Find the integers

Question

The sum of three consecutive integers is 21 larger than twice the smallest integer.

Answer 1

The sum of three consecutive integers is 21 larger than twice the smallest integer.

So, if you subtract 21 from the sum of the three consecutive integers it will equal twice the smallest integer.

Set up the consecutive integers as follows: Let (x) equal the value of the first integer, (x+1) the value of the second integer, and (x+2) the value of the third integer.

The smallest integer is (x) so, twice the smallest integer = 2x

(x) + (x + 1) + (x + 2) - 21 = 2x              Set up equation

3x + 3 - 21 = 2x                                     Simplify equation to show relationship

3x - 18 = 2x                                           Subtract integers

x = 18                                                   Subtract 2x from both sides

                                                             Add 18 to both sides

Check:

(18) + (19) + (20) -21 = 2 * 18

                     57 – 21 = 36

                            36 = 36

Answer 2

hi chelsea--

 

think of the 3 consecutive integers as n, n + 1, and n + 2

 

added together, they are 3n + 3, and 

 

3n + 3 = 2n + 21

 

now combine terms by subtracting 2n from both sides, and subtracting 3 from both sides:

 

n = 18, which means the 3 integers are 18, 19, and 20.

 

Their sum is 57.

 

Twice the smallest is 36, and add 21 to that and it's 57!

 

Hope this helps.

 

Matt in NY