How to solve

Question

find 3 consecutive integers such that the 3 times the third is 102 more than the sum of the first and second

Answer 1

Hi, Stormie,

 

This problem looks more complicated than it is to solve. The main thing is to make sure that you set up the equation correctly.

 

First you need to set up you integers you are looking for:

 

let x = the first integer

 

let x+1 = the second integer

 

let x+2 = equal the third integer

 

3 times the third integer means 3(x+2)

 

is means equals

 

more than means addition, so

 

3 times the third is 102 more than the sum of the first and second  means 3(x+2) = 102 + x + (x+1)

 

now we simplify and solve for x:

 

3(x+2) = 102 + x + (x+1): distribute the 3 on the left and combine like terms on the right

3x + 6 = 103 + 2x: subtract 2x from both sides

x + 6 = 103: subtract 6 from both sides

x = 97

therefore;

x + 1 = 98 and 

x + 2 = 99, so the three consecutive integers are 97, 98, and 99

 

Check

3(x+2) = 102 + x + (x+1): substitute 97 for x

3(97+2) = 102 + 97 + (97+1)

297 = 297