking all three classes. So if you try to use this method, when you do x+y+z, the people who are taking all three classes are counted three times. But we don't want to count these people at all, so we have to subtract them 3 times, which would give us 19-3(3)=10.
Analyzing your method step-by-step: Let's say we define x,y, and z to represent the number of people taking EXACTLY two of the classes, which I assume was your intent if you expected x+y+z to be the right answer. When you add 25+25+34 you count the people who are taking exactly one class once, the people who are taking exactly two classes twice, and the people who are taking exactly three classes three times. When you subtract x,y,z you are now counting the people who are taking exactly one class once, the people who are taking exactly two classes once, but you are still counting the number of people taking exactly three classes three times because when we subtract x,y,z, we're not subtracting ANY of the people taking all three classes, because we defined the variables to only include those taking EXACTLY two classes. When you add the number of people taking all three classes at the end, now you're counting the people taking exactly one class once, the people taking exactly two classes once, and the people taking exactly three classes FOUR times. These people are being vastly overcounted, so you have no reasonable expectation that this should all add up to the total number of people, 68, making your equation invalid. Instead of adding three, you should have subtracted six, because you WERE counting them 3 times and to get everything to add to 68 you should only be counting everybody once, so you need to subtract 3 from this total twice. This would make x+y+z=10
Remember that the point of adding in the number of people who are in all three groups in a problem like this is because they usually get undercounted once you subtract out the people who are in at least two sets. Normally you add everyone in at least one set and the people in exactly one, two, and three sets are counted 1,2 and 3 times respectively. Then you subtract out the people in at least two sets and the people are counted 1, 1, and 0 times. Then you have to add in the people in all the sets to get it back to 1, 1, 1 so that everybody is counted exactly once. But here the people in all three sets are not undercounted because they weren't part of the groups we were subtracting out. And if they're not undercounted, it makes no sense to add them back in.
