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- \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
- \mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
- \mathrm{Bob's\:age\:is\:twice\:that\:of\:Barry's.\:Five\:years\:ago,\:Bob\:was\:three\:times\:older\:than\:Barry.\:Find\:the\:age\:of\:both.}
- \mathrm{Two\:men\:who\:are\:traveling\:in\:opposite\:directions\:at\:the\:rate\:of\:18\:and\:22\:mph\:respectively\:started\:at\:the\:same\:time\:at\:the\:same\:place.\:In\:how\:many\:hours\:will\:they\:be\:250\:apart?}
- \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
- How do you solve word problems?
- To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
- How do you identify word problems in math?
- Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
- Is there a calculator that can solve word problems?
- Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
- What is an age problem?
- An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

word-problems-calculator

- High School Math Solutions – Inequalities Calculator, Exponential Inequalities Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential... Read More

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## How to Solve Word Problems Involving the One-Step Equation

In this article, we will guide you through the process of understanding and solving word problems that involve one-step equations.

## A step-by-step guide to solving word problems involving the one-step equation

One-step equations are equations that can be solved in a single step. They are equations of the form ax = b or x/a = b, where a and b are numbers and x is the variable we are solving for.

Word problems that involve one-step equations often ask us to solve for a missing variable given some other information.

Here is a step-by-step guide to solving word problems involving one-step equations:

Step 1: Read the problem carefully and identify the unknown variable.

It is important to understand what the problem is asking and what we are trying to find. Identify the variable that is unknown and needs to be solved.

Step 2: Translate the problem into an equation.

Use the given information in the problem to set up an equation that can be solved for the unknown variable. Make sure to use the correct mathematical operation for the given problem. For example, if the problem involves addition, subtraction, multiplication, or division, use the corresponding mathematical symbol to write the equation.

Step 3: Solve the equation.

Solve the equation by isolating the unknown variable on one side of the equation. If the variable is on the left side, move it to the right side by performing the opposite mathematical operation. If the variable is on the right side, move it to the left side by performing the opposite mathematical operation. Perform the same operation on both sides of the equation to maintain balance.

Step 4: Check the solution.

Plug the solution obtained in Step 3 into the original equation and check whether the equation holds true. If the equation holds true, the solution is correct.

Step 5: Write the final answer.

Write the final answer in a sentence or in the appropriate units, depending on what the problem is asking for.

Note that the steps may need to be adapted depending on the specific problem. However, following this guide can help make the process of solving one-step equation word problems more straightforward.

## Word Problems Involving the One-Step Equation – Example 1

How many packs of popcorn (p) can you buy for $52 if one package costs $4? Solution: List keywords and phrases in the problem to write the equation. Each package costs $4. The total money is $52. How many packs of popcorn is p. The model is a multiplication equation. \(4.p=52\) Solve it by the inverse operation. \(4.p=52→p=52÷4=13\) You can buy 13 packs of popcorn for $52.

## Word Problems Involving the One-Step Equation – Example 2

At a restaurant, Maria and her three friends decided to divide the bill evenly. If each person paid $21 then what was the total bill (b)? Solution: List keywords and phrases in the problem to write the equation. There were 4 people. Each person paid $21. What was the total bill b. The model is a division equation. \(\frac{b}{4}=21\) Solve it by the inverse operation. \(\frac{b}{4}=21→b=21×4=84\) The total bill was $84.

by: Effortless Math Team about 7 months ago (category: Articles )

## Effortless Math Team

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## Mastering Grade 5 Math Word Problems The Ultimate Guide to Tackling 5th Grade Math Word Problems

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## Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

## Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

## Turning English into Algebra

To turn the English into Algebra it helps to:

- Read the whole thing first
- Do a sketch if possible
- Assign letters for the values
- Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

## Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

## Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

- Let S = dollars Sam has
- Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

## Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

- Let D = number of dogs
- Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

## Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

- Use w for width of rectangle: w = 12m
- Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

## Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

- Use S for how many games Sam played
- Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

## Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

- Use a for Alex's work rate
- Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

## Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

- The number of "5 hour" days: d
- The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

## Example: A circle has an area of 12 mm 2 , what is its radius?

- Use A for Area: A = 12 mm 2
- Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

## Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

- Use V for Volume
- Use A for Area
- Use s for side length of cube
- Volume of a cube: V = s 3
- Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

## Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

- Joel's normal rate of pay: $N per hour
- Joel works for 40 hours at $N per hour = $40N
- When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
- Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
- And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

## Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

- Present Value PV = $2,000
- Interest Rate (as a decimal): r = 0.11
- Number of Periods: n = 3
- Future Value (the value we want): FV

We are being asked for the Future Value: FV

## Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

- Present Value PV = $1,000
- Interest Rate (the value we want): r
- Number of Periods: n = 9
- Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

## Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

- Number of boys now: b
- Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

## Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

- We could try, say, n=10: 10(12) = 120 NO (too small)
- Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

## Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side. Your client has 56 square meters of beautiful marble tiles to cover the whole area. What should the length of the room be?

Let's first make a sketch so we get things right!:

- the length of the room: L
- the width of the room: W
- the total Area including veranda: A
- the width of the room is half its length: W = ½L
- the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

This is a quadratic equation , there are many ways to solve it, this time let's use factoring :

And so L = 8 or −14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?

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I wrote ____ in the place of total marbles since that is what the problem is asking for (the unknown).

All of this may look oversimplified, but helping children to see the underlying relationship between the quantities is important. Consider now this problem:

Example: Jenny and Kenny together have 37 marbles, and Kenny has 15. How many does Jenny have? Many teachers might try to explain this as a subtraction problem, but in the most fundamental level it is about addition! It still talks about two people having certain amount of marbles together . The relationship between the quantities is the SAME as above, so we still need to write an addition equation. Relationship: Jenny's marbles + Kenny's marbles = Total marbles Equation: _____ + 15 = 37 Then, we can solve the equation ____ + 15 = 37 by subtracting. Using this kind of approach in the elementary grades will help children to set up equations in algebra story problems later.

Example : Jenny, Kenny, and Penny together have 51 marbles. Kenny has double as many marbles as Jenny has, and Penny has 12. How many does Jenny have? The relationship between the quantities is the same, so it is solved the same way: by writing an addition equation. However, we need to denote the number of Jenny's and Kenny's marbles with something. Jenny's marbles are unknown, so we can denote that with the variable n . Then Kenny has 2 n marbles. Relationship: Jenny's marbles + Kenny's marbles + Penny's marbles = Total marbles Equation: n + 2 n + 12 = 51

Example: Jane is on page 79 of her book. The book has 254 pages. How many pages does she still have to read? This time the word " still " clues us in to an additive relationship where one of the addends is missing. You can initially write an empty line for what is not known, and later replace that with a variable. pages already read + pages still to read = pages total + = This equation is of course is then solved by subtraction, but it is better if you view it as an addition situation and write an addition equation for it.

Example: The number of hours that were left in the day was one-third of the number of hours already passed. How many hours were left in the day? (From Grade 5 word problems for kids ) Can you see the general principle governing this problem? It talks about the hours of the day where some hours already passed and some hours left. This, of course, points to addition once again: we have one part of the day, another part, and a total. The only quantity we know is the total hours for the day. We don't know the hours already passed nor the hours left, so initially you can use two empty lines in the equation that shows the basic relationship between the quantities: hours already passed + hours left = total hours = Then, the information in the first sentence gives us another relationship: "The number of hours that were left in the day was one-third of the number of hours already passed." We don't know the amount of hours passed nor the hours left. So let's use the variable p for the hours passed. Then we can write an expression involving p for the hours left, because "hours left is one-third of the hours passed," or hours left = 1/3 p Then writing 1/3 p for the "hours left" in the first equation will give us: hours already passed + hours left = total hours p + 1/3 p = 24 This can be solved using basic algebra or by guess & check.

