Question

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 1

9,11 and 13

Explanation:

X+(x+2)+(x+4)= ?

3x+6= ?

70/2=35 35-2=33

3x+6=33

3x=27

x=9

9+22+39=70

Answer 2

the three consecutive odd integers are 9, 11, and 13.

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)=70

x+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 13

hope this helps :)