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Fraction word problems

Here you will learn about fraction word problems, including solving math word problems within a real-world context involving adding fractions, subtracting fractions, multiplying fractions, and dividing fractions.

Students will first learn about fraction word problems as part of number and operations—fractions in 4 th grade.

What are fraction word problems?

Fraction word problems are math word problems involving fractions that require students to use problem-solving skills within the context of a real-world situation.

To solve a fraction word problem, you must understand the context of the word problem, what the unknown information is, and what operation is needed to solve it. Fraction word problems may require addition, subtraction, multiplication, or division of fractions.

After determining what operation is needed to solve the problem, you can apply the rules of adding, subtracting, multiplying, or dividing fractions to find the solution.

For example,

Natalie is baking 2 different batches of cookies. One batch needs \cfrac{3}{4} cup of sugar and the other batch needs \cfrac{2}{4} cup of sugar. How much sugar is needed to bake both batches of cookies?

You can follow these steps to solve the problem:

Step-by-step guide: Adding and subtracting fractions

Step-by-step guide: Adding fractions

Step-by-step guide: Subtracting fractions

Step-by-step guide: Multiplying and dividing fractions

Step-by-step guide: Multiplying fractions

Step-by-step guide: Dividing fractions

Common Core State Standards

How does this relate to 4 th grade math to 6 th grade math?

• Grade 4: Number and Operations—Fractions (4.NF.B.3d) Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
• Grade 4: Number and Operations—Fractions (4.NF.B.4c) Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat \cfrac{3}{8} of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie?
• Grade 5: Number and Operations—Fractions (5.NF.A.2) Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result \cfrac{2}{5}+\cfrac{1}{2}=\cfrac{3}{7} by observing that \cfrac{3}{7}<\cfrac{1}{2} .
• Grade 5: Number and Operations—Fractions (5.NF.B.6) Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
• Grade 5: Number and Operations—Fractions (5.NF.B.7c) Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if 3 people share \cfrac{1}{2} \: lb of chocolate equally? How many \cfrac{1}{3} cup servings are in 2 cups of raisins?
• Grade 6: The Number System (6.NS.A.1) Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for \cfrac{2}{3} \div \cfrac{4}{5} and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that \cfrac{2}{3} \div \cfrac{4}{5}=\cfrac{8}{9} because \cfrac{3}{4} of \cfrac{8}{9} is \cfrac{2}{3}. (In general, \cfrac{a}{b} \div \cfrac{c}{d}=\cfrac{a d}{b c} \, ) How much chocolate will each person get if 3 people share \cfrac{1}{2} \: lb of chocolate equally? How many \cfrac{3}{4} cup servings are in \cfrac{2}{3} of a cup of yogurt? How wide is a rectangular strip of land with length \cfrac{3}{4} \: m and area \cfrac{1}{2} \: m^2?

[FREE] Fraction Operations Worksheet (Grade 4 to 6)

Use this quiz to check your grade 4 to 6 students’ understanding of fraction operations. 10+ questions with answers covering a range of 4th to 6th grade fraction operations topics to identify areas of strength and support!

How to solve fraction word problems

In order to solve fraction word problems:

Determine what operation is needed to solve.

Write an equation.

Solve the equation.

State your answer in a sentence.

Fraction word problem examples

Example 1: adding fractions (like denominators).

Julia ate \cfrac{3}{8} of a pizza and her brother ate \cfrac{2}{8} of the same pizza. How much of the pizza did they eat altogether?

The problem states how much pizza Julia ate and how much her brother ate. You need to find how much pizza Julia and her brother ate altogether , which means you need to add.

2 Write an equation.

3 Solve the equation.

To add fractions with like denominators, add the numerators and keep the denominators the same.

4 State your answer in a sentence.

The last step is to go back to the word problem and write a sentence to clearly say what the solution represents in the context of the problem.

Julia and her brother ate \cfrac{5}{8} of the pizza altogether.

Example 2: adding fractions (unlike denominators)

Tim ran \cfrac{5}{6} of a mile in the morning and \cfrac{1}{3} of a mile in the afternoon. How far did Tim run in total?

The problem states how far Tim ran in the morning and how far he ran in the afternoon. You need to find how far Tim ran in total , which means you need to add.

To add fractions with unlike denominators, first find a common denominator and then change the fractions accordingly before adding.

\cfrac{5}{6}+\cfrac{1}{3}= \, ?

The least common multiple of 6 and 3 is 6, so 6 can be the common denominator.

That means \cfrac{1}{3} will need to be changed so that its denominator is 6. To do this, multiply the numerator and the denominator by 2.

\cfrac{1 \times 2}{3 \times 2}=\cfrac{2}{6}

Now you can add the fractions and simplify the answer.

\cfrac{5}{6}+\cfrac{2}{6}=\cfrac{7}{6}=1 \cfrac{1}{6}

Tim ran a total of 1 \cfrac{1}{6} miles.

