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

First integer = x+2

Second integer = x+4

Third integer = x+6

Total = 70

.

So,

(x+2)+2(x+4)+3(x+6)=70

x+2+2x+8+3x+18=70

6x+28=70

 

6x=70-28

6x=42

x=7

.

Plugging the values back in:

x+2=7+2=9

x+4=7+4=11

x+6=7+6=13

.

and

9+(2)11)+3(13)=70

70=70 [checks out]

Answer 2

Answer:

Step-by-step explanation:

21,23,25 is the answer