SOLUTION: Find 4 consecutive even integers where the product of the two smaller numbers is 56 less than the product of the two larger numbers.

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Find 4 consecutive even integers where the product of the two smaller numbers is 56 less than the product of the two larger numbers.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 202378: Find 4 consecutive even integers where the product of the two smaller numbers is 56 less than the product of the two larger numbers.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
smallest number is x
next larger number is x + 2
then x + 4
then x + 6
-----
they go up by 2 because they are even integers.
if they were odd integers they would still go up by 2.
your answer, in this case, however, should be an even integer.
-----
problem states that product of 2 smaller numbers = product of 2 larger numbers minus 56.
x * (x+2) = (x+4)*(x+6) - 56
expanding this makes it:
x%5E2+%2B+2%2Ax = x%5E2+%2B+10%2Ax+%2B+24+-+56
subtract x%5E2+%2B+2%2Ax from both sides of the equation to get:
0 = x%5E2+%2B+10%2Ax+%2B+24+-+56+-+x%5E2+-+2%2Ax
which becomes:
0 = 8%2Ax+%2B+24+-+56
add 56 and subtract 24 from both sides of the equation to get:
56+-+24+=+8%2Ax
simplify to get:
32+=+8%2Ax
divide both sides of the equation by 8 to get:
x+=+4
-----
smallest number is 4.
next largest number is 6
then 8
then 10
-----
4*6 = 24
8*10 = 80 - 56 = 24
product of the smallest numbers = product of the largest numbers - 56 so your answer is good.
answer is:
the 4 consecutive even numbers are: 4,6,8,10
-----