Example 3: subtracting fractions (like denominators)

Pia walked \cfrac{4}{7} of a mile to the park and \cfrac{3}{7} of a mile back home. How much farther did she walk to the park than back home?

The problem states how far Pia walked to the park and how far she walked home. Since you need to find the difference ( how much farther ) between the two distances, you need to subtract.

To subtract fractions with like denominators, subtract the numerators and keep the denominators the same.

\cfrac{4}{7}-\cfrac{3}{7}=\cfrac{1}{7}

Pia walked \cfrac{1}{7} of a mile farther to the park than back home.

Example 4: subtracting fractions (unlike denominators)

Henry bought \cfrac{7}{8} pound of beef from the grocery store. He used \cfrac{1}{3} of a pound of beef to make a hamburger. How much of the beef does he have left?

The problem states how much beef Henry started with and how much he used. Since you need to find how much he has left , you need to subtract.

To subtract fractions with unlike denominators, first find a common denominator and then change the fractions accordingly before subtracting.

\cfrac{7}{8}-\cfrac{1}{3}= \, ?

The least common multiple of 8 and 3 is 24, so 24 can be the common denominator.

That means both fractions will need to be changed so that their denominator is 24.

To do this, multiply the numerator and the denominator of each fraction by the same number so that it results in a denominator of 24. This will give you an equivalent fraction for each fraction in the problem.

\begin{aligned}&\cfrac{7 \times 3}{8 \times 3}=\cfrac{21}{24} \\\\ &\cfrac{1 \times 8}{3 \times 8}=\cfrac{8}{24} \end{aligned}

Now you can subtract the fractions.

\cfrac{21}{24}-\cfrac{8}{24}=\cfrac{13}{24}

Henry has \cfrac{13}{24} of a pound of beef left.

Example 5: multiplying fractions

Andre has \cfrac{3}{4} of a candy bar left. He gives \cfrac{1}{2} of the remaining bit of the candy bar to his sister. What fraction of the whole candy bar does Andre have left now?

It could be challenging to determine the operation needed for this problem; many students may automatically assume it is subtraction since you need to find how much of the candy bar is left.

However, since you know Andre started with a fraction of the candy bar and you need to find a fraction OF a fraction, you need to multiply.

The difference here is that Andre did NOT give his sister \cfrac{1}{2} of the candy bar, but he gave her \cfrac{1}{2} of \cfrac{3}{4} of a candy bar.

To solve the word problem, you can ask, “What is \cfrac{1}{2} of \cfrac{3}{4}? ” and set up the equation accordingly. Think of the multiplication sign as meaning “of.”

\cfrac{1}{2} \times \cfrac{3}{4}= \, ?

To multiply fractions, multiply the numerators and multiply the denominators.

\cfrac{1}{2} \times \cfrac{3}{4}=\cfrac{3}{8}

Andre gave \cfrac{1}{2} of \cfrac{3}{4} of a candy bar to his sister, which means he has \cfrac{1}{2} of \cfrac{3}{4} left. Therefore, Andre has \cfrac{3}{8} of the whole candy bar left.

Example 6: dividing fractions

Nia has \cfrac{7}{8} cup of trail mix. How many \cfrac{1}{4} cup servings can she make?

The problem states the total amount of trail mix Nia has and asks how many servings can be made from it.

To solve, you need to divide the total amount of trail mix (which is \cfrac{7}{8} cup) by the amount in each serving ( \cfrac{1}{4} cup) to find out how many servings she can make.

To divide fractions, multiply the dividend by the reciprocal of the divisor.

\begin{aligned}& \cfrac{7}{8} \div \cfrac{1}{4}= \, ? \\\\ & \downarrow \downarrow \downarrow \\\\ &\cfrac{7}{8} \times \cfrac{4}{1}=\cfrac{28}{8} \end{aligned}

You can simplify \cfrac{28}{8} to \cfrac{7}{2} and then 3 \cfrac{1}{2}.

Nia can make 3 \cfrac{1}{2} cup servings.

Teaching tips for fraction word problems

• Encourage students to look for key words to help determine the operation needed to solve the problem. For example, subtracting fractions word problems might ask students to find “how much is left” or “how much more” one fraction is than another.
• Provide students with an answer key to word problem worksheets to allow them to obtain immediate feedback on their solutions. Encourage students to attempt the problems independently first, then check their answers against the key to identify any mistakes and learn from them. This helps reinforce problem-solving skills and confidence.
• Be sure to incorporate real-world situations into your math lessons. Doing so allows students to better understand the relevance of fractions in everyday life.
• As students progress and build a strong foundational understanding of one-step fraction word problems, provide them with multi-step word problems that involve more than one operation to solve.
• Take note that students will not divide a fraction by a fraction as shown above until 6 th grade (middle school), but they will divide a unit fraction by a whole number and a whole number by a fraction in 5 th grade (elementary school), where the same mathematical rules apply to solving.
• There are many alternatives you can use in place of printable math worksheets to make practicing fraction word problems more engaging. Some examples are online math games and digital workbooks.

