Translate into an algebraic equation and solve. Find three consecutive odd integers such that the difference between four times the first in

CPhill is correct; a  slightly different approach is:

Consecutive odd integers are like:  3, 5, 7, 9, ...   (they are 2 apart from each other)

So, three consecutive odd integers can be:  x, x + 2, and x + 4

"four times the first integer"  --->  4·x

"twice the largest integer"     --->  2(x + 4)  =  2x + 8

"difference is two" means subtract and get an answer of two; so subtact the smallest from the largest and have that answer be two:  (2x + 8) - (4x)  =  2

--->                              2x + 8 - 4x  =  2

--->   simplify   --->               8 - 2x  = 2

--->   subtract 8   --->              -2x  =  -6

--->   divide by -2   --->              x  =  3      --->  x + 2  =  5      and     x + 4  =  7

Translate into an algebraic equation and solve.  Find three consecutive odd integers such that the difference between four times the first integer and twice the largest integer is two. 

Let the integers be 2n+ 1, 2n + 3 and 2n +5

So

4(2n + 1) - 2(2n + 5)  = 2    simplify

8n + 4  - 4n -10  = 2

4n - 6  = 2    add 6 to both sides

4n = 8     divide both sides by 4

n = 2

So....the first integer is 2(2)+ 1 = 5    and the last integer  is 2(2) + 5 = 9

Thanks, geno.....yours "works," too!!!