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## Frequently Asked Questions (FAQ)

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## 1.20: Word Problems for Linear Equations

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable \(x .\) Twice the variable is \(2 x .\) Adding \(2 x\) to 4 gives:

b) Three times a number is subtracted from 7.

Solution: Three times a number is \(3 x .\) We need to subtract \(3 x\) from 7. This means:\

For example, 8 less than 10 is \(10-8=2\).

d) Subtract \(5 p^{2}-7 p+2\) from \(3 p^{2}+4 p\) and simplify.

Solution: We need to calculate \(3 p^{2}+4 p\) minus \(5 p^{2}-7 p+2:\)

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

e) The amount of money given by \(x\) dimes and \(y\) quarters.

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is \(13 .\) Find the number.

Solution: Translating the problem into an algebraic equation gives:

We solve this for \(x\). First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives \(x=\frac{18}{2}=9\).

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

We solve this as follows. First, add \(x\) :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is \(x=\frac{9}{3}=3\)

Solution: We have the equation:

Dividing both sides by 4 gives the answer: \(x=3\).

We solve for \(x\) by adding 7 on both sides of the equation:

After dividing by \(2,\) we obtain the answer \(x=8\)

Solution: We calculate the price increase as \(5 \% \cdot \$ 2.40 .\) We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

The new price is therefore \(\$ 2.52\).

\[12 \cdot 3 \cdot x=180\nonumber\]

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

Combining the like terms on the left, we get

\[x=\frac{300}{5}=60\nonumber\]

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Distributing and collecting like terms give

f) If a circle has circumference 4in, what is its radius?

Dividing both sides by \(2 \pi\) gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Multiplying both sides by 10 gives

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

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## REAL WORLD PROBLEMS: How to Write Equations Based on Algebra Word Problems

Ok... let's put all this newly learned knowledge to work.

Click here if you need to review how to solve equations.

There are a few rules to remember when writing Algebra equations:

## Writing Equations For Word Problems

- First, you want to identify the unknown, which is your variable. What are you trying to solve for? Identify the variable: Use the statement, Let x = _____. You can replace the x with whatever variable you are using.
- Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem.
- The following key words will help you write equations for Algebra word problems:

## subtraction

Let's look at an example of an algebra word problem.

## Example 1: Algebra Word Problems

- Let x represent the number of children's tickets sold.
- Write an expression to represent the number of adult tickets sold.
- Write an expression to represent the number of senior tickets sold.
- Adult tickets cost $5, children's tickets cost $2, and senior tickets cost $3. Linda made $700. Write an equation to represent the total ticket sales.
- How many children's tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?

## A few notes about this problem

Let x = the number of children’s tickets sold

## Where Can You Find More Algebra Word Problems to Practice?

Click here for more information.

The next example shows how to identify a constant within a word problem.

## Example 2 - Identifying a Constant

1. Write an equation that models this situation.

2. How many minutes were charged on this bill?

## Notes For Example 2

- $12.95 is a monthly rate. Since this is a set fee for each month, I know that this is a constant. The rate does not change; therefore, it is not associated with a variable.
- $0.25 per minute per call requires a variable because the total amount will change based on the number of minutes. Therefore, we use the expression 0.25m
- You must solve the equation to determine the value for m, which is the number of minutes charged.

The last example is a word problem that requires an equation with variables on both sides.

## Example 3 - Equations with Variables on Both Sides

## Notes for Example 3

- $60 and $120 are constants because this is the amount of money that they each have to begin with. This amount does not change.
- $7 per week and $5 per week are rates. They key word "per" in this situation means to multiply.
- The key word "same" in this problem means that I am going to set my two expressions equal to each other.
- When we set the two expressions equal, we now have an equation with variables on both sides.
- After solving the equation, you find that x = 30, which means that after 30 weeks, you and your sister will have the same amount of money.

I'm hoping that these three examples will help you as you solve real world problems in Algebra!

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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?

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

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

## Word problems in algebra

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:

## Step 1: Read through the problem carefully.

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

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.

## 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.

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

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

$30 ⋅ day + $.50 ⋅ mile = $360

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

## Step 4: Solve the problem.

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

## Step 5: Check the problem.

$30 per day and $0.50 per mile

## Problem 1 Answer

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

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

## Step 2: Represent the unknown numbers with variables

## Step 3: Translate the rest of the problem

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

## Step 4: Solve the problem

- 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

- 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 .

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

## Problem 2 Answer

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

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

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

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

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

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:

m plus three times m equals $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 .