Easy mistakes to make

• Misinterpreting the problem Misreading or misunderstanding the word problem can lead to solving for the wrong quantity or using the wrong operation.
• Not finding common denominators When adding or subtracting fractions with unlike denominators, students may forget to find a common denominator, leading to an incorrect answer.
• Forgetting to simplify Unless a problem specifically says not to simplify, fractional answers should always be written in simplest form.

Related fractions operations lessons

• Fractions operations
• Multiplicative inverse
• Reciprocal math
• Fractions as divisions

Practice fraction word problem questions

1. Malia spent \cfrac{5}{6} of an hour studying for a math test. Then she spent \cfrac{1}{3} of an hour reading. How much longer did she spend studying for her math test than reading?

Malia spent \cfrac{1}{2} of an hour longer studying for her math test than reading.

Malia spent \cfrac{5}{18} of an hour longer studying for her math test than reading.

Malia spent \cfrac{1}{2} of an hour longer reading than studying for her math test.

Malia spent 1 \cfrac{1}{6} of an hour longer studying for her math test than reading.

To find the difference between the amount of time Malia spent studying for her math test than reading, you need to subtract. Since the fractions have unlike denominators, you need to find a common denominator first.

You can use 6 as the common denominator, so \cfrac{1}{3} becomes \cfrac{3}{6}. Then you can subtract.

\cfrac{3}{6} can then be simplified to \cfrac{1}{2}.

Finally, you need to choose the answer that correctly answers the question within the context of the situation. Therefore, the correct answer is “Malia spent \cfrac{1}{2} of an hour longer studying for her math test than reading.”

2. A square garden is \cfrac{3}{4} of a meter wide and \cfrac{8}{9} of a meter long. What is its area?

The area of the garden is 1\cfrac{23}{36} square meters.

The area of the garden is \cfrac{27}{32} square meters.

The area of the garden is \cfrac{2}{3} square meters.

The perimeter of the garden is \cfrac{2}{3} meters.

To find the area of a square, you multiply the length and width. So to solve, you multiply the fractional lengths by mulitplying the numerators and multiplying the denominators.

\cfrac{24}{36} can be simplified to \cfrac{2}{3}. 

Therefore, the correct answer is “The area of the garden is \cfrac{2}{3} square meters.”

3. Zoe ate \cfrac{3}{8} of a small cake. Liam ate \cfrac{1}{8} of the same cake. How much more of the cake did Zoe eat than Liam?

Zoe ate \cfrac{3}{64} more of the cake than Liam.

Zoe ate \cfrac{1}{4} more of the cake than Liam.

Zoe ate \cfrac{1}{8} more of the cake than Liam.

Liam ate \cfrac{1}{4} more of the cake than Zoe.

To find how much more cake Zoe ate than Liam, you subtract. Since the fractions have the same denominator, you subtract the numerators and keep the denominator the same.

\cfrac{2}{8} can be simplified to \cfrac{1}{4}. 

Therefore, the correct answer is “Zoe ate \cfrac{1}{4} more of the cake than Liam.”

4. Lila poured \cfrac{11}{12} cup of pineapple and \cfrac{2}{3} cup of mango juice in a bottle. How many cups of juice did she pour into the bottle altogether?

Lila poured 1 \cfrac{7}{12} cups of juice in the bottle altogether.

Lila poured \cfrac{1}{4} cups of juice in the bottle altogether.

Lila poured \cfrac{11}{18} cups of juice in the bottle altogether.

Lila poured 1 \cfrac{3}{8} cups of juice in the bottle altogether.

To find the total amount of juice that Lila poured into the bottle, you need to add. Since the fractions have unlike denominators, you need to find a common denominator first.

You can use 12 as the common denominator, so \cfrac{2}{3} becomes \cfrac{8}{12}.  Then you can add.

\cfrac{19}{12} can be simplified to 1 \cfrac{7}{12}. 

Therefore, the correct answer is “Lila poured 1 \cfrac{7}{12} cups of juice in the bottle altogether.”

5. Killian used \cfrac{9}{10} of a gallon of paint to paint his living room and \cfrac{7}{10} of a gallon to paint his bedroom. How much paint did Killian use in all?

Killian used \cfrac{2}{10} gallons of paint in all.

Killian used \cfrac{1}{5} gallons of paint in all.

Killian used \cfrac{63}{100} gallons of paint in all.

Killian used 1 \cfrac{3}{5} gallons of paint in all.

To find the total amount of paint Killian used, you add the amount he used for the living room and the amount he used for the kitchen. Since the fractions have the same denominator, you add the numerators and keep the denominators the same.

\cfrac{16}{10} can be simplified to 1 \cfrac{6}{10} and then further simplified to 1 \cfrac{3}{5}.

Therefore, the correct answer is “Killian used 1 \cfrac{3}{5} gallons of paint in all.”

