Math Is Easy - Use Algebra To Solve Word Problem 9!

 

Find three consecutive even integers such that four times the first minus the third is 6 more than twice the second. (Note 1)

 

Since even consecutive integers differ by 2 (2,4,6), 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:

4(n) - (n + 4)  = 2 (n + 2) + 6

Solve this equation

 4(n) - (n + 4)  =  2(n + 2) + 6

 4n – n – 4       =  2n + 4 + 6  Remove parentheses

 3n - 4             =  2n + 10       Add the common terms

 3n -  4            =  2n + 10

      + 4            =       +   4       Add 4 to each side

 3n + 0            =  2n + 14

 3n                  =  2n + 14

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

 1n                  =  0n + 14

  n                   =          14

 

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

4(n) - (n + 4)  = 2 (n + 2) + 6

4(14) - (14 + 4)  = 2 (14 + 2) + 6

4(14) - (18)  = 2 (16) + 6

56 - 18  = 32 + 6

38 = 38

 

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

Find three consecutive even integers such that four times the first minus the third is 6 more than twice the second.

Find three consecutive even integers (14,16,18 are consecutive even integers) such that four times the first (4 X 14) minus the third (18) is 6 more than twice the second (2 X 16).

Find three consecutive even integers such that four times the first (56) minus the third (18) is 6 more than twice the second (32).

Find three consecutive even integers such that 56 minus 18 is 6 more than 32.

56 – 18 = 6 + 32

38 = 38

 

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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