Question

The sum of three consecutive integers is 31 more than the largest integer. Find the integers

Answer 1

Let's express this in algebraic terms:

`(x-2)+(x-1)+x=x+31\\x-2+x-1+x=x+31\\3x-3=x+31\\3x=x+31+3\\3x=x+34\\3x-x=34\\2x=34\\x=34/2\\x=17`

The expressions for the integers and respective values are expressed below:

`x=17\\\\Integer 1 => x =17\\Integer 2 => x-1=17-1=16\\Integer 3 => x-2=17-2=15`

Now, to check, we apply the original problem and the values we discovered:

`(x-2)+(x-1)+x=x+31\\((17)-2)+((17)-1)+(17)=(17)+31\\(15)+(16)+17=48\\48=48`

We are right, so:

`x=17\\\\Integer 1 => 17\\Integer 2 => 16\\Integer 3 => 15`

Answer 2

To solve with an equation, let's call the three consecutive integers g, g+1, and g+2.

So, (g)+(g+1)+(g+2)=(g+2)+31

3g + 3 = g + 33

2g = 30

g = 15

The integers are 15, 16, and 17.

To check:

The sum of 15, 16, and 17 is 48. 17 + 31 is also 48.