6. Evan pours \cfrac{4}{5} of a liter of orange juice evenly among some cups.

He put \cfrac{1}{10} of a liter into each cup. How many cups did Evan fill?

Evan filled \cfrac{2}{25} cups.

Evan filled 8 cups.

Evan filled \cfrac{9}{10} cups.

Evan filled 7 cups.

To find the number of cups Evan filled, you need to divide the total amount of orange juice by the amount being poured into each cup. To divide fractions, you mulitply the first fraction (the dividend) by the reciprocal of the second fraction (the divisor).

\cfrac{40}{5} can be simplifed to 8.

Therefore, the correct answer is “Evan filled 8 cups.”

Fraction word problems FAQs

Fraction word problems are math word problems involving fractions that require students to use problem-solving skills within the context of a real-world situation. Fraction word problems may involve addition, subtraction, multiplication, or division of fractions.

To solve fraction word problems, first you need to determine the operation. Then you can write an equation and solve the equation based on the arithmetic rules for that operation.

Fraction word problems and decimal word problems are similar because they both involve solving math problems within real-world contexts. Both types of problems require understanding the problem, determining the operation needed to solve it (addition, subtraction, multiplication, division), and solving it based on the arithmetic rules for that operation.

The next lessons are

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How to Solve Fraction Questions in Math

Last Updated: April 14, 2024 Fact Checked

This article was co-authored by Mario Banuelos, PhD and by wikiHow staff writer, Sophia Latorre . Mario Banuelos is an Associate Professor of Mathematics at California State University, Fresno. With over eight years of teaching experience, Mario specializes in mathematical biology, optimization, statistical models for genome evolution, and data science. Mario holds a BA in Mathematics from California State University, Fresno, and a Ph.D. in Applied Mathematics from the University of California, Merced. Mario has taught at both the high school and collegiate levels. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,201,367 times.

Fraction questions can look tricky at first, but they become easier with practice and know-how. Start by learning the terminology and fundamentals, then pratice adding, subtracting, multiplying, and dividing fractions. [1] X Research source Once you understand what fractions are and how to manipulate them, you'll be breezing through fraction problems in no time.

Doing Calculations with Fractions

• For instance, to solve 5/9 + 1/9, just add 5 + 1, which equals 6. The answer, then, is 6/9 which can be reduced to 2/3.

• For instance, to solve 6/8 - 2/8, all you do is take away 2 from 6. The answer is 4/8, which can be reduced to 1/2.

• For example, if you need to add 1/2 and 2/3, start by determining a common multiple. In this case, the common multiple is 6 since both 2 and 3 can be converted to 6. To turn 1/2 into a fraction with a denominator of 6, multiply both the numerator and denominator by 3: 1 x 3 = 3 and 2 x 3 = 6, so the new fraction is 3/6. To turn 2/3 into a fraction with a denominator of 6, multiply both the numerator and denominator by 2: 2 x 2 = 4 and 3 x 2 = 6, so the new fraction is 4/6. Now, you can add the numerators: 3/6 + 4/6 = 7/6. Since this is an improper fraction, you can convert it to the mixed number 1 1/6.
• On the other hand, say you're working on the problem 7/10 - 1/5. The common multiple in this case is 10, since 1/5 can be converted into a fraction with a denominator of 10 by multiplying it by 2: 1 x 2 = 2 and 5 x 2 = 10, so the new fraction is 2/10. You don't need to convert the other fraction at all. Just subtract 2 from 7, which is 5. The answer is 5/10, which can also be reduced to 1/2.

• For instance, to multiply 2/3 and 7/8, find the new numerator by multiplying 2 by 7, which is 14. Then, multiply 3 by 8, which is 24. Therefore, the answer is 14/24, which can be reduced to 7/12 by dividing both the numerator and denominator by 2.

• For example, to solve 1/2 ÷ 1/6, flip 1/6 upside down so it becomes 6/1. Then just multiply 1 x 6 to find the numerator (which is 6) and 2 x 1 to find the denominator (which is 2). So, the answer is 6/2 which is equal to 3.

Joseph Meyer

Think about fractions as portions of a whole. Imagine dividing objects like pizzas or cakes into equal parts. Visualizing fractions this way improves comprehension, compared to relying solely on memorization. This approach can be helpful when adding, subtracting, and comparing fractions.

Practicing the Basics

• For instance, in 3/5, 3 is the numerator so there are 3 parts and 5 is the denominator so there are 5 total parts. In 7/8, 7 is the numerator and 8 is the denominator.

• If you need to turn 7 into a fraction, for instance, write it as 7/1.

• For example, if you have the fraction 15/45, the greatest common factor is 15, since both 15 and 45 can be divided by 15. Divide 15 by 15, which is 1, so that's your new numerator. Divide 45 by 15, which is 3, so that's your new denominator. This means that 15/45 can be reduced to 1/3.

• Say you have the mixed number 1 2/3. Stary by multiplying 3 by 1, which is 3. Add 3 to 2, the existing numerator. The new numerator is 5, so the mixed fraction is 5/3.