## Subtraction word problems

One situation that indicates subtraction is difference or how many/much more . However, the presence of the word "more" can indicate either an addition or subtraction, so be careful.

Example: Ted read 17 pages today, and Fred read 28. How many more pages did Fred read? The solution is of course 28 − 17 = 11, but it's not enough to simply announce that – children need also to understand that difference is the result of a subtraction and tells the answer to how many more . Relationship: Pages Fred read − pages Ted read = difference Equation: 28 − 17 = __ Example: Greg has 17 more marbles than Jack. Jack has 15. How many does Greg have? Here the word more has a different meaning. This problem is not about the difference. The question asks how many does Greg have – not what is the difference in the amounts of marbles. It simply states Greg has 17 more compared to Jack, so here the word more simply indicates addition: Greg has as many as Jack AND 17 more, so Greg has 15 + 17 marbles.

Example: The mass of the Great Pyramid is 557t greater than that of the Leaning Tower of Pisa. Stone Henge has a mass of 2695t which is 95t less than the Leaning Tower of Pisa. There once was a Greater Pyramid which had a mass twice that of the Great Pyramid. What was the mass of the Greater Pyramid? (From Grade 5 word problems for kids ) Each of the first three sentences give information that can be translated into an equation. The question is not about how many more so it's not about difference. One thing being greater than another implies you add. One thing being less than another implies you subtract. And one thing being twice something indicates multiplying by 2. When I read this problem, I could immediately see that I could write equations from the different sentences in the problem, but I couldn't see the answer right away. I figured that after writing the equations I would see some way forwad; probably one equation is solved and gives an answer to another equation. The first sentence says, "The mass of the Great Pyramid is 557t greater than that of the Leaning Tower of Pisa". What are the quantities and the relationship between them here? mass of Great Pyramid = mass of the Leaning Tower of Pisa + 557t The second sentence says "Stonehenge has a mass of 2695t which is 95t less than the Leaning Tower of Pisa." Here it gives you a relationship similar to the one above, and it actually spells out the mass of Stonehenge. It's like two separate pieces of information: "Stonehenge weighs 95t less than the tower. Stonehenge weighs 2695t." Less means you subtract. If you have trouble deciding which is subtracted from which, you can think in your mind which is heavier: Stonehenge or the tower? either mass of Stonehenge = mass of tower − 95t or mass of tower = mass of Stonehenge − 95t Now since the mass of Stonehenge is given, you can solve this equation, and from that knowledge you can solve the first equation, and from that go on to the mass of the " Greater Pyramid ".

If the teacher just jumps directly to the number sentences when solving word problems, then the students won't see the step that happens in the mind before that. The quantities and the relationship between them have to be made clear and written down before fiddling with the actual numbers. Finding this relationship should be the most important part of the word problems. One could even omit the actual calculations and concentrate just finding the quantities and relationships.

## Problem of Helen's hair length

Problem. Helen has 2 inches of hair cut off each time she goes to the hair salon. If h equals the length of hair before she cuts it and c equals the length of hair after she cuts it, which equation would you use to find the length of Helen's hair after she visit the hair salon? a. h = 2 − c c. c = h − 2 b. c = 2 − h d. h = c − 2

Solution. Ignoring the letters c and h for now, what are the quantities? What principle or relationship is there between them? Which possibility of the ones listed below is right? Which do you take away from which?

SIMPLE, isn't it?? In the original problem, the equations are given with the help of h and c instead of the long phrases "hair length before cutting" and "hair length after cutting". You can substitute the c , h , and 2 into the relationships above, and then match the equations (1) - (4) with the equations (a) through (d).

