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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
The moduli will give out only non-negative outputs, and since we are to consider only integer values of x and y, this drastically reduces the possible cases.
We can get 2 from either 2 + 0 or 1 + 1
We get a 2 + 0 form when either the first term or the second term is 0
The second term is 0; this is when x = y, in this case, |2x|= 2, where x can be 1 or – 1; therefore, the two cases are (1, 1) and (– 1, – 1)
The first term is 0; this is the case when x = – y, in this case, |x – (– x)| = 2, giving x = 1 or – 1 yet again, here the two cases are (1, – 1) and (– 1, 1)
The other way we can get 2 is through 1 + 1
This is possible when one of the terms is 0; if y = 0, |x|+|x|= 2, where x can be 1 or – 1, giving two cases (1, 0) and (– 1, 0)
Similarly, y for x = 0, we get two cases, (0, 1) and (0, – 1)
Therefore, there are 8 pairs of (x, y) that satisfy the given equation.