Tip: Typically, you'll need to convert mixed numbers to improper fractions if you're multiplying or dividing them.

• Say that you have the improper fraction 17/4. Set up the problem as 17 ÷ 4. The number 4 goes into 17 a total of 4 times, so the whole number is 4. Then, multiply 4 by 4, which is equal to 16. Subtract 16 from 17, which is equal to 1, so that's the remainder. This means that 17/4 is the same as 4 1/4.

Fraction Calculator, Practice Problems, and Answers

Community Q&A

• Check with your teacher to find out if you need to convert improper fractions into mixed numbers and/or reduce fractions to their lowest terms to get full marks. Thanks Helpful 2 Not Helpful 1
• Take the time to carefully read through the problem at least twice so you can be sure you know what it's asking you to do. Thanks Helpful 2 Not Helpful 2
• To take the reciprocal of a whole number, just put a 1 over it. For example, 5 becomes 1/5. Thanks Helpful 1 Not Helpful 1

You Might Also Like

• ↑ https://www.sparknotes.com/math/prealgebra/fractions/terms/
• ↑ https://www.bbc.co.uk/bitesize/articles/z9n4k7h
• ↑ https://www.mathsisfun.com/fractions_multiplication.html
• ↑ https://www.mathsisfun.com/fractions_division.html
• ↑ https://medium.com/i-math/the-no-nonsense-straightforward-da76a4849ec
• ↑ https://www.youtube.com/watch?v=PcEwj5_v75g
• ↑ https://sciencing.com/solve-math-problems-fractions-7964895.html

About This Article

To solve a fraction multiplication question in math, line up the 2 fractions next to each other. Multiply the top of the left fraction by the top of the right fraction and write that answer on top, then multiply the bottom of each fraction and write that answer on the bottom. Simplify the new fraction as much as possible. To divide fractions, flip one of the fractions upside-down and multiply them the same way. If you need to add or subtract fractions, keep reading! Did this summary help you? Yes No

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Fraction Word Problems (Difficult)

Here are some examples of more difficult fraction word problems. We will illustrate how block models (tape diagrams) can be used to help you to visualize the fraction word problems in terms of the information given and the data that needs to be found.

Related Pages Fraction Word Problems Singapore Math Lessons Fraction Problems Using Algebra Algebra Word Problems

Block modeling (also known as tape diagrams or bar models) are widely used in Singapore Math and the Common Core to help students visualize and understand math word problems.

Example: 2/9 of the people on a restaurant are adults. If there are 95 more children than adults, how many children are there in the restaurant?

Solution: Draw a diagram with 9 equal parts: 2 parts to represent the adults and 7 parts to represent the children.

5 units = 95 1 unit = 95 ÷ 5 = 19 7 units = 7 × 19 = 133

Answer: There are 133 children in the restaurant.

Example: Gary and Henry brought an equal amount of money for shopping. Gary spent $95 and Henry spent $350. After that Henry had 4/7 of what Gary had left. How much money did Gary have left after shopping?

350 – 95 = 255 3 units = 255 1 unit = 255 ÷ 3 = 85 7 units = 85 × 7 = 595

Answer: Gary has $595 after shopping.

Example: 1/9 of the shirts sold at Peter’s shop are striped. 5/8 of the remainder are printed. The rest of the shirts are plain colored shirts. If Peter’s shop has 81 plain colored shirts, how many more printed shirts than plain colored shirts does the shop have?

Solution: Draw a diagram with 9 parts. One part represents striped shirts. Out of the remaining 8 parts: 5 parts represent the printed shirts and 3 parts represent plain colored shirts.

3 units = 81 1 unit = 81 ÷ 3 = 27 Printed shirts have 2 parts more than plain shirts. 2 units = 27 × 2 = 54

Answer: Peter’s shop has 54 more printed colored shirts than plain shirts.

Solve a problem involving fractions of fractions and fractions of remaining parts

Example: 1/4 of my trail mix recipe is raisins and the rest is nuts. 3/5 of the nuts are peanuts and the rest are almonds. What fraction of my trail mix is almonds?

How to solve fraction word problem that involves addition, subtraction and multiplication using a tape diagram or block model

Example: Jenny’s mom says she has an hour before it’s bedtime. Jenny spends 3/5 of the hour texting a friend and 3/8 of the remaining time brushing her teeth and putting on her pajamas. She spends the rest of the time reading her book. How long did Jenny read?

How to solve a four step fraction word problem using tape diagrams?

Example: In an auditorium, 1/6 of the students are fifth graders, 1/3 are fourth graders, and 1/4 of the remaining students are second graders. If there are 96 students in the auditorium, how many second graders are there?

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Undoing Subtraction

We can still add the same amount to both sides of an equation without changing the solution.

Solve for x : \(x - \frac{5}{6} = \frac{1}{3}\).

