Find two consecutive integers whose sum is eleven less than three times the smaller integer.

If you call the smaller integer "x" and the greater integer "y", then:

x+y=3x-11             The sum of the two is 11 less than 3 times the smaller

y=x+1                    Because they are consecutive, the greater must be 1 more than the smaller

Then you can solve for x:

x+(x+1)=3x-11       Substitute x+1 for y

2x+1=3x-11            Combine like terms

2x+12=3x               Add 11 to both sides

12=x                      Subtract 2x from both sides

The smaller integer is 12, thus the greater integer is 1 more or 13.

Check:

12+13=3(12)-11

25=36-11

25=25                       Yes, so the integers are 12 and 13

Step-by-step explanation:

thats pretty much it

B) 8in

Step-by-step explanation: