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Alenkasestr [34]
3 years ago
5

find three consecutive negative integers such that six times the largest is equal to the twice sum of the two smaller integers.

Mathematics
1 answer:
NISA [10]3 years ago
3 0

The three consecutive negative integers are -5 , -4 , -3

Step-by-step explanation:

Consecutive numbers are numbers that follow each other in order.

They have a difference of 1 between every two numbers

Examples:

  • 2 , 3 , 4, .......
  • -5 , -4 , -3 , .......
  • n , n + 1 , n + 2 , ........

Assume that the smallest negative number is -n

∵ There are three negative consecutive numbers

∵ The smallest one = -n

∴ The three numbers are (-n) , (-n + 1) , (-n + 2)

∵ The largest number is (-n + 2)

∵ Six times the largest is equal to the twice sum of the two smaller

  integers

- Multiply (-n + 2) by 6 and multiply the the sum of (-n) and (-n + 1) by 2

∴ 6(-n + 2) = 2[(-n) + (-n + 1)]

- Simplify the equation

∵ 6(-n) + 6(2) = 2[-2n + 1]

∴ -6n + 12 = 2(-2n) + 2(1)

∴ -6n + 12 = -4n + 2

- Subtract 12 from both sides

∴ -6n = -4n - 10

- Add 4n to both sides

∴ -2n = -10

- Divide both sides by -2

∴ n = 5

∵ The numbers are (-n) , (-n + 1) , (-n + 2)

∴ The numbers are (-5) , (-5 + 1) , (-5 + 2)

∴ The numbers are (-5) , (-4) , (-3)

The three consecutive negative integers are -5 , -4 , -3

Learn more:

You can learn more about consecutive numbers in brainly.com/question/5496711

#LearnwithBrainly

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