To “undo” subtracting 5/6, add 5/6 to both sides of the equation and simplify.

\[ \begin{aligned} x - \frac{5}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Original equation.}} \\ x - \frac{5}{6} + \frac{5}{6} = \frac{1}{3} + \frac{5}{6} ~ & \textcolor{red}{ \text{ Add } \frac{5}{6} \text{ to both sides.}} \\ x = \frac{1 \cdot 2}{3 \cdot 2} + \frac{5}{6} ~ & \textcolor{red}{ \text{ Equivalent fractions, LCD = 6.}} \\ x = \frac{2}{6} + \frac{5}{6} ~ & \textcolor{red}{ \text{ Simplify.}} \\ x = \frac{7}{6} ~ & \textcolor{red}{ \text{ Add.}} \end{aligned}\nonumber \]

It is perfectly acceptable to leave your answer as an improper fraction. If you desire, or if you are instructed to do so, you can change your answer to a mixed fraction (7 divided by 6 is 1 with a remainder of 1). That is \(x = 1 \frac{1}{6}\).

Checking the Solution

Substitute 7/6 for x in the original equation and simplify.

\[ \begin{aligned} x - \frac{5}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{7}{6} - \frac{5}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Substitute 7/6 for } x.} \\ \frac{2}{6} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Subtract.}} \\ \frac{1}{3} = \frac{1}{3} ~ & \textcolor{red}{ \text{ Reduce.}} \end{aligned}\nonumber \]

Because the last statement is true, we conclude that 7/6 is a solution of the equation x − 5/6 = 1/3.

Undoing Addition

You can still subtract the same amount from both sides of an equation without changing the solution.

Solve for x : \(x + \frac{2}{3} = - \frac{3}{5}\).

To “undo” adding 2/3, subtract 2/3 from both sides of the equation and simplify.

\[ \begin{aligned} x + \frac{2}{3} = - \frac{3}{5} ~ & \textcolor{red}{ \text{ Original equation.}} \\ x + \frac{2}{3} - \frac{2}{3} = - \frac{3}{5} - \frac{2}{3} ~ & \textcolor{red}{ \text{ Subtract } \frac{2}{3} \text{ from both sides.}} \\ x = - \frac{3 \cdot 3}{5 \cdot 3} - \frac{2 \cdot 5}{3 \cdot 5} ~ & \textcolor{red}{ \text{ Equivalent fractions, LCD = 15.}} \\ x = - \frac{9}{15} - \frac{10}{15} ~ & \textcolor{red}{ \text{ Simplify.}} \\ x = - \frac{19}{15} ~ & \textcolor{red}{ \text{ Subtract.}} \end{aligned}\nonumber \]

Readers are encouraged to check this solution in the original equation.

Solve for x : \(x + \frac{3}{4} = - \frac{1}{2}\)

Undoing Multiplication

We “undo” multiplication by dividing. For example, to solve the equation 2 x = 6, we would divide both sides of the equation by 2. In similar fashion, we could divide both sides of the equation

\[ \frac{3}{5} x = \frac{4}{10}\nonumber \]

by 3/5. However, it is more efficient to take advantage of reciprocals. For convenience, we remind readers of the Multiplicative Inverse Property .

Multiplicative Inverse Property

Let a / b be any fraction. The number b / a is called the multiplicative inverse or reciprocal of a / b . The product of reciprocals is 1.

\[ \frac{a}{b} \cdot \frac{b}{a} = 1.\nonumber \]

Let’s put our knowledge of reciprocals to work.

Solve for x : \(\frac{3}{5}x = \frac{4}{10}\).

To “undo” multiplying by 3/5, multiply both sides by the reciprocal 5/3 and simplify.

\[ \begin{aligned} \frac{3}{5} x = \frac{4}{10} ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{5}{3} \left( \frac{3}{5} x \right) = \frac{5}{3} \left( \frac{4}{10} \right) & ~ \textcolor{red}{ \text{ Multiply both sides by 5/3.}} \\ \left( \frac{5}{3} \cdot \frac{3}{5} \right) x = \frac{20}{30} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, use the associative property to regroup.} \\ \text{ On the right, multiply.} \end{array}} \\ 1x = \frac{2}{3} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, } \frac{5}{3} \cdot \frac{3}{5} = 1. \\ \text{ On the right, reduce: } \frac{20}{30} = \frac{2}{3}. \end{array}} \\ x = \frac{2}{3} ~ & \textcolor{red}{ \text{ On the left, } 1x = x.} \end{aligned}\nonumber \]

Substitute 2/3 for x in the original equation and simplify.

\[ \begin{aligned} \frac{3}{5} x = \frac{4}{10} ~ & \textcolor{red}{ \text{ Original equation.}} \\ \frac{3}{5} \left( \frac{2}{3} \right) = \frac{4}{10} ~ & \textcolor{red}{ \text{ Substitute 2/3 for }x.} \\ \frac{6}{15} = \frac{4}{10} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \\ \frac{2}{5} = \frac{2}{5} ~ & \textcolor{red}{ \text{ Reduce both sides to lowest terms.}} \end{aligned}\nonumber \]

Because this last statement is true, we conclude that 2/3 is a solution of the equation (3/5) x = 4/10.

