Then the three consecutive integers are: #n-1, n, n+1# and there sum is #3n# We are told #color(white)("XXX")3n=264# Dividing both sides by #3#, we find #color(white)("XXX")n=88# So the three numbers are #(n-1,n,n+1) = (87,88,89)# Consider the three numbers to be x, (x+2) and (x+4) According to the question, 3.x = 2(x+4) + 3 3x = 2x + 8 +3 3x - 2x = 11 x = 11 The next two integers are 11 + 2 = 13 and 11 + 4 = 15.
If the first and third of three odd consecutive integers are added, the result is 87 less than five times the second integer. Find the third integer. Solution Let 2n + 1 be the first odd integer Let 2n + 3 be the second odd integer Let 2n + 5 be the third odd integer Adding the first and the third gives the following expression. 2n + 1 + 2n + 5
Correct answers: 3 question: Find three consecutive even integers such that the sum of the first and second equals the sur of the third and -10.
The left-hand side, the 3's cancel out-- that was the whole point behind dividing by 3-- we get just an x is being equal to-- and 225 divided by 3. Let me do it over here. So 3 goes into 225. It goes into 22 7 times. 7 times 3 is 21. 22 minus 21 is 1. Bring down the 5. 3 goes into 15 5 times. 5 times 3 is 15. Subtract, no remainder. Sep 01, 2016 · Find three consecutive integers whose sum of the first and the third is –88. would it be x=-45
The sum of three consecutive integers is 312. Answers (1) ... Let x be first number. X+1 To be second number. x+2be third number. X + (x+1) + (x+2) = 312.