SOLUTION: Find five consecutive integers such that: “The sum of the first and 4 times the third is equal to 56 less than 3 times the sum of the second, fourth, and fifth.”

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find five consecutive integers such that: “The sum of the first and 4 times the third is equal to 56 less than 3 times the sum of the second, fourth, and fifth.”       Log On


   

Question 1203707: Find five consecutive integers such that:
“The sum of the first and 4 times the third is equal to 56 less than 3 times the sum of the second, fourth, and fifth.”
Found 3 solutions by josgarithmetic, MathLover1, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Mostly you write the worded description as symbolism literally.

Not the only way to assign, but:
n-2
n-1
n
n+1
n+2

%28n-2%29%2B4n=-56%2B3%28n-1%2Bn%2B1%2Bn%2B2%29
Solve this....

5n-2=-56%2B3%283n%2B2%29
5n-2=9n%2B6-56
-2=4n-50
4n=48
highlight%28n=12%29---------------the integer in the middle







**************************this is wrong******************************
5n-2=-56%2B3%283n%2B2%29
5n-2=9n%2B6-56
-2=4n-50
cross%28n=13%29-----------NO. This does not check.
MISTAKE SOMEWHERE....

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The formula for a sequence of consecutive integers is n, n%2B1, n%2B2,+n%2B3, n%2B4,..., n%2Bk.
The sum of the first and 4+times the third is equal to 56 less than 3+times the sum of the second, fourth, and fifth.
n%2B4%28n%2B2%29=3%28%28n%2B1%29%2B%28n%2B3%29%2B%28n%2B4%29%29-56
n%2B4n%2B8=3%283n%2B8%29-56
5n%2B8=9n%2B24-56
56%2B8-24=9n-5n
40=4n
n=10
five consecutive integers are:10, 11, 12, 14, 14
check:
10%2B4%2812%29=3%28%2811%29%2B%2813%29%2B%2814%29%29-56
10%2B48=3%2838%29-56
58=58

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find five consecutive integers such that:
“The sum of the first and 4 times the third is equal to 56 less than 3 times
the sum of the second, fourth, and fifth.”
~~~~~~~~~~~~~~~~ 

Let the numbers n, (n+1), (n+2), (n+3) and (n+4) be five consecutive integer numbers.


Write equation as you read the problem

    n + 4*(n+2) = 3*((n+1) + (n+3) + (n+4)) - 56.


Simplify step by step

    n + 4n + 8 = 3*(3n+8) - 56

     5n    + 8 = 9n + 24 - 56

     8 - 24  + 56 = 9n - 5n

        40     =    4n

         n     =   40/4 = 10.


ANSWER.  The numbers are  10, 11, 12, 13, 14.

Solved.