SOLUTION: Bobby and Joan found four consecutive integers such that 5 times the sum ofbthe second and third was 6 less than 7 times the first. What were their integers?

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Bobby and Joan found four consecutive integers such that 5 times the sum ofbthe second and third was 6 less than 7 times the first. What were their integers?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 442056: Bobby and Joan found four consecutive integers such that 5 times the sum ofbthe second and third was 6 less than 7 times the first. What were their integers?
Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
Let's represent the 4 integers as: x (x+1) (x+2) (x+3) Now we can set up the following equation:
5(x+1+x+2)=7(x)-6
Rewritten as:
10x+15=7x-6
Subtract 7x and 15 from both sides to get:
3x=-21
Divide by 3:
x=-7
So now we know the integers are:
-7, -6,-5,-4
Make sense?
RJ
www.math-unlock.com