Solution :
y = 3[x – 2] + 5 = 3[x] – 1
so 3[x] – 1 = 2[x] + 3
[x] = 4 \(\implies\) 4 \(\le\) x < 5
then y = 11
so x + y will lie in the interval [15, 16)
so [x + y] = 15
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