## Simple Word Problems Resulting In Linear Equations

Read this article for another example of this type of a general statement problem. In this problem, it looks like there are two variables. However, we can relate the quantity of one variable to that of the other. This allows us to write the equation in terms of only one variable.

At the bottom of the page, try a few practice problems and check your answers. Try a couple of these until you feel comfortable writing and solving equations from general word problems.

Many word problems, upon translation, result in two equations involving two variables (two "unknowns"). In mathematics, a collection of more than one equation being studied together is called a system of equations.

The systems in this section are fairly simple, and can be solved by substituting information from one equation into the other. The procedure is illustrated in the following example:

### STEP 1: Name Your Unknowns

Decide what piece(s) of information are unknown, and give name(s) to these things.

Let n be the number of night tickets (evening shows).

Let d be the number of day tickets (matinee shows).

### STEP 2: What Can We Write Down that is True?

The English words will translate into mathematical sentences involving your unknowns.

 English Words Translation into Math Notes/Conventions "Antonio went to see a total of $12$ movies" $n+d=12$ There are many real-number choices for n and d that make this equation true. Here are a few: $0+12=12$ $1+11=12$ $1.3+10.7=12$ $(-2)+14=12$ Of course, we want whole number solutions, and we also need something else to be true. "... and spent $86.00$" $8n+6d=86$ Each night movie costs $8.00$, so n night movies cost $8n$ dollars. Each day movie costs $6.00$, so d day movies cost $6d$ dollars. Both 8n and 6d have units of dollars.Also, the number $86$ has units of dollars.It is important that you have the same units on both sides of the equal sign. Here, we have: dollars plus dollars equals dollars. CONVENTIONWrite $8n$, not $8.00n$, $8n$ or $8.00n$ for example. Convince yourself that there are also infinitely many real-number choices for n and d that make this equation true. We want a choice for n, and a choice for d, that make BOTH equations true at the same time.

### STEP 3: Choose a Simplest Equation, and Solve for One Variable in Terms of the Other

Remember that to solve for a variable means to get it all by itself, on one side of the equation, with none of that variable on the other side.

Here, you are getting a new name for one of your variables that is helpful for finding the solution.

Clearly, the equation $n+d=12$ is simpler than $8n+6d=86$.

We could solve the equation $n+d=12$ for either $n$ or $d$;

hmmm…… think I will choose to solve for $n$. (It does not matter!)

Subtracting dd from both sides, we get: $n=12−d$

### STEP 4: Use Your New Name in the Other Equation

Take your new name from the previous step, and substitute it into the remaining equation.

This will give you an equation that has only one unknown.

Substituting $n=12−d$ into the equation $8n+6d=86$ gives: $8\overset{n}{\overbrace{(12-d)}}+6d=86$

### STEP 5: Solve the Equation for One Unknown

Solve the resulting equation in one variable. Be sure to write a nice, clean list of equivalent equations.

 $8(12−d)+6d=86$ original equation $96−8d+6d=86$ distributive law $96−2d=86$ combine like terms $−2d=−10$ subtract $96$ from both sides $d=5$ divide both sides by $−2$

### STEP 6: Use the Known Variable to Find the Remaining Variable

Go back to the simplest equation, substitute in your new information, and solve for the remaining variable.

Make sure you understand the logic being used:

If both $n+d=12$ and $8n+6d=86$ are true, then d must equal 5.

Substitute $d=5$ into the simple equation $n+d=12$ and solve:

 $n+d=12$ the simple equation $n+5=12$ substitute in the known information $n=7$ subtract 55 from both sides

Check that the numbers you have found make both of the equations true.

 Equations Check Is it True $n+d=12$ Does $7+5=12$ Yes! $8n+6d=86$ Does $8(7)+6(5)=86$ Yes!

The original problem asked how many night movies Antonio attended, so here is what you would report as your answer:

### The Good News!

Even though this explanation was very long, you will actually be writing down very little!

Here is the word problem again, and what I ask my students to write down:

 Antonio loves to go to the movies. He goes both at night and during the day. The cost of a matinee is $6.00$. The cost of an evening show is $8.00$. If Antonio went to see a total of 12 movies and spent $86.00$, how many night movies did he attend? Let n=# night tickets. Let d=# day tickets. $n+d=12$ $8n+6d=86$ $n=12−d$ $8(12−d)+6d=86$ $96−8d+6d=86$ $96−2d=86$ $−2d=−10$ $d=5$ (circle this) $n+5=12$ $n=7$ (circle this) Does $7+5=12$ Does $8(7)+6(5)=86$

#### Antonio Attended 7 Night Movies

Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/simple_word_probs.htm