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

/en/algebra-topics/distance-word-problems/content/

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## How to Solve Systems of Equations Word Problems? (+FREE Worksheet!)

Learn how to create and solve systems of equations word problems by the elimination method.

## Related Topics

- How to Solve One-Step Equations
- How to Solve One-Step Inequalities
- How to Solve Multi-Step Inequalities
- How to Solve Systems of Equations
- How to Graph Single–Variable Inequalities

## Step by step guide to solve systems of equations word Problems

- Find the key information in the word problem that can help you define the variables.
- Define two variables: \(x\) and \(y\)
- Write two equations.
- Use the elimination method for solving systems of equations.
- Check the solution by substituting the ordered pair into the original equations.

## Systems of Equations Word Problems – Example 1:

## Exercises for Solving Systems of Equations Word Problems

- A farmhouse shelters \(10\) animals, some are pigs and some are ducks. Altogether there are \(36\) legs. How many of each animal are there?
- A class of \(195\) students went on a field trip. They took vehicles, some cars and some buses. Find the number of cars and the number of buses they took if each car holds \(5\) students and each bus hold \(45\) students.
- The difference of two numbers is \(6\). Their sum is \(14\). Find the numbers.
- The sum of the digits of a certain two–digit number is \(7\). Reversing its increasing the number by \(9\). What is the number?
- The difference of two numbers is \(18\). Their sum is \(66\). Find the numbers.

## Download Systems of Equations Word Problems Worksheet

- There are \(8\) pigs and \(2\) ducks.
- There are \(3\) cars and \(4\) buses.
- \(10\) and \(4\).
- \(24\) and \(42\).

by: Reza about 3 years ago (category: Articles )

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## How to Solve Word Problems in Algebra

Last Updated: December 19, 2022 References

## 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.

- 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

## 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. ⧼thumbs_response⧽ Helpful 2 Not Helpful 1
- The number of variables is always equal to the number of unknowns. ⧼thumbs_response⧽ Helpful 1 Not Helpful 0
- While solving word problems you should always read every sentence carefully and try to extract all the numerical information. ⧼thumbs_response⧽ 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

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## Math Lesson 9.2.5 - Equations for solving Word Problems

## Using Other Types of Equations to Solve Word Problems

## Second - order equations with one variable

Let's consider an example to clarify this point.

Hence, the side length of the square is

## First - order equations with two variables

First - order equations with two variables have the general form

Now, we have a first - order equation with one variable, which can be easily solved. Thus,

Hence, the number of girls in the class is

Hence, since x + y = 30 , we have

Hence, the length of the rectangle is 16 cm and the width is 14 cm.

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- Check your calculations for Equations questions with our excellent Equations calculators which contain full equations and calculations clearly displayed line by line. See the Equations Calculators by iCalculator™ below.
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## Basics on the topic Solving Equations: Word Problems

Explain steps to solve algebraic equations. CCSS.MATH.CONTENT.HSA.REI.A.1

## Transcript Solving Equations: Word Problems

## Word Problem Example

## Translating the Word Problem and Writing the Equation

## Solving the Word Problem Example 1

- Here, multiplication is first . 50 · 1 = 50 and 5 · 20 = 100.
- Now we have to add : 50 + 100 = 150.
- 150 = 10x. To isolate x , we use the opposite operation here. We divide both sides by 10.
- x = 15. The ring for Mommy could have 15 diamonds.

## Solving the Word Problem Example 2

- Again we start with multiplication here: 50 · 1 = 50 and 10 · 60 = 600.
- Now let's get rid of the 50 to isolate x. We subtract 50 on both sides. So 5x = 550.
- Now isolate x by dividing both sides by 5.
- x = 110. Dr. Evil will need to sell 110 rocket crocs.

## Solving the Word Problem Example 3

Set this equal to the value and the number of diamonds on the ring. Simplify and solve.

- Multiplication first : 50 · 1 = 50, x · 55 = 55x, 10 · 60 = 600.
- Now to get rid of this 50, we use opposite operations by subtracting 50 from both sides. So 55x = 550.
- To isolate x here we divide both sides by 55.
- x = 10. This means he must sell the rocket crocs for 10 gold bars each.

## Solving Equations: Word Problems exercise

Examine the number of diamonds dr. evil can buy for his mom's ring..

- P aranthesis
- M ultiplication
- S ubtraction.
- $+~\longleftrightarrow~-$ and
- $\times~\longleftrightarrow~\div$.

The solution is a natural number.

After he sells the island and rocket crocs, how many diamonds can Dr. Evil buy for his mom?