## Helping students to write the algebraic equations

One idea that came to mind is to go through the examples above, and more, based on the typical word problems in the math books, and then turn the whole thing around and have students do exercises such as:

money earned − money spent on this - money spent on that = money left

original price − discount percent x original price = discounted price

money earned each month − expenses/taxes each month = money to use each month AND money to use each month × number of months = money to use over a period of time

speed × time = distance AND distance from A to B + distance from B to C = distance from A to C

Why are math word problems SO difficult for elementary school children? Hint: it has to do with a "recipe" that many math lessons follow.

The do's and don'ts of teaching problem solving in math General advice on how you can teach problem solving in elementary, middle, and high school math.

How I Teach Word Problems by Andre Toom (PDF) This article is written by a Russian who immigrated to US and noticed how COLLEGE LEVEL students have difficulties even with the simplest word problems! He describes his ideas on how to fill in the gap formed when students haven't learned how to solve word problems in earlier education.

A list of websites focusing on word problems and problem solving Use these sites to find good word problems to solve. Most are free!

When solving word problems, students must first decide what quantity represents x and then must write all the other quuantities in terms of x. I teach the students to set up arrows according to the language in the problem. All arrows point to x. Example. Harry had 10 less toys than Mark. Sue has twice as many toys as Harry. Set up arrows: Sue--- Harry---Mark Therefore Mark is x, Harry is x-10 and Sue is 2(x-10). Students find this very helpful. Sandy Denny

My idea is the math teacher might teach and understand the students at the same time and everyone would have a sense of humor. So I think she/he will know if the students are listening or not, when after the class, talk to the student and ask what is wrong. Don't hurt the student's feelings. lorence

## How Do You Write an Equation from a Word Problem?

Having difficulty turning a word problem into an algebra equation? Then this tutorial is for you! With this tutorial, you'll learn how to break down word problems and translate them into mathematical equations.

- word problem
- words to math
- multiple variables
- translate words to algebra
- more than one variable

## Background Tutorials

Introduction to algebraic expressions.

## What is a Variable?

You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!

## Translating Between Words and Math

## What Are Some Words You Can Use To Write Word Problems?

Knowing the mathematical meaning of words allows you to decipher word problems and gives you the power to write your own word problems, too! Take a look at these words and learn their mathematical translations.

## Further Exploration

## How Do You Solve a Word Problem by Writing an Equation and Plugging in Values?

Setting up and solving an equation from a word problem can be tricky, but this tutorial can help. See all the steps, from defining variables to getting that final answer, and everything in between! With this tutorial, you'll learn what it takes to solve a word problem.

## Simplifying Expressions

## How Do You Simplify an Expression?

Simplifying an algebraic expression is a fundamental part of solving math problems. Get some practice putting an expression in simplest form by following along with this tutorial.

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## Algebra Topics - Introduction to Word Problems

Algebra topics -, introduction to word problems, algebra topics introduction to word problems.

## Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

/en/algebra-topics/solving-equations/content/

## What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you've ever taken a math class, you've probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you're supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 - 4 = 8 , so you know Johnny has 8 apples left.

## Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

- Read through the problem carefully, and figure out what it's about.
- Represent unknown numbers with variables.
- Translate the rest of the problem into a mathematical expression.
- Solve the problem.
- Check your work.

We'll work through an algebra word problem using these steps. Here's a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let's go through it step by step.

## Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you're reading, consider:

- What question is the problem asking?
- What information do you already have?

Let's take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There's only one question here. We're trying to find out how many miles Jada drove . Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

- The van cost $30 per day.
- In addition to paying a daily charge, Jada paid $0.50 per mile.
- Jada had the van for 2 days.
- The total cost was $360 .

## Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables . (To learn more about variables, see our lesson on reading algebraic expressions .) You can use a variable in the place of any amount you don't know. Looking at our problem, do you see a quantity we should represent with a variable? It's often the number we're trying to find out.

Since we're trying to find the total number of miles Jada drove, we'll represent that amount with a variable—at least until we know it. We'll use the variable m for miles . Of course, we could use any variable, but m should be easy to remember.

