Q:

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

Accepted Solution

A:
The three consecutive negative integers are -5 , -4 , -3Step-by-step explanation:Consecutive numbers are numbers that follow each other in order. They have a difference of 1 between every two numbersExamples: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 , -3Learn more:You can learn more about consecutive numbers in brainly.com/question/5496711#LearnwithBrainly