Find three consecutive odd integers such that the sum of the first is two less than the second,and three more than the third is 70

Accepted Solution

Answer:the three consecutive odd integers are 9, 11, and 13.Step-by-step explanation:Let's say that the first odd integer is x. The second consecutive odd integer would have to be x+2. (It would not be x+1  because that would result in an even integer. The sum of any two odd numbers is even.) The third consecutive odd integer would be (x+2) +2 or x+4.The sum of the first, twice the second, and three times the third can be written as:x+2(x+2)+3(x+4)This equals 70. We can now distribute and solve for x:x+2(x+2)+3(x+4)=70x+2x+4+3x+12=70(distribute)6x+16=70(combine like terms)6x=54(subtract 16 from both sides)x=9(divide by 9)Thus, the three consecutive odd integers are 9, 11, and 13hope this helps :)