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| Question 157772:  Find four consecutive integers such that twice the first subtracted from the sum of the other three integers is sixteen.
 Found 3 solutions by  checkley77, jojo14344, Electrified_Levi:
 Answer by checkley77(12844)
      (Show Source): 
You can put this solution on YOUR website! Integers=x,x+1,x+2,x+3 (x+1+x+2+x+3)-2x=16
 3x+6-2x=16
 x=16-6
 x=10 answer for the first integer.
 10,11,12,13 are all the integers.
 proof:
 11+12+13-2*10=16
 36-20=16
 16=16
 
Answer by jojo14344(1513)
      (Show Source): 
You can put this solution on YOUR website! Let ¨x= 1st integer x+1 = 2nd integer
 x+2 = 3rd integer
 x+3 = 4th integer
 Then, (x+1+x+2+x+3)-2x=16 ----------------------- condition, twice the 1st is subtracted from the sum of 2nd, 3rd & 4th:
 continuing;
 3x-2x=16-6
 x=10 ------------------------------------- 1st integer
 10+1=11 ---------------------------------- 2nd integer
 10+2=12 ---------------------------------- 3rd integer
 10+3=13 ---------------------------------- 4th integer
 To check, go back to the condition:
 11+12+13-2(10)=16
 36-20=16
 16=16
 Thank you,
 Jojo
Answer by Electrified_Levi(103)
      (Show Source): 
You can put this solution on YOUR website! Hi,Hope I can help .
 Find four consecutive integers such that twice the first subtracted from the sum of the other three integers is sixteen.
 .
 consecutive means " in order"(examples: 45,46,47,48 . Sunday, Monday, Tuesday)
 .
 To find the 4 consecutive numbers, we have to find the variables for all four numbers
 .
 Since consecutive means " in order "(we are trying to find 4 numbers that come right after the other(example of consecutive numbers: 22,23,24,25,26)
 .
 We can always name the first number of any "consecutive numbers" - "x"
 .
 "x" is our first number, to find the other numbers, we will add "1" each time(we are trying to find numbers that come right after the other)
 .
 2nd number = (add "1" to our first number "x") = "x+1"
 3rd number = (add "1" to the second number "x + 1", ("x + 1 + 1")) = "x+2"
 4th number = (add "1" to the third number "x+2",("x+2+1")) = "x+3"
 .
 (If there was a fifth number you would add "1" to the fourth number "x+3" = "x+4", the sixth number would be "1" added to the fifth number "x+4" = "x+5", you get the idea)(If you were solving for consecutive odd or even numbers, you would add "2" every time, consecutive even/odd number = ( x, x + 2, x + 4, x + 6, x + 8, ...)
 (examples of consecutive even/odd number: 3, 5, 7, 9 : 16, 18, 20, 22))
 .
 Since we are solving for just consecutive numbers, our numbers are
 .
 1st number = "
  " 2nd number = "
  " 3rd number = "
  " 4th number = "
  " .
 Now we can replace these variables with the problem, and solve for the numbers
 .
 Find four consecutive integers such that twice the first subtracted from the sum of the other three integers is sixteen.
 .
 If we replace the variables in the problem, and put the word problem in an equation, the equation =
  .
 Now we just solve for "x"
 .
 
  .
 We will remove the first set of parentheses,we will multiply 2(x)
 .
 
  =  .
 We will remove the other sets of parentheses
 .
 
  =  .
 We will rearrange the equation, so we can add/subtract like terms
 .
 
  =  .
 We will add/subtract like terms
 .
 
  =, (x+x+x-2x)(+1+2+3) = 16, =  .
 To solve for "x" we will move "6" to the right side
 .
 
  =  =  =  .
 "x" = "10" , we can check by replacing "x" with our equation
 .
 
  =  .
 
  =  .
 
  =  .
 
  =  =  =  =  (True) .
 Here is our number variables again (To find the numbers, replace "x" with "10")
 .
 1st number = "
  ", "  " 2nd number = "
  ", "  ", "  ", "  " 3rd number = "
  ", "  ", "  ", "  " 4th number = "
  ", "  ", "  ", "  " .
 Find four consecutive integers such that twice the first subtracted from the sum of the other three integers is sixteen. (
  =  =  =  (True)) .
 1st number =
  2nd number =
  3rd number =
  4th number =
  .
 Hope I helped, Levi
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