## Step 3: Translate the rest of the problem.

Let's take another look at the problem, with the facts we'll use to solve it highlighted.

The rate to rent a small moving van is $30 per day , plus $0.50 per mile . Jada rented a van to drive to her new home. It took 2 days , and the van cost $360 . How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It's $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360 . The shorter version will be easier to translate into a mathematical expression.

Let's start by translating $30 per day . To calculate the cost of something that costs a certain amount per day, you'd multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅ days, or 30 times the number of days . (Not sure why you'd translate it this way? Check out our lesson on writing algebraic expressions .)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50 , $.50 per mile became $.50 ⋅ mile , and is became = .

Next, we'll add in the numbers and variables we already know. We already know the number of days Jada drove, 2 , so we can replace that. We've also already said we'll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that's left to do is solve it.

## Step 4: Solve the problem.

This problem will take a few steps to solve. (If you're not sure how to do the math in this section, you might want to review our lesson on simplifying expressions .) First, let's simplify the expression as much as possible. We can multiply 30 and 2, so let's go ahead and do that. We can also write .5 ⋅ m as 0.5 m .

60 + .5m = 360

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we'll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides .

The only thing left to get rid of is .5 . Since it's being multiplied with m , we'll do the reverse and divide both sides of the equation with it.

.5 m / .5 is m and 300 / 0.50 is 600 , so m = 600 . In other words, the answer to our problem is 600 —we now know Jada drove 600 miles.

## Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got— 600 —and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada's distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let's take another look at the problem.

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We're done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Let's practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Try completing this problem on your own. When you're done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Here's another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

## Problem 1 Answer

Here's Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Answer: $29

Let's solve this problem step by step. We'll solve it the same way we solved the problem on page 1.

## Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it . Let's look at the problem again. The question is right there in plain sight:

So is the information we'll need to answer the question:

- A single ticket costs $8 .
- The family pass costs $25 more than half the price of the single ticket.

## Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass . We'll represent it with the variable f .

## Step 3: Translate the rest of the problem

Let's look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8 . A family pass costs $25 more than half that . How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25 . To turn this into a problem we can solve, we'll have to translate it into math. Here's how:

- First, replace the cost of a family pass with our variable f .

f equals half of $8 plus $25

- Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

- Finally, translate the rest of the problem. Half of can be written as 1/2 times , or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

## Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

- f is already alone on the left side of the equation, so all we have to do is calculate the right side.
- First, multiply 1/2 by 8 . 1/2 ⋅ 8 is 4 .
- Next, add 4 and 25. 4 + 25 equals 29 .

That's it! f is equal to 29. In other words, the cost of a family pass is $29 .

## Step 5: Check your work

Finally, let's check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let's look at the original problem again.

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

- We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 - 25

- Let's work on the right side first. 29 - 25 is 4 .
- To find the value of s , we have to get it alone on the left side of the equation. This means getting rid of 1/2 . To do this, we'll multiply each side by the inverse of 1/2: 2 .

According to our math, s = 8 . In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that's correct!

So now we're sure about the answer to our problem: The cost of a family pass is $29 .

## Problem 2 Answer

Here's Problem 2:

Answer: $70

Let's go through this problem one step at a time.

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it . What's the question here?

To solve the problem, you'll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

- The amount Flor donated is three times as much the amount Mo donated
- Flor and Mo's donations add up to $280 total

The unknown number we're trying to identify in this problem is Mo's donation . We'll represent it with the variable m .

Here's the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo . Between the two of them, they donated $280 . How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo's donation plus Flor's donation equals $280

Because we know that Flor's donation is three times as much as Mo's donation, we could go even further and say:

Mo's donation plus three times Mo's donation equals $280

We can translate this into a math problem in only a few steps. Here's how:

- Because we've already said we'll represent the amount of Mo's donation with the variable m , let's start by replacing Mo's donation with m .

m plus three times m equals $280

- Next, we can put in mathematical operators in place of certain words. We'll also take out the dollar sign.

m + three times m = 280

- Finally, let's write three times mathematically. Three times m can also be written as 3 ⋅ m , or just 3 m .

m + 3m = 280

It will only take a few steps to solve this problem.

