audio read-through Simple Word Problems Resulting in Linear Equations

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.

This section can be included in a high-level Algebra I curriculum. It is also available in the Algebra II curriculum, where systems are studied in much more detail.

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:

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?

Step 1:  Name Your Unknowns

Decide what piece(s) of information are UNKNOWN, and give name(s) to these things. Choose names that help you to remember what they represent!

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?

Re-read the word problem. The English words will translate into mathematical sentences involving your unknowns. You may need additional mathematical concepts in making your translation.
English Words Translation into Math Notes/Conventions
‘Antonio went to see a total of 12 movies’ $n+d = 12$

Note: There are many real-number choices for $\,n\,$ and $\,d\,$ that make this equation true. Here are a few:

$$ \begin{gather} \cssId{s32}{0 + 12 = 12}\cr \cssId{s33}{1 + 11 = 12}\cr \cssId{s34}{1.3 + 10.7 = 12}\cr \cssId{s35}{(-2) + 14 = 12}\cr \end{gather} $$

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's important that you have the same units on both sides of the equal sign. Here, we have: dollars plus dollars is dollars.

Convention: Write $\,8n\,,$ not (say) $\,8.00n\,$ or $\,\$8n\,$ or $\,\$8.00n\,.$

Note: 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're 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$\,\ldots\,$ think I'll choose to solve for $\,n\,$. (It doesn't matter!)

Subtracting $\,d\,$ 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:

$$ \cssId{s71}{8(\overset{n}{\overbrace{12 - d}}) + 6d = 86} $$

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 $\,5\,$ from both sides

Step 7:  Check, and Report Your Answers

Check that the numbers you've found make both of the equations true. Then, report your answer(s) using a complete English sentence.
Equations Check True?
$n + d = 12$ $7 + 5 \,\,\overset{\text{?}}{ = }\,\, 12$ Yes!
$8n + 6d = 86$ $8(7) + 6(5) \,\,\overset{\text{?}}{ = }\,\, 86$ Yes! (Feel free to use your calculator.)

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

Antonio attended  7  night movies.

The Good News!

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

Here's 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?
$$ \begin{gather} \cssId{s120}{\text{Let $\,n = \text{# night tickets}\,$.}}\cr \cssId{s121}{\text{Let $\,d = \text{# day tickets}\,$.}}\cr \cssId{s122}{n + d = 12}\cr \cssId{s123}{8n + 6d = 86}\cr \cssId{s124}{n = 12 - d}\cr \cssId{s125}{8(12-d) + 6d = 86}\cr \cssId{s126}{96 - 8d + 6d = 86}\cr \cssId{s127}{96 - 2d = 86}\cr \cssId{s128}{-2d = -10}\cr \cssId{s129}{d = 5}\ \ \cssId{s130}{\text{(circle this)}}\cr \cssId{s131}{n + 5 = 12}\cr \cssId{s132}{n = 7}\ \ \cssId{s133}{\text{(circle this)}}\cr \cssId{s134}{7 + 5 \,\,\overset{\text{?}}{=}\,\,12}\ \ \cssId{s135}{\text{☺}}\cr \cssId{s136}{8(7) + 6(5) \,\,\overset{\text{?}}{=}\,\,86}\ \ \cssId{s137}{\text{☺}} \end{gather} $$

Antonio attended  7  night movies.

Concept Practice

Choose Your Own Names!

These practice problems will be MORE FUN if they use people you know!

So... take a minute and put in some names!

  • Think of a name. Type it in the name box below.
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  • Click the ‘Add this name!’ button.
  • Put in as many or as few as you want. (We'll throw in some of our own, just to spice things up.)
  • Refresh this page if you want to throw everything away and start over.

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