1.

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

Answer»

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)=70

x+2x+4+3x+12=70 →distribute

6x+16=70 → gather like terms

6x=54 → subtract 16 from both sides

x=9 → divide both sides by 6

Thus, the three consecutive odd INTEGERS are 9,11, and 13.


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