# Let’s write down three consecutive integers such that 5 less than the sum of the number is equal to 11 more than twice the second number. Let’s find the three consecutive integers.

Let the three consecutive numbers be x, (x+1), (x+2).

Now according to the given criteria,

5 less than the sum of the number = x+(x+1)+(x+2)-5…..(i)

11 more than twice the second number = 2(x+1)+11…….(ii)

So from question equation (i) and equation (ii) are equal,

Hence

x+(x+1)+(x+2)-5 = 2(x+1)+11

3x+3-5=2x+2+11

3x-2=2x+13

3x-2x=13+2

x=15

x+1=15+1=16

x+2=15+2=17

Hence the three consecutive numbers are

15, 16 and 17

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