I think 5X+1x. Sorry if it's wrong.
We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.
Since 200 km is "halfway", the total distance must be 400 km.
time = distance / speed
total time = (time for first half) + (delay) + (time for second half)
400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40
The speed of the bus before the traffic holdup was 40 km/h.
You need to know the slope and the y intercept
Where’s the question I don’t get it
I hope this is what you're looking for. The concept of regrouping<span> (borrowing) is so that students can see what actually happens in it. For example, to </span>subtract <span>52 − 38, we write 52 as 50 + 2 (breaking it down into its tens and ones). Then, </span>regrouping<span> means that 50 + 2 becomes 40 + 12.</span>
