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Find 4 consecutive integers such that, 5 times the 3rd number decrease by the 2nd number is 41

Sagot :

Consecutive integers differ by 1.

Let x represent the first integer.

The second integer would be x + 1

The third integer would be x + 1 + 1 = x + 2

The fourth integer would be x + 1 + 1 + 1 = x + 3

Given that 5 times the third number decreases te second by 41, it means that

5(x + 2) - (x + 1) = 41

5x + 10 - x - 1 = 41

5x - x + 10 - 1 = 41

4x + 9 = 41

4x = 32

x = 32/4

x = 8

The numbers are 8, 8 + 1, 8 + 2 and 8 + 3

The numbers are 8, 9, 10 and 11

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