Math Is Easy - Use Algebra To Solve Word Problem 2!

 

The sum of three consecutive integers is 13 greater than twice the smallest of the three integers. Find the numbers. (Note 1)

 

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

let n represent the smallest of the three consecutive integers; then n + 1 represents the second largest and , n + 2 represents the largest .

 

We can now create the following equation:

n + (n + 1) + ( n + 2)  =  2n + 13

Solve this equation

n + (n + 1) + ( n + 2)  =   2n + 13

n + n + 1 +  n + 2       =   2n + 13  Remove parentheses

 3n + 3                        =   2n + 13  Add the common terms

 3n + 3                        =   2n + 13 

-2n                              = -2n           Subtract 2n from each side

 1n + 3                        =   0n + 13

   n + 3                        =           13

      - 3                         =           -3 Subtract 3 from each side

 n +  0                         =           10

 n                                =           10

 

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

n + (n + 1) + ( n + 2)  =  2n + 13

10 + (10 + 1) + (10 + 2) = 2(10) + 13

10 + 11 + 12 = 20 + 13

33 = 33

 

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

The sum of three consecutive integers is 13 greater than twice the smallest of the three integers.

The sum of three consecutive integers (10,11,12 are consecutive integers) is 13 greater than twice the smallest of the three integers (2 X 10).

The sum of three consecutive integers (10 + 11 + 12 = 33) is 13 greater than twice the smallest of the three integers (2 X 10 = 20).

The sum of three consecutive integers (33) is 13 greater than twice the smallest of the three integers (20).

33 = 13 + 20

33 = 33

 

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

 


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