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

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.

Answer 2

= x+2(x+4)+3(x+4)=70

=x+2x+4+3x+12=70

6x+16=70

6x=70-16

6x=54

x=54/6

x=9

The second consecutive odd integers =x+2

9+2=11

The third consecutive odd integers =x+4

9+4=13

x=13

therefore, three consecutive odds integers x=9,x=11,x=13

Hence, the three consecutive Odd integers are 9,11 and 13