Posted by Alan on Monday, October 29, 2007 at 5:02pm.
You can solve it algebraically or you can set up a chart:
First train:
1 hr: 75 mi
2 hr: 150 mi
3 hr: 225 mi
-- Continue this chart
Second train:
1 hr: 100 mi
2 hr: 200 mi
3 hr: 300 mi
-- Continue this chart.
Compare the charts to see when they've both traveled the same distance, but the second train reached it in two hours less.
i needhelp with solving systems of equetions algebrically
distance = rate x time
A train rate = 75 km/hr
B train rate = 100 km/hr
distance traveled by A train =
75 x (t+2 hrs)
distance traveled by B train =
100 x t
Distance is the same so set them equal to each other.
75(t+2) = 100t
solve for t = time.
Use the matrix method to solve:
x + 4y = 8
2x + y = 9
x = a0
y = a1
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