V vondava3 New member Joined Mar 5, 2009 Messages 1 Mar 6, 2009 #1 Find four consecutive integers such that twice the largest plus the smallest is 20 more than three times the sum of the middle two integers. Im lost. 2(n+3)+n=20+3(n+1+n+2) that is how (one of many) I had it set up. Can anyone help?
Find four consecutive integers such that twice the largest plus the smallest is 20 more than three times the sum of the middle two integers. Im lost. 2(n+3)+n=20+3(n+1+n+2) that is how (one of many) I had it set up. Can anyone help?
stapel Elite Member Joined Feb 4, 2004 Messages 16,543 Mar 6, 2009 #2 Your set-up is fine; you've done the hard part. :wink: Now solve the linear equation "3n + 3 = 6n + 29" for the value of n. Back-solve to find the other three numbers.
Your set-up is fine; you've done the hard part. :wink: Now solve the linear equation "3n + 3 = 6n + 29" for the value of n. Back-solve to find the other three numbers.