Solve for y : \( \frac{2}{3} y = \frac{4}{5}\)

Solve for x : \(- \frac{8}{9} x = \frac{5}{18}\).

To “undo” multiplying by −8/9, multiply both sides by the reciprocal −9/8 and simplify.

\[ \begin{aligned} - \frac{8}{9} x = \frac{5}{18} ~ & \textcolor{red}{ \text{ Original equation.}} \\ - \frac{9}{8} \left( - \frac{8}{9} x \right) = - \frac{9}{8} \left( \frac{5}{18} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by } -9/8.} \\ \left[ - \frac{9}{8} \cdot \left( - \frac{8}{9} \right) \right] x = - \frac{3 \cdot 3}{2 \cdot 2 \cdot 2} \cdot \frac{5}{2 \cdot 3 \cdot 3} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, use the associative property to regroup.} \\ \text{ On the right, prime factor.} \end{array}} 1x = \frac{ \cancel{3} \cdot \cancel{3}}{2 \cdot 2 \cdot 2} \cdot \frac{5}{2 \cdot \cancel{3} \cdot \cancel{3}} ~ & \textcolor{red}{ \begin{array}{l} \text{ On the left, } - \frac{9}{8} \cdot \left( - \frac{8}{9} \right) = 1. \\ \text{ On the right, cancel common factors.} \end{array}} \\ x = - \frac{5}{16} ~ & \textcolor{red}{ \text{ On the left, } 1x = x. \text{ Multiply on the right.}} \end{aligned}\nonumber \]

Solve for z: \(− \frac{2}{7} z = \frac{4}{21}\)

Clearing Fractions from the Equation

Although the technique demonstrated in the previous examples is a solid mathematical technique, working with fractions in an equation is not always the most efficient use of your time.

To clear all fractions from an equation, multiply both sides of the equation by the least common denominator of the fractions that appear in the equation.

Let’s put this idea to work.

In Example 1, we were asked to solve the following equation for x :

\[x - \frac{5}{6} = \frac{1}{3}.\nonumber \]

Take a moment to review the solution technique in Example 1. We will now solve this equation by first clearing all fractions from the equation.

Multiply both sides of the equation by the least common denominator for the fractions appearing in the equation.

\[ \begin{aligned} x - \frac{5}{6}= \frac{1}{3} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 6 \left( x - \frac{5}{6} \right) = 6 \left( \frac{1}{3} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 6.}} \\ 6x - 6 \left( \frac{5}{6} \right) = 6 \left( \frac{1}{3} \right) ~ & \textcolor{red}{ \text{ Distribute the 6.}} \\ 6x-5 = 2 ~ & \textcolor{red}{ \text{ On each side, multiply first.}} \\ ~ & \textcolor{red}{6 \left( \frac{5}{6} \right) = 5 \text{ and } 6 \left( \frac{1}{3} \right) = 2.} \end{aligned}\nonumber \]

Note that the equation is now entirely clear of fractions, making it a much simpler equation to solve.

\[ \begin{aligned} 6x - 5 + 5 = 2 + 5 ~ & \textcolor{red}{ \text{ Add 5 to both sides.}} \\ 6x = 7 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \\ \frac{6x}{6} = \frac{7}{6} ~ & \textcolor{red}{ \text{ Divide both sides by 6.}} \\ x = \frac{7}{6} ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber \]

Note that this is the same solution found in Example 1.

Solve for t : \(t - \frac{2}{7} = - \frac{1}{4}\)

In Example 4, we were asked to solve the following equation for x .

\[- \frac{8}{9}x = \frac{5}{18}\nonumber \]

Take a moment to review the solution in Example 4. We will now solve this equation by first clearing all fractions from the equation.

Multiply both sides of the equation by the least common denominator for the fractions that appear in the equation.

\[ \begin{aligned} - \frac{8}{9} x = \frac{5}{18} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 18 \left( - \frac{8}{9} x \right) = 18 \left( \frac{5}{18} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 18.}} \\ -16x=5 ~ & \textcolor{red}{ \text{ On each side, cancel and multiply.}} \\ ~ & \textcolor{red}{ 18 \left( - \frac{8}{9} \right) = -16 \text{ and } 18 \left( \frac{5}{18} \right) = 5.} \end{aligned}\nonumber \]

Note that the equation is now entirely free of fractions. Continuing,

\[ \begin{aligned} \frac{-16x}{-16} = \frac{5}{-16} ~ & \textcolor{red}{ \text{ Divide both sides by } -16.} \\ x = - \frac{5}{16} ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber \]

Note that this is the same as the solution found in Example 4.

