Systems of Linear Equations Word Problems - Practice

(Videos) Set up Word Problems Using a System

by mathman1024

This video by mathman1024 works through word problems with systems of equations.

$\large{\text{Summary}}$

The general guideline to follow when doing these word problems is:

1. Define your variables. Since this is a system, there will be two or more variables.
2. Write your equations. Again, since this is a system, there will be two or more equations. This is also the tricky part which will be focused on in the videos.
3. Solve the system. This can be done using the elimination method, substitution method, or graphing.
4. State your answer.

Problem 1

The first problem that he goes over in the video is, "The sum of two numbers is $79$, and their difference is $23$. What are the two numbers?"

We want to write algebraic expressions for this system.

Step one is defining the variables. The question asks for two numbers so we can see that these are the variables. $$\begin{align}\text{Let }x&=\text{one number}\\ y&=\text{the other number} \end{align}$$ The first sentence is, "The sum of two numbers is 79." Since sum means addition, we write, $$x+y=79$$ The next part states, "The difference of those two numbers is 23." Since difference means subtraction, we write, $$x-y=23$$ And now we have our system of equations: $\begin{cases}x+y=79\\x-y=23\end{cases}$. This can be solved with substitution but you might notice it would be easier with elimination. $$\begin{align} x+y&=79\\ +\quad x-y&=23\\ \hline 2x+0y&=102\\ 2x&=102\\ \frac{2x}{\color{#8925AE}{2}}&=\frac{102}{\color{#8925AE}{2}}\\ x&=51\end{align}$$ Finally, plug this back in to either equation to find the $y$ value. $$\begin{align}{\color{#8925AE}{(51)}}+y&=79\\ 51{\color{#8925AE}{-51}}+y&=79{\color{#8925AE}{-51}}\\ y&=28 \end{align}$$ The solution to this word problem's system of equations is $(51,28)$.

Problem 2

The next problem is a little trickier. It isn't as obvious how to set up the equations. It says, "A minor league ballpark attracts $88$ fans and draws in $\$553$ in revenue from ticket sales. A child's ticket costs $\$4$ and an adult's ticket is $\$7$. How many of each type of ticket were sold?"

First, we define our variables. The problem asks for how many of each type of ticket. Since there are two types of tickets we can write, The first snippet is, "A minor league ballpark attracts $88$ fans". We know there's a total of $88$ fans which are made up of children and adults. So we can write, $$C+A=88$$ Next we have, "and draws in $\$553$ in revenue from ticket sales. A child's ticket costs $\$4$ and an adult's ticket is $\$7$." So we have to factor in the money. We see that we get $\$4$ from each child and we get $\$7$ from each adult. Total, we have $\$553$. So we can write, $$4C+7A=553$$ Now we have the system: $\begin{cases}C+A=88\\4C+7A=553\end{cases}$. This can be solved by either elimination or substitution. For this one, I personally would choose substitution, but either works.

Let's solve the first equation for $C$. $$\begin{align} C+A&=88\\ C+A\color{#8925AE}{-A}&=88\color{#8925AE}{-A}\\ C&=88-A \end{align}$$ Plug this back into the other equation to solve for $A$. $$\begin{align}4{\color{#8925AE}{(88-A)}}+7A&=553\\ 352-4A+7A&=553\\ 352+3A&=553\\ 352{\color{#8925AE}{-352}}+3A&=553{\color{#8925AE}{-352}}\\ 3A&=201\\ \frac{3A}{\color{#8925AE}{3}}&=\frac{201}{\color{#8925AE}{3}}\\ A&=67\end{align}$$ Finally, plug this back into either equation to solve for $C$. $$\begin{align}C&=88-{\color{#8925AE}{(67)}}\\ C&=21 \end{align}$$ Our solution is $(21,67)$.

Remember, the best way to double check your answer to either of these example problems is to plug in your solved coordinates back into the original equations.

As with most word problems, the most effective approach to take is translation. The goal is to take written information and translate into a new language: math equations.

Let's work through an example to see how this translation works.

The key to translating a word problem is to identify the given information. Here's what we know from reading the problem:

• Cupcakes cost $\$2$
• Pies cost $\$7.50$
• William bought $20$ items
• Total cost (pies + cupcakes) = $\$73$

We can get two equations from this information.

Our first equation is $C + P = 20.$

Our second equation is $2C + 7.5P = 73.$

This equation helps us take into account the number of cupcakes and pies William bought as well as the cost of each.

And just like that, we have a system of equations!

$$\begin{align} C + P &= 20\\ 2C + 7.5P &= 73 \end{align}$$

The biggest advantage of using a system of equations to solve this word problem is that we can manipulate both equations to solve for the variables $C$ and $P$. Let's take a look at how we do that.

We're going to start with the first equation because it's simpler, but if you wanted to start with the second one, that would work too. And we're going to solve for $P$ first because we like that letter more than $C$.

Right! $P = 20 - C.$

$$\begin{align} 2C + 7.5(20-C) &= 73 \\\ 2C + 150 - 7.5C &=73 \\\ -5.5C &= -77 \\\ C &= 14 \end{align}$$

This tells us that William bought $14$ cupcakes. As you can see, once we set up the equation, there was nothing super fancy about solving!

Plug $C=14$ into either of the equations in the system to solve for $P$.

$$\begin{align} C + P &= 20 \\\ 14 + P &= 20 \\\ P &= 6 \end{align}$$

This means that William bought $6$ pies. So, to answer the original question (How many cupcakes and pies did William buy?), we could say:

William bought $14$ cupcakes and $6$ pies.

Note: It's always a good idea to check your work by plugging your values into both equations.

Systems of Linear Equations: Word Problems

Sometimes, it is helpful to translate a word problem into a system of linear equations and solve the system. After translating words to math, isolate one variable in one equation, then use its corresponding expression to solve for the other variable in the other equation. Finally, solve for the first variable! Here is a graphic with an example.