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## Writing Systems of Linear Equations from Word Problems

As we learn about math, we often ask ourselves, "How is any of this knowledge useful in life? When will I ever need to use algebra?" Here's the thing: When you know how to "translate" word problems into algebraic equations, you'll immediately see how useful math really is. With a few simple steps, you can turn real questions about the world around you into equations, allowing you to calculate things that you never thought possible. It's one of the most interesting things about math -- so let's get started!

## How to tell when a word problem can become a linear equation

First, we need to keep our eyes open for a number of clues. These clues tell us that we can turn our word problem into a linear equation:

- There are different quantities of things, such as a specific number of people, objects, hours, and so on.
- Each quantity has a clear value. Instead of saying, "There are a few boxes," the word problem needs to tell us how many boxes there are. Are there five? Six? Seven?
- We need to know at least some of these values to build our linear equation. The unknowns can become variables, like "x" or "y."

## How to turn word problems into linear equations

If we follow a few simple steps, we can turn certain word problems into linear equations:

- Take a second to think about the "problem." What are we trying to find out? What is the "variable" in this word problem? Define all of the words carefully. If we can't define the words properly, our equation won't be accurate.
- Turn the word problem into an equation. Plug in all of your known values and use letters like "x" and "y" to represent the unknown variables. Make sure you write out what each variable represents below the equation so you don't forget.
- Solve the equation. Using our math skills, we can now solve the problem and find the values of our variables. We can use a wide range of strategies to solve the equation, including substitution, elimination, or graphing.

## An example of a word problem translated into a linear equation

Now let's see how this all works with an example. Here's our word problem:

We decided to go to a music concert with all our friends, including 12 children and 3 adults. We paid for everyone's tickets for a total of $162. Another group of friends paid $122 for 8 children and 3 adults. How much does a child's ticket cost, and how much does an adult's ticket cost?

1. Understand the problem

We know two values: 12 children and 3 adults cost $162, while 8 children and 3 adults cost $122.

What we don't know is how much an adult's ticket costs, and how much a child's ticket costs.

Let's create variables for those unknowns:

x = the cost of one child's ticket

y = the cost of one adult's ticket

2. Translate the problem into an equation

We know that 12 children and 3 adults cost $162. Let's plug in our variables and create an equation based on this:

12x + 3y = 162

We can do the same for the other group of concert-goers:

8x + 3y = 122

3. Solve the equation

On one weekend they sold a total of 12 adult tickets and 3 child tickets for a total of 162 dollars, and the next weekend they sold 8 adult tickets and 3 child tickets for 122 dollars, find the price for a child's and adult's ticket.

When we have two very similar equations like this, we can simply subtract them from each other to get the values we need:

Now that we know the value of x, we can use it to find y.

12(10) + 3y = 162

120 + 3y = 162

Now we know that a child's ticket costs $10, while an adult's ticket costs $14.

We can now check our work by plugging our solutions back into our original equations:

12(10) + 3(14) = 162

8(10) + 3(14) = 122

You can use this strategy to solve all kinds of everyday math problems you encounter in life!

## Topics related to the Writing Systems of Linear Equations from Word Problems

Consistent and Dependent Systems

Word Problems

Cramer's Rule

## Flashcards covering the Writing Systems of Linear Equations from Word Problems

Algebra 1 Flashcards

College Algebra Flashcards

## Practice tests covering the Writing Systems of Linear Equations from Word Problems

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

## Get more help with linear equation word problems

Does your child need a little extra help? Are they craving new challenges? Contact Varsity Tutors today, and we will find them a professional math tutor whose skills match your student's unique needs. Whether they need more help identifying elements from a word problem to plug into a potential linear equation or they need more challenging problems, a tutor can provide exactly the type of assistance your student needs.

## Download our free learning tools apps and test prep books

## Solving Word Problems with Linear Systems

Description.

Read this article and watch the video. The article describes examples in which systems of equations can be used to solve real-world quantities. After you review, complete problems 1 to 4 and check your answers.

## Table of contents

Applications of linear systems, real-world application: yearly membership , real-world application: comparing options , real-world application: price of fruit , review (answers).

In this section, we'll see how consistent, inconsistent, and dependent systems might arise in applications.

Source: CK-12, https://www.ck12.org/algebra/Applications-of-Linear-Systems/lesson/Applications-of-Linear-Systems-ALG-I/

The movie rental store CineStar offers customers two choices. Customers can pay a yearly membership of $45 and then rent each movie for $2 or they can choose not to pay the membership fee and rent each movie for $3.50. How many movies would you have to rent before the membership becomes the cheaper option?

Let's translate this problem into algebra. Since there are two different options to consider, we can write two different equations and form a system.

Our system of equations is:

Here's a graph of the system:

You would have to rent 30 movies per year before the membership becomes the better option.

This example shows a real situation where a consistent system of equations is useful in finding a solution. Remember that for a consistent system, the lines that make up the system intersect at single point. In other words, the lines are not parallel or the slopes are different.

In this case, the slopes of the lines represent the price of a rental per movie. The lines cross because the price of rental per movie is different for the two options in the problem

Two movie rental stores are in competition. Movie House charges an annual membership of $30 and charges $3 per movie rental. Flicks for Cheap charges an annual membership of $15 and charges $3 per movie rental. After how many movie rentals would Movie House become the better option?

