Math Is Easy - Use Algebra To Solve Word Problem 8!

 

Find three consecutive odd integers such that three times the second minus the third is 11 more than the first. (Note 1)

 

Since odd consecutive integers differ by 2 (1,3,5), we can represent them as follows:

let n represent the smallest of the three consecutive integers; then n + 2 represents the second largest and, n + 4 represents the largest .

 

We can now create the following equation:

3(n + 2) - (n + 4)  =  n + 11

Solve this equation

3(n + 2) - (n + 4)  =  n + 11

3n + 6 – n – 4       =  n + 11  Remove parentheses

2n + 2                   =   n + 11 Add the common terms

2n + 2                   =   n + 11 

      - 2                   =         -2  Subtract 2 from each side

 2n + 0                  =    n + 9

 2n                        =    n + 9

- n                        =  - n        Subtract n from each side

 1n                       =   0n + 9

  n                        =           9

 

Now we substitute the answer for n into the original equation to see if it works.

3(n + 2) - (n + 4)  =  n + 11

3(9 + 2) - (9 + 4)  =  9 + 11

3(11) - (13)  =  20

33 - 13  =  20

20 = 20

 

We must also see if the solution works in the original problem statement.

Find three consecutive odd integers such that three times the second minus the third is 11 more than the first.

Find three consecutive odd integers (9,11,13 are consecutive odd integers) such that three times the second (3 X 11) minus the third (13) is 11 more than the first (9).

Find three consecutive odd integers such that three times the second (33) minus the third (13) is 11 more than the first. (9).

Find three consecutive odd integers such that 33 minus 13 is 11 more than  9.

33 - 13 = 11 + 9

20 = 20

 

We have just solved a first degree equation.

 

(Note 1)

The source for the problem statement is:

“Intermediate Algebra  For College Students” by Jerome E. Kaufman

 


Close Window