- $50\times 1$ gold bars for his island
- $5\times 20$ gold bars for $20$ rocket crocodiles
- All together, he has $50\times 1+5\times 20$ gold bars.

$50\times 1+5\times 20=10\times x$

- First, we multiply: $50+100=10x$,
- Next, we Combine Like Terms on the left side: $150=10x$.
- Finally, we divide both sides of the equation by $10$ to get the solution $x=15$.

## Decide which equation best describes each of the given situations.

You just have to multiply the number of rocket crocodiles by the price per crocodile.

Don't forget Dr. Evil sells his island for $50$ gold bars.

To solve word problems, you must recognize what values are given and what is unknown.

To write an equation, we assign the the variable $x$ to our unknown value.

- The number of diamonds is unknown, but we know the price. So $x$ diamonds cost $10\times x$ gold bars. Write this term on the right side of the equation.
- By selling one island and 20 rocket crocodiles, Dr. Evil earns $50\times 1+5\times 20$ gold bars.
- This leads to the equation $50\times 1+5\times 20= 10\times x$.
- This time, we assign the variable $x$ to the number of rocket crocodiles. Together with the price of the island, we get $50\times 1+5\times x$.
- On the other side of the equation, we have the cost for $60$ diamonds at $10$ gold bars each: $10\times 60$.
- This leads to the equation $50\times 1+5\times x=10\times 60$.
- On the left side we have $50\times 1$ for the island plus $x\times 55$ for the rocket crocodiles, giving us $50\times 1+x\times 55$.
- This leads to the equation $50\times 1+x\times 55=10\times 60$.

## Determine how many hours Dr. Evil's assistant can work.

- Simplify the equation by using PEMDAS.
- Isolate the variable using opposite operations.
- $\text{addition}~\longleftrightarrow~\text{subtraction}$
- $\text{multiplication}~\longleftrightarrow~\text{division}$

Unfortunately, one of the rocket crocodiles exploded, and Dr. Evil's lab is completely destroyed.

- his beach hut for $400$ gold coins
- $40$ piranhas for $3$ gold coins each
- $400\times 1$ gold coins for his beach hut
- $3\times 40$ gold coins for $40$ piranhas
- All together he has $400\times 1+3\times 40$ gold coins.
- $400\times 1+3\times 40=4\times x$
- By multiplying, we get: $400+120=4x$
- Next, we Combine Like Terms on the left side: $520=4x$.
- Last but not least, we divide both sides of the equation by $4$ to get $x=130$.

## Set up equations and solve for the unknown values.

First, assign the variable $x$ to the unknown.

Use PEMDAS to simplify the equation.

To isolate the variable $x$, you have to perform opposite operations.

- First, we assign the variable $x$ to the unknown.
- Next, we formulate an equation.
- Finally, we solve the equation using opposite operations.

$x=5$ means Lara can buy $5$ new teacups.

If Lara sells each book for $\$4.50$, she can buy the $6$ new teacups she wants.

$x=35$ means Lara has to sell $35$ books.

## Summarize what you know about opposite operations.

To solve an equation like $x+3=7$, you must isolate the variable:

Think of an equation like a scale in balance.

We can use Opposite Operations to isolate the variable $x$.

- The opposite operation of addition is subtraction, and vice versa.
- The opposite operation of multiplication is division, and vice versa.

## Solve the following equations.

- Simplify the equation using PEMDAS or the Distributive Property
- Isolate the Variable using Opposite Operations *
- $\text{multiplication}~\longleftrightarrow~\text{division}$.

Some solutions do not equal $-1$, $1$ or $2$.

The other two equations have solutions that do not equal $-1$, $1$ or $2$.

Solving Multi-Step Equations with Variables on One Side

Solving Multi-Step Equations with Variables on Both Sides

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Algebraic word problems are questions that require translating sentences to equations, then solving those equations. The equations we need to write will only involve basic arithmetic operations and a single variable. Usually, the variable represents an unknown quantity in a real-life scenario.

Two-Step Equation Word Problems: Integers Interpret this set of word problems that require two-step operations to solve the equations. Each printable worksheet has five word problems ideal for 6th grade, 7th grade, and 8th grade students. Two-Step Equation Word Problems: Fractions and Decimals

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.

Two-step equations word problems CCSS.Math: 7.EE.B.4, 7.EE.B.4a Google Classroom You might need: Calculator Mindy and Troy combined ate 9 9 pieces of the wedding cake. Mindy ate 3 3 pieces of cake and Troy had \dfrac14 41 of the total cake. Write an equation to determine how many pieces of cake (c) (c) there were in total.

