Question
I don’t quite understand how to go about solving this problem, can a further explanation be provided?
Answer
We set up this problem's solution with a chart because it allows us to easily test each possibility of x and y being even or odd. Essentially, we've got 4 possibilities here:
x is even and y is even
x is even and y is odd
x is odd and y is even
x is odd and y is odd
Any numbers we choose for x and y will fit one of these possibilities. Since we know the properties of even/odd numbers hold true no matter what the numbers are, it doesn't matter which numbers we choose as long as they fit our specifications. So without even worrying about the answer choices just yet, we can plug in numbers to get every possible scenario for w. Here we pick 1 for our representative odd number and 0 for our even number (yes, 0 is even!), we can find out whether w is even or odd, fairly simply. Again, we don't even need to look at the answer choices just yet. Here's the first possibility for example:
x is even and y is even
x = 0, y = 0
w = x^2y + x + 3y
w=0^2 * 0 + 0 + 3 (0)
w = 0
w = even
So every time that x is even and y is even, then w is even. We could continue to do this for the other possibilities:
x is even and y is odd ==> therefore w is odd
x is odd and y is even ==> therefore w is odd
x is odd and y is odd ==> therefore w is odd
Now we can use this information to decide which answers are true. If you're struggling a bit to understand even and odd integers, then check out the even and odd integers video lesson.
Comments
1 comment
Hi,
I'm not sure about this claim "Since we know the properties of even/odd numbers hold true no matter what the numbers are..."
Could you give me a better explanation?
Thanks
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