It’s Professor Dave, let’s look at some word problems.
Algebra brings us our first example of how math can be applied in abstract ways to solve
real world problems.
A good way to demonstrate this is by constructing algebraic equations that are meant to give
us a solution to a question we might have about a person, or any number of things.
Take the following example.
There are two sisters, Sally and Gabby.
Sally is three years younger than twice the age of Gabby.
If Gabby is twelve, how old is Sally?
Let’s find an equation that can describe this scenario.
The first thing we have to do is assign some variables.
What do we know, and what do we want to know?
Well there’s Gabby’s age.
Let’s call that G.
Then there’s Sally’s age, let’s call that S. How are these two values related?
Well we know something about S, we know that it is three less than twice G.
This means that S equals two G minus three.
This equation says exactly the same thing as this sentence, just with symbols instead
of words.
Now we have a very simple equation to solve.
We just plug in twelve for G, since that is Gabby’s age, double it, and subtract three.
Sally must be twenty-one.
That was an easy one, because we didn’t have to do any algebraic manipulation, we
just plugged in a number and evaluated.
What about something a little harder?
Say there are three brothers, Huey, Louie, and Dewey.
Their ages are consecutive multiples of three, and the sum of their ages is thirty-six.
How old are they?
Here, we start to see the benefit of constructing an equation, because while it is certainly
possible to just guess and check, once a problem gets sufficiently complex, solving an equation
becomes much easier than guessing the answer.
Let’s say the youngest brother has an age equal to X.
What would be the other two ages in terms of X?
Well, if the ages are consecutive multiples of three, and each multiple is three more
than the last, then the other two ages must be X plus three and X plus three plus three,
or X plus six.
Then it says that the sum is thirty-six.
Sum means we add them up, so X plus X plus three plus X plus six equals thirty-six.
Now all we have to do is simplify.
Three X’s make three X, and three plus six is nine.
Now we have three X plus nine equals thirty-six.
Subtract nine from both sides to get twenty-seven, divide both sides by three, and we get X equals
nine.
Since X is the age of the youngest, the ages of the brothers are nine, twelve, and fifteen.
Let’s try one more.
Rover and Jasper are dogs, and they each have spots.
Jasper has six more spots than rover does, and twice the number of Rover’s spots plus
three times the number of Jasper’s spots equals sixty-eight.
How many spots do they have?
Well first, let’s assign the variables.
Rover’s spots are represented by R, and Jasper’s by J. How do these variables relate
to one another?
Well if Jasper has six more spots than rover, then J equals R plus six.
Then we have this other bit of information, whereby two R plus three J equals sixty-eight.What
can we do from here?
Well if J equals R plus six, we can put R plus six in this other equation instead of
J, because they are equivalent.
This term has to go inside parentheses, and we can immediately distribute this three.
That gives us two R plus three R plus eighteen.
We can combine the R’s to get five R, subtract eighteen from both sides to get fifty, and
then divide everything by five to get R equals ten.
If R is ten, then J is sixteen, and that’s all the spots.
So we can see that in these scenarios, we just have to decide on some variables, and
construct whatever equations we can from the information we have.
That’s the trickier part, because it requires some abstract thought.
Once we’ve done that, solving them is the easy part, as that will be the same old math
every time.
I think you’ve got the hang of it, so let’s check comprehension.