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find four consecutive odd integers such that the sum of twice the smallest and the largest is 11 more than the sum of the middle two integers

Respuesta :

amakaolive

Answer:

x+(2+x)+(4+x)+(6+x)= x+12

2x + 6+x = 3x +6

2+ x + 4+x= 2x + 6

2x + 6 + 11 = 3x+6

2x + 17 = 3x +6

3x-2x = 17-6

x = 11

therefore the four consecutive odd integers are

11, 13, 15 and 17

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