- To get the correct answer, we'll have to get m alone on one side of the equation.
- To start, let's add m and 3 m . That's 4 m .
- We can get rid of the 4 next to the m by dividing both sides by 4. 4 m / 4 is m , and 280 / 4 is 70 .

We've got our answer: m = 70 . In other words, Mo donated $70 .

The answer to our problem is $70 , but we should check just to be sure. Let's look at our problem again.

If our answer is correct, $70 and three times $70 should add up to $280 .

- We can write our new equation like this:

70 + 3 ⋅ 70 = 280

- The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

- The last step is to add 70 and 210. 70 plus 210 equals 280 .

280 is the combined cost of the tickets in our original problem. Our answer is correct : Mo gave $70 to charity.

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## Word Problems – System Of Equations

After you finish this lesson, view all of our Pre-Algebra lessons and practice problems. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

## Solving Word Problems

To solve a word problem using a system of equations , it is important to; – Identify what we don’t know – Declare variables – Use sentences to create equations

An example on how to do this:

Mary and Jose each bought plants from the same store. Mary spent $188 on 7 cherry trees and 11 rose bushes. Jose spent $236 on 13 cherry trees and 11 rose bushes. Find the cost of one cherry tree and the cost of one rose bush.

The y-values cancel each other out, so now you are left with only x-values and real numbers.

Then, you plug in your x-value into an original equation in order to find the y-value.

Cost of a cherry tree: $8 Cost of a rose bush: $12

## Examples of Word Problems – System Of Equations

Let’s solve this by following steps.

1. What we don’t know:

Cost of a single coffee

Cost of single cupcake

2. Declare variables :

3. Use sentences to create equations.

Now, we have a system of equations:

Let’s solve for one of the variables in one of the equations and then use that to substitute into the other.

Combine like terms

And we’ll have

Students in each van

Students in each bus

2. Declare variables:

High School A rented and filled 8 vans and 8 buses with 240 students.

High School B rented and filled 4 vans and 1 bus with 54 students.

## Video-Lesson Transcript

To solve a word problem using system of equations, it is important to: 1. Identify what we don’t know 2. Declare variables. 3. Use sentences to create equations.

Let’s have an example:

Let’s solve this by following steps above.

1. What we don’t know: cost of a cherry tree cost of a rose bush

Now, we have a system of equations

We can solve this by process of substitution, elimination or fraction.

Here we’ll have a negated equation

Let’s do the process of elimination now

We’ll have

Now, we have

Remember, our declared variable?

Now we can say that

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## How to Solve Word Problems in Algebra

Last Updated: December 19, 2022 References

This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been viewed 63,477 times.

You can solve many real world problems with the help of math. In order to familiarize students with these kinds of problems, teachers include word problems in their math curriculum. However, word problems can present a real challenge if you don't know how to break them down and find the numbers underneath the story. Solving word problems is an art of transforming the words and sentences into mathematical expressions and then applying conventional algebraic techniques to solve the problem.

## Assessing the Problem

- For example, you might have the following problem: Jane went to a book shop and bought a book. While at the store Jane found a second interesting book and bought it for $80. The price of the second book was $10 less than three times the price of he first book. What was the price of the first book?
- In this problem, you are asked to find the price of the first book Jane purchased.

- For example, you know that Jane bought two books. You know that the second book was $80. You also know that the second book cost $10 less than 3 times the price of the first book. You don't know the price of the first book.

