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September 2, 2014

September 2, 2014

Posted by **Alan** on Monday, October 29, 2007 at 5:02pm.

- Algebra II -
**Ms. Sue**, Monday, October 29, 2007 at 5:14pmYou 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.

- Algebra II -
**robby**, Monday, October 29, 2007 at 5:29pmi needhelp with solving systems of equetions algebrically

- Algebra II -
- Algebra II -
**DrBob222**, Monday, October 29, 2007 at 5:18pmdistance = 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.

- Algebra II -
**nicole**, Friday, March 30, 2012 at 2:21pmUse the matrix method to solve:

x + 4y = 8

2x + y = 9

x = a0

y = a1

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