Word Problems.  Everyone’s favorite. From time to time, I will blog on how to solve them and other algebra problems. Today’s blog will be on coin problems. You know, the ones where you wonder why don’t the people simply look in their pockets and count? But they hone your algebra skills and allow you to understand principles involved in systems of equations and other mathematical procedures.

Here’s a typical one:

Jack and Diane have 28 coins that are nickels and dimes. If the value of the coins is $1.95, how many of each type do they have?

As with many math problems, once you simplify, the answer is a lot easier to come by. We have two variables. Ah, if we only had one.  But alas, we have two.

The first step is to get rid of that pesky fact and define one in terms of the other. This way, we only have one. Don’t worry. We will later bring back the other variable.

The variables will stand for our two coins—nickels and dimes. So our variables are N and D. (Pretty clever, right? I could have picked X and Y, but N and D are so much more representative of nickels and dimes, I think.)

The problem tells us that the number of nickels and dimes adds to 28. So our first equation:

N +D = 28.

The number of nickels, N, and the number of dimes, D equals 28. With me so far?  Good.

But we need another equation to solve our problem. The solution is to define the variables and the result in the equation (in this case, the $1.95) by elements they all have in common.  They are all composed of pennies. Hey, that leads us to:

5N +10D =198.

As there are 5 pennies in each nickel, ten in each dime, and 198 altogether. Beautiful. Now, all we have to do is to define one variable in terms or the other and we are all set.

Let’s define nickels in terms of dimes. We could do it the other way, but I like nickels. But maybe you like dimes. Then, you can do it for dimes.

N + D =28.

We solve for N.

N + D-D=28-D

We add the -D to D on the left, and what happens when you add the additive inverse to a number? You cancel the number. As we are in an equation and have to do to the other side what we do to the side we are working, we get the result 28-D.

Think of an equation as like a teeter-totter. You start with your friend who equals your weight. Then Donald Trump sets on your friend’s side. You go flying. Unless, of course, Chris Christy sits on your side. You are back in balance. All’s well, the world, or at least the equation is in balance.

So, N =28-D.

We substitute that for N in our equation:

5(28-D) +10D =195

We now multiply both terms of the parenthetical expression by 5 and the result is:

140-5D+10D =195

We combine like terms.

140 +5D =195

We now have your classic two-step equation. You solve it in, well, two steps.

Step 1

140-140 +5D =195-140

There’s that teeter-totter again. Maybe a less controversial example would be Nero Wolf and Cannon, two fine but large detectives. In any case, however you imagine it, you get:

5D =55

Step 2

Now, to cancel a multiplication, you do a division. To cancel five times D, you divide by 5. Again, you do it to both sides and you get:

D =11.

And Since N+D =28, and you can now substitute 11 for D, you get:

N+11 =28

You subtract the 11 from both sides, and you get 17

Next time, we might try something really hard, like how to balance your checkbook.

Dr. Fred Young








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