Problem
From the figure: AC stretches from A to C, BD from B to D, and these two arcs overlap on segment BC. AE = AC + CD + DE, where CD is the part of BD past C. Equivalently, AE = AC + BD + DE - BC, because adding AC and BD counts the overlap BC twice.
Drawing the line shows the two arcs share the middle piece BC, so it must be removed once.
4.OA.A.3Draw A DiagramThe distance from A to E equals AC + BD + DE with the shared overlap BC subtracted once: AE = AC + BD + DE - BC.
Because A, B, C, D, E sit on the line in that order with no gaps, the distance from A to E is exactly its touching pieces added up: AE = AB + BC + CD + DE.
A length cut at points lying on it splits into pieces that leave no gap and no overlap, so those pieces add back to the whole length.
Because B lies inside AC and C lies inside BD, the span AC equals AB + BC and the span BD equals BC + CD, so adding AC and BD lands on the middle piece BC two times.
Each span is cut by an inside point into two pieces that fill it with no gap or overlap, so those piece lengths add up to the whole span.
Because BC was added in twice, subtracting BC once takes back the extra copy and leaves every piece AB, BC, CD, DE counted exactly one time, which is AE.
Taking a length away undoes adding that same length, so one subtraction cancels one of the two times BC was added.
Combine the whole numbers: 17 + 19 + 4 - 6 = 34.
Handling the whole parts first keeps the fraction work small.
4.OA.A.3Identify SubproblemsAll fractions share denominator 5: 2/5 + 4/5 + 3/5 - 1/5 = (2+4+3-1)/5 = 8/5 = 1 3/5.
Like denominators let you add and subtract numerators, then regroup 8/5 as 1 3/5.
4.NF.B.3Identify SubproblemsAdd the whole-part total and the fraction-part total: 34 + 1 3/5 = 35 3/5.
Putting the pieces back together gives the full A-to-E distance.
4.NF.B.3Count The ComplementThis only needs Grade 4 fraction add/subtract — draw the line and remember the overlap gets counted twice, so subtract it once!