Algebra 3 5 Consecutive Integer Problems Consider integers

Algebra 3. 5 Consecutive Integer Problems

Consider integers on a number line……. . -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 n n + 1 n + 2 n + 3

Wherever you start, consecutive integers go up 1 at a time……. . -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 n n + 1 n + 3 n + 2 Etc.

consecutive even integers go up 2 at a time……. . -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 n n + 4 n + 2 n + 6 Etc.

and consecutive odd integers go up 2 at a time……. . -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 n n + 4 n + 2 n + 6 Etc.

Let’s try this problem The sum of three consecutive integers is 39. Find the integers. a + b + c = 39 n + 1 + n + 2 = 39 3 n + 3 = 39 3 n = 36 n = 12 So a = 12, b = 13 and c = 14. They add to 39.

These are easy! The sum of three consecutive odd integers is 87. Find the integers. a + b + c = 87 n + 2 + n + 4 = 87 3 n + 6 = 87 3 n = 81 n = 27 So a = 27, b = 29 and c = 31. They add to 87.

Still Easy! The sum of four consecutive even integers is -4. Find the integers. a + b + c + d = -4 n + 2 + n + 4 + n + 6 = -4 4 n + 12 = -4 4 n = -16 n = -4 So a = -4, b = -2, c = 0, and d = 2. They add to -4.

Piece of Cake! Find four consecutive integers such that three times the second integer when decreased by 10 is equal to the sum of the third and fourth integers. Using a, b, c and d: 3 b - 10 = 3 ( n + 1 ) - 10 = 3 n + 3 - 10 3 n - 7 c + d n+2 + n+3 = 2 n + 5 = 2 n +5 n = 12 So a = 12, b = 13, c = 14, and d = 15. ü 3(13) – 10 = 14 + 15

Try this one……. . Find three consecutive odd integers such that when twice the second integer is increased by 10 it is equal to 7 less than three times the third integer. Set-up: 2 b + 10 = 3 c - 7 Equation: 2 (n + 2) + 10 = 3 (n+ 4) – 7 Solution: The numbers are 9, 11 and 13.

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