- Multiplication keywords include times, of, and f actor. [9] X Research source
- Division keywords include per, out of, and percent. [10] X Research source
- Addition keywords include some, more, and together. [11] X Research source
- Subtraction keywords include difference, fewer, and decreased. [12] X Research source

## Finding the Solution

## Completing a Sample Problem

- Robyn and Billy run a lemonade stand. They are giving all the money that they make to a cat shelter. They will combine their profits from selling lemonade with their tips. They sell cups of lemonade for 75 cents. Their mom and dad have agreed to double whatever amount they receive in tips. Write an equation that describes the amount of money Robyn and Billy will give to the shelter.

- Since you are combining their profits and tips, you will be adding two terms. So, x = __ + __.

## Expert Q&A

## Video . By using this service, some information may be shared with YouTube.

- Word problems can have more than one unknown and more the one variable. Thanks Helpful 2 Not Helpful 1
- The number of variables is always equal to the number of unknowns. Thanks Helpful 1 Not Helpful 0
- While solving word problems you should always read every sentence carefully and try to extract all the numerical information. Thanks Helpful 1 Not Helpful 0

## You Might Also Like

- ↑ Daron Cam. Academic Tutor. Expert Interview. 29 May 2020.
- ↑ http://www.purplemath.com/modules/translat.htm
- ↑ https://www.mathsisfun.com/algebra/word-questions-solving.html
- ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut8_probsol.htm
- ↑ http://www.virtualnerd.com/algebra-1/algebra-foundations/word-problem-equation-writing.php
- ↑ https://www.khanacademy.org/test-prep/praxis-math/praxis-math-lessons/praxis-math-algebra/a/gtp--praxis-math--article--algebraic-word-problems--lesson

## About This Article

To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Next, summarize what information you know and what you need to know. Then, assign variables to the unknown quantities. For example, if you know that Jane bought 2 books, and the second book cost $80, which was $10 less than 3 times the price of the first book, assign x to the price of the 1st book. Use this information to write your equation, which is 80 = 3x - 10. To learn how to solve an equation with multiple variables, keep reading! Did this summary help you? Yes No

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## How to Solve Word Problems Requiring Quadratic Equations

Last Updated: December 27, 2020

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Some word problems require quadratic equations in order to be solved. In this article, you will learn how to solve those types of problems. Once you get the hang of it, it will be very easy.

## Quadratic Equations

- For the real life scenarios, factoring method is better.
- In geometric problems, it is good to use the quadratic formula.

## Real Life Scenario

- In this problem, it asks for Kenny's birthday.

- Since negative month does not exist, 3 is the only one that makes sense.
- Because the problem asks for both the month and the date, the answer would be March 18th. (Use the value for the other variable that you found in step 3.)

## Geometric Problems

- In the problem above, it asks you only for the height of the triangle.

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Solving algebraic word problems requires us to combine our ability to create equations and solve them. To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer.

How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform.

Often, word problems appear confusing, and it is difficult to know where to begin. Here are some steps that will make solving word problems easier: Read the problem. Determine what is known and what needs to be found (what is unknown). Try a few numbers to get a general idea of what the solution could be. Write an equation. Solve the equation ...

How to write word problems as equations — Krista King Math | Online math help turning word problems into equations (KristaKingMath) Share Watch on Word problems can seem to be tricky at first. What is the problem actually asking you to do? There are certain phrases that always mean the same operation in math.

Step 1: Read the problem carefully and identify the unknown variable. It is important to understand what the problem is asking and what we are trying to find. Identify the variable that is unknown and needs to be solved. Step 2: Translate the problem into an equation.

How do we solve them? The trick is to break the solution into two parts: Turn the English into Algebra. Then use Algebra to solve. Turning English into Algebra To turn the English into Algebra it helps to: Read the whole thing first Do a sketchif possible Assign lettersfor the values Find or work out formulas

original price − discount percent x original price = discounted price. Write 3 different story problems where the solution is based on the relationships. money earned each month − expenses/taxes each month = money to use each month AND. money to use each month × number of months = money to use over a period of time.

