In my last blog I described what master equations are and how you can use them to solve word problems. I then promised to show you how to use master equations to actually solve word problems.

Here is the blog that shows how you use them to solve equations.

The two master equations I described were:

1) Distance 1 = Distance 2

and

2) Distance 1 + Distance 2 = Distance Total

Let’s see how you solve a word problem with one of these master equations.

Here’s the word problem that we will be solving:

Tino and Gino get into an argument and drive away from one another. Tino leaves first, heading north at 65 kilometers per hour. Two hours later Gino heads south, traveling 45 kilometers per hour. The question: at what time will Tino and Gino be 460 kilometers apart?

Step 1: Decide which master equation to use. Since Tino and Gino are traveling in opposite directions, they are covering different distances. Since their distances are different, we would not use the master equation Distance 1 = Distance 2 . The only other option at this time is Distance 1 + Distance 2 = Distance Total.

We can use this master equation, calling the distance that Tino travels Distance 1, and calling the distance that Gino travels Distance 2. Then the Distance Total would be the 460 kilometers.

At this point we can specify the master equation for this problem like this:

(Distance of Tino) + (Distance of Gino) = 460

The next step is extremely useful, and it makes everything start coming together. To use this step we rely on the fact that distance = (rate) x (time). That being the case, we can express (Distance of Tino) as (rate Tino) x (time Tino), and we can similarly express (Distance of Gino) as (rate Gino) x (time Gino). Using this step, the original master equation morphs into:

(rate Tino) x (time Tino) + (rate Gino) x (time Gino) = 460

Once we reach this level of specificity, we can start filling in the blanks, as follows:

(rate Tino) = 65

(rate Gino) = 45

Of course we also need to come up with expressions for (time Tino) and (time Gino), and this is a bit more tricky, but not too bad. Notice that the problem says that Tino leaves first, and that Gino leaves two hours later. That means that Gino drives two hours LESS than Tino. In algebra-ese, we can express this idea by letting t = the time Gino drives, and then (t + 2) for the time that Tino drives.

So now we have:

(time Tino) = t + 2

(time Gino) = t

Putting it all together we make this grand substitution:

(rate Tino) x (time Tino) + (rate Gino) x (time Gino) = 460

(65) x (t + 2) + (45) x (t) = 460

Do you see what is great about this equation? We have one equation and just one variable. In the world of algebra that means “Hallelujah” because it tells us that we can solve for the variable — which we do as follows:

65t + 130 + 45t = 460

110t + 130 = 460

110t = 330

t = 3

Since we let the variable t stand for Gino’s time driving, this means that Gino has been on the road for three hours when he is 460 kilometers from Gino.

Since Tino drove two hours more than Gino, Tino must have been on the road for five hours when he and Gino were 460 kilometers apart.

Problem solved.

Again, the main point is simply that understanding master equations gives you a guideline that makes it simple to understand problems that otherwise would have left us scratching our heads.

I’ll probably write a bit more about master equations, as they are so useful that everyone should really know what they are and how to use them.

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