Systems of Equations

We will be solving problems in Real Life for solving systems of equations.  Most everyone has a fear of word problems because you haven’t had much experience in solving them.  If you just take them one step at a time, you will find them manageable and build your confidence in your ability to successfully tackle any word problem you meet.

Please read the prompt above to see additional instructions for your Initial Post.

Here is an example to get us started.  I am working out the entire problem for you, but in your initial post you only need to set up the situation and leave solving the problem to a classmate (see prompt given above).

APPLICATION PROBLEM:       DISTANCE = RATE X TIME A boat’s crew rowed 12 miles downstream, with the current, in 1.5 hours.  The return trip, against the current, covered the same distance but took 4 hours.  Find the crews rowing rate in still water and the rate of the current.

The relationship we need to use is Distance = Rate x Time.  We will have two equations: one for the trip downstream and one for the trip upstream.  We have two variables: crews rowing rate in still water and rate of the current.

Let w = crews rowing rate in still water

Let c = rate of the current

Downstream: Distance = Rate x Time    Rate will be w+c (rate in still water + current)

(1)                     12     = (w+c) 1.5 = 1.5w+1.5c

Upstream: Distance = Rate x Time    Rate will be w-c (rate in still water – current)

(2)                    12      = (w-c) 4 = 4w-4c

One approach is to simplify the equations before solving the system.

(1)     12=1.5w+1.5c divide both sides by 1.5

8 = w + c

(2)     12 = 4w=4c   divide both sides by 4

3 = w – c

Use Elimination on the two equations           8 = w+c
3 = w-c      After adding the two equations we have          11 = 2w     divide by 2

w = 11/2 = 5.5 Use this value to solve for c.

w+c = 8

5.5+c = 8

c = 8 – 5.5

c=2.5

Therefore, the rate of the current is 2.5 and the rate of the crew rowing in still water is 5.5.