Solution: Adding 9 to a number is written as x + 9, while subtracting 7 from three times the number is written as 3x − 7. We therefore get the equation: x + 9 = 3x − 7. We solve for x by adding 7 on both sides of the equation: x + 16 = 3x. Then we subtract x: 16 = 2x. After dividing by 2, we obtain the answer x = 8.

Word problems are a great way to see math in action! See how to translate a word problem into an equation, solve to find the answer, and check your found answer all in this tutorial. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 ...

0:00 / 6:47 Season 2 Solving Word Problem Using Equations MathTalk 3.58K subscribers Subscribe 344 27K views 4 years ago Are those word problems not adding up? Solve the word...

We also know that the highest grade added to the lowest grade is 138. So, (highest grade) + (lowest grade) = 142. In terms of our variable, Step 5: Solve the equation. Step 6: Answer the question in the problem. The problem asks us to find the lowest grade. We decided that l would be the number, so we have l = 48.

Simplifying an algebraic expression is a fundamental part of solving math problems. Get some practice putting an expression in simplest form by following along with this tutorial. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes ...

Here's how: First, replace the cost of a family pass with our variable f. f equals half of $8 plus $25. Next, take out the dollar signs and replace words like plus and equals with operators. f = half of 8 + 25. Finally, translate the rest of the problem. Half of can be written as 1/2 times, or 1/2 ⋅ : f = 1/2 ⋅ 8 + 25.

To solve a word problem using a system of equations, it is important to; - Identify what we don't know - Declare variables - Use sentences to create equations An example on how to do this: Mary and Jose each bought plants from the same store. Mary spent $188 on 7 cherry trees and 11 rose bushes.

Determine what you are asked to find. In many problems, what you are asked to find is presented in the last sentence. This is not always true, however, so you need to read the entire problem carefully. Write down what you need to find, or else underline it in the problem, so that you do not forget what your final answer means. In an algebra word problem, you will likely be asked to find a ...

The three examples in this section present simple word problems and explain how to solve word problems in algebra. Algebra Word Problem Example 1 Josh and Jeremy bought a total of 12 apples.

But first they'll need to use their reading comprehension skills to decide on the important pieces of information needed to solve each problem. This sixth-grade math worksheet provides essential practice writing and solving one-step equations with multi-digit numbers, and even includes a few problems with decimals. Download Free Worksheet.

Quadratic equations word problem | Algebra (video) | Khan Academy Algebra 1 Course: Algebra 1 > Unit 14 Lesson 10: Quadratic standard form Finding the vertex of a parabola in standard form Graphing quadratics: standard form Graph quadratics in standard form Quadratic word problem: ball Quadratic word problems (standard form) Math > Algebra 1 >

This algebra video tutorial explains how to solve system of equations with 3 variables and with word problems. It contains two example word problems on inve...

Substituting 1/x and 1/y for A and B in the equation, we get: 8 = 1/x + 1/y If we solve this equation for x and y, we will get different values than the correct answer. Therefore, it's important to use the correct units when defining the variables and writing the equation to ensure that the units cancel out correctly and we get the correct answer.

11 years ago No, this would result in the right numbers, but the wrong inequality. You would have gotten w <= 152.5 and w <= 147.5. Which is equal to W being less than 147.5 to -infinity. ( 6 votes)

Since the equation is already given in a basic word problem, the tricky part of solving this type of problem is knowing how to solve an equation involving a radical. The following...

Try it! If you need to use an equation, add or write it in Word. Select Insert > Equation or press Alt + =. To use a built-in formula, select Design > Equation. To create your own, select Design > Equation > Ink Equation. Use your finger, stylus, or mouse to write your equation. Select Insert to bring your equation into the file.

5. Write down any geometric formula that you need to solve the problem. Since the problem gives us the base, the height, and the area of a triangle, we can use the formula. a = b h 2 {\displaystyle a= {\frac {bh} {2}}} 6. Plug the values in the formula. Be sure to use the relationship you got in step three.