It should already be clear to see that Movie House will never become the better option, since its membership is more expensive and it charges the same amount per movie as Flicks for Cheap.

The system of equations that describes this problem is:

Let's solve this system by substituting the second equation into the first equation:

This means that the system is inconsistent.

Peter buys two apples and three bananas for $4. Nadia buys four apples and six bananas for $8 from the same store. How much does one banana and one apple costs?

We must write two equations: one for Peter's purchase and one for Nadia's purchase.

Let's solve this system by multiplying the first equation by -2 and adding the two equations:

This statement is always true. This means that the system is dependent.

Looking at the problem again, we can see that we were given exactly the same information in both statements. If Peter buys two apples and three bananas for $4, it makes sense that if Nadia buys twice as many apples (four apples) and twice as many bananas (six bananas) she will pay twice the price ($8). Since the second equation doesn't give us any new information, it doesn't make it possible to find out the price of each fruit.

A baker sells plain cakes for $7 and decorated cakes for $11. On a busy Saturday the baker started with 120 cakes, and sold all but three. His takings for the day were $991. How many plain cakes did he sell that day, and how many were decorated before they were sold?

The baker sold 74 plain cakes and 43 decorated cakes.

- Twice John's age plus five times Claire's age is 204. Nine times John's age minus three times Claire's age is also 204. How old are John and Claire?
- Juan is considering two cell phone plans. The first company charges $120 for the phone and $30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone but charges $45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same?
- Jamal placed two orders with an internet clothing store. The first order was for 13 ties and 4 pairs of suspenders, and totaled $487. The second order was for 6 ties and 2 pairs of suspenders, and totaled $232. The bill does not list the per-item price, but all ties have the same price and all suspenders have the same price. What is the cost of one tie and of one pair of suspenders?
- An airplane took four hours to fly 2400 miles in the direction of the jet-stream. The return trip against the jet-stream took five hours. What were the airplane's speed in still air and the jet-stream's speed?

For questions 5-7, a movie theater charges $4.50 for children and $8.00 for adults.

- On a certain day, 1200 people enter the theater and $8375 is collected. How many children and how many adults attended?
- The next day, the manager announces that she wants to see them take in $10000 in tickets. If there are 240 seats in the house and only five movie showings planned that day, is it possible to meet that goal?
- At the same theater, a 16-ounce soda costs $3 and a 32-ounce soda costs $5. If the theater sells 12,480 ounces of soda for $2100, how many people bought soda? ( Note: Be careful in setting up this problem!)

For questions 8-10, consider the situation: Nadia told Peter that she went to the farmer's market and bought two apples and one banana, and that it cost her $2.50. She thought that Peter might like some fruit, so she went back to the seller and bought four more apples and two more bananas. Peter thanked Nadia, but told her that he did not like bananas, so he would only pay her for four apples. Nadia told him that the second time she paid $6.00 for the fruit.

- What did Peter find when he tried to figure out the price of four apples?
- Nadia then told Peter she had made a mistake, and she actually paid $5.00 on her second trip. Now what answer did Peter get when he tried to figure out how much to pay her?
- Alicia then showed up and told them she had just bought 3 apples and 2 bananas from the same seller for $4.25. Now how much should Peter pay Nadia for four apples?
- Claire is 28 and John is 32.
- Ties are $23, and suspenders are $47.
- The plane flies at 540mph in still air, and the jet stream is 60mph.
- 350 children and 850 adults
- Not possible, the maximum take is $9,600.
- 300 16-oz sodas, and 240 32-oz sodas
- Peter discovered that Nadia would have had to pay a different price for each apple or banana (or both) each time for the second purchase to have been $6.
- Apples cost less than $1.25 each, and bananas cost less than $2.50 each.
- Four apples cost $3.00. Apples cost $0.75 each and bananas $1.00 each.

## Systems of Linear Equations Word Problems - Practice - Expii

Systems of linear equations word problems - practice, explanations (3).

## (Videos) Set up Word Problems Using a System

by mathman1024

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

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

- Define your variables . Since this is a system, there will be two or more variables.
- 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.
- Solve the system. This can be done using the elimination method , substitution method , or graphing .
- State your answer.

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. Let x=one numbery=the other number 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: {x+y=79x−y=23. This can be solved with substitution but you might notice it would be easier with elimination. x+y=79+x−y=232x+0y=1022x=1022x2=1022x=51 Finally, plug this back in to either equation to find the y value. (51)+y=7951−51+y=79−51y=28 The solution to this word problem's system of equations is (51,28).

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, Let C=# of childrenA=# of adults 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: {C+A=884C+7A=553. 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. C+A=88C+A−A=88−AC=88−A Plug this back into the other equation to solve for A. 4(88−A)+7A=553352−4A+7A=553352+3A=553352−352+3A=553−3523A=2013A3=2013A=67 Finally, plug this back into either equation to solve for C. C=88−(67)C=21 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.

## Related Lessons

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.

Image by Clker-Free-Vector-Images via Pixabay ( CC0 )

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.

First, we know that, all together, William bought 20 items. If we say that the number of cupcakes is C and the number of pies is P, we can write the equation:

## 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.

Image source: By Caroline Kulczycky

## IMAGES

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