Linear equations word problems. CCSS.Math: HSF.LE.B.5. Google Classroom. Ever since Renata moved to her new home, she's been keeping track of the height of the tree outside her window. H H represents the height of the tree (in centimeters), t t years since Renata moved in. H=210+33t H = 210+ 33t.

Students will draw on their algebra knowledge to answer word problems by writing and then solving two-step equations. 7th grade. Math. Worksheet. Solve One-Step Addition and Subtraction Equations ... Write and Solve Equations From Word Problems. Help learners understand and solve real-world problems using algebraic reasoning with these mixed ...

Solve the following word problems: a) Five times an unknown number is equal to 60. Find the number. Solution: We translate the problem to algebra: 5x = 60 We solve this for x : x = 60 5 = 12 b) If 5 is subtracted from twice an unknown number, the difference is 13. Find the number. Solution: Translating the problem into an algebraic equation gives:

Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem. The following key words will help you write equations for Algebra word problems: Addition altogether increase more plus sum total combine subtraction difference decrease less fewer reduce minus multiplication per times

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.

These word problems involve variables and equations. Students write out a full equation with the suggested variable representing the unknown, then solve the equation. Underlying operations include addition, subtraction, multiplication and division of whole numbers and fractions. Worksheet #1 Worksheet #2 Worksheet #3 Worksheet #4.

Algebra - Solving Equations (word problem) yaymath 243K subscribers Subscribe 127K views 6 years ago Yay Math in Studio presents a single word problem, the likes of which students around...

Name_____ SYSTEM OF EQUATIONS-WORD PROBLEMS #2 . Directions: Find the answers to each situation below by setting up and solving a system of equations.. 1) Ben bought a total of 11 pizzas for his class party. The amount of large pizzas was one less than twice the amount of medium pizzas. How many large and how many medium pizzas did he buy?

Find the key information in the word problem that can help you define the variables. Define two variables: x x and y y Write two equations. Use the elimination method for solving systems of equations. Check the solution by substituting the ordered pair into the original equations. Systems of Equations Word Problems

Systems of equations word problems (with zero and infinite solutions) Systems of equations with elimination: TV & DVD Systems of equations with elimination: apples and oranges Systems of equations with substitution: coins Systems of equations with elimination: coffee and croissants Systems of equations: FAQ Math > Algebra 1 > Systems of equations >

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 ...

Problem 6 sent by Κυριάκος There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original's minus one, and whose last two digits are the same and equal to the half of the original's.

Now we can use the inverse property of multiplication (which is division) to solve for W. We can divide 6 times W by 6; and we also have to do that to the other side so we have 6W/6 = 60/6; then we get the value of W which equals 10 ft. Now we can substitute what we know about length (L). We know that L is equal to 2W.

Learn the steps used in solving word problems which include visualizing the problem, writing the equation, and solving the equation with examples of how it's done. Updated: 10/05/2021 Create an ...

For the larger hose 1/55 of the pool will be filled in a minute. The smaller hose would then be 1/x per minute. Now we can add both hoses together (1/x + 1/55) and set it equal to how much of the pool is filled when both are working (1/35) 1/x +1/55 = 1/35 and solve. 1/x=1/35-1/55. 1/x=4/385.

Equations and Word Problems Task Cards: Use these task cards instead of worksheets to get students practicing solving equations (8 problems), writing equations from real world situations (12 problems), and generating their own story problems represented by a given equation (4 problems). These task cards come in both printable AND digital formats.

Microsoft Math Solver - Math Problem Solver & Calculator Get step-by-step solutions to your math problems Try Math Solver Type a math problem Solve Quadratic equation x2 − 4x − 5 = 0 Trigonometry 4sinθ cosθ = 2sinθ Linear equation y = 3x + 4 Arithmetic 699 ∗533 Matrix [ 2 5 3 4][ 2 −1 0 1 3 5] Simultaneous equation { 8x + 2y = 46 7x + 3y = 47

Using Other Types of Equations to Solve Word Problems There are many types of equations besides the first - order equations with one variable. We will discuss a few examples where two other types of equations (second - order equations with one variable and first - order equations with two variables) are used in word problems.

Solving the Word Problem Example 1 Now, to solve the equation for x, you have to simplify the equation using PEMDAS and opposite operations. Here, multiplication is first. 50 · 1 = 50 and 5 · 20 = 100. Now we have to add: 50 + 100 = 150. 150 = 10x. To isolate x, we use the opposite operation here. We divide both sides by 10. x = 15.