Solve for u :

\[ - \frac{7}{9} u = \frac{14}{27}\nonumber \]

Solve for x : \(\frac{2}{3}x + \frac{3}{4} = \frac{1}{2}\).

\[ \begin{aligned} \frac{2}{3} x + \frac{3}{4} = \frac{1}{2} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 12 \left( \frac{2}{3} x + \frac{3}{4} = \right) = 12 \left( \frac{1}{2} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 12.}} \\ 12 \left( \frac{2}{3}x \right) + 12 \left( \frac{3}{4} \right) = 12 \left( \frac{1}{2} \right) ~ & \textcolor{red}{ \text{ On the left, distribute 12.}} \\ 8x + 9 = 6 ~ & \textcolor{red}{ \text{ Multiply: } 12 \left( \frac{2}{3} x \right) = 8x, ~ 12 \left( \frac{3}{4} \right) = 9,} \\ ~ & \textcolor{red}{ \text{ and } 12 \left( \frac{1}{2} \right) = 6.} \end{aligned}\nonumber \]

Note that the equation is now entirely free of fractions. We need to isolate the terms containing x on one side of the equation.

\[ \begin{aligned} 8x + 9 - 9 = 6 - 9 ~ & \textcolor{red}{ \text{ Subtract 9 from both sides.}} \\ 8x = - 3 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \\ \frac{8x}{8} = \frac{-3}{8} ~ & \textcolor{red}{ \text{ Divide both sides by 8.}} \\ x = - \frac{3}{8} ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]

Solve for r : \(\frac{3}{4} r + \frac{2}{3} = \frac{1}{2}\)

Solve for x : \( \frac{2}{3} - \frac{3x}{4} = \frac{x}{2} - \frac{1}{8}.\)

Multiply both sides of the equation by the least common denominator for the fractions in the equation.

\[ \begin{aligned} \frac{2}{3} - \frac{3x}{4} = \frac{x}{2} - \frac{1}{8} ~ & \textcolor{red}{ \text{ Original equation.}} \\ 24 \left( \frac{2}{3} - \frac{3x}{4} \right) = 24 \left( \frac{x}{2} - \frac{1}{8} \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 24.}} \\ 24 \left( \frac{2}{3} \right) - 24 \left( \frac{3x}{4} \right) = 24 \left( \frac{x}{2} \right) - 24 \left( \frac{1}{8} \right) ~ & \textcolor{red}{ \text{ On both sides, distribute 24.}} \\ 16 - 18x = 12x - 3 ~ & \textcolor{red}{ \text{ Left: } 24 \left( \frac{2}{3} \right) = 16, ~ 24 \left( \frac{3x}{4} \right) = 18x.} \\ ~ & \textcolor{red}{ \text{ Right: } 24 \left( \frac{x}{2} \right) = 12x, ~ 24 \left( \frac{1}{8} \right) = 3.} \end{aligned}\nonumber \]

\[ \begin{aligned} 16 - 18x - 12x = 12x - 3 - 12x ~ & \textcolor{red}{ \text{ Subtract } 12x \text{ from both sides.}} \\ 16 - 30x = -3 ~ & \textcolor{red}{ \begin{aligned} \text{ Left: } -18x - 12x = -30x. \\ \text{ Right: } 12x - 12x = 0. \end{aligned}} \\ 16 - 30x - 16 = -3 - 16 ~ & \textcolor{red}{ \text{ Subtract 16 from both sides.}} \\ -30x = -19 ~ & \textcolor{red}{ \begin{aligned} \text{ Left: } 16-16=0. \\ \text{ Right: } -3 - 16 = -19. \end{aligned}} \\ \frac{-30x}{-30} = \frac{-19}{-30} ~ & \textcolor{red}{ \text{ Divide both sides by } -30.} \\ x = \frac{19}{30} ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]

Solve for s : \( \frac{3}{2} - \frac{2s}{5} = \frac{s}{3} - \frac{1}{5}\).

Add texts here. Do not delete this text first.

Applications

Let’s look at some applications that involve equations containing fractions. For convenience, we repeat the Requirements for Word Problem Solutions .

Requirements for Word Problem Solutions

• Statements such as “Let P represent the perimeter of the rectangle.”
• Labeling unknown values with variables in a table.
• Labeling unknown quantities in a sketch or diagram.
• Set up an Equation . Every solution to a word problem must include a carefully crafted equation that accurately describes the constraints in the problem statement.
• Solve the Equation . You must always solve the equation set up in the previous step.
• Answer the Question . This step is easily overlooked. For example, the problem might ask for Jane’s age, but your equation’s solution gives the age of Jane’s sister Liz. Make sure you answer the original question asked in the problem. Your solution should be written in a sentence with appropriate units. 5. Look Back. It is important to note that this step does not imply that you should simply check your solution in your equation. After all, it’s possible that your equation incorrectly models the problem’s situation, so you could have a valid solution to an incorrect equation. The important question is: “Does your answer make sense based on the words in the original problem statement.”

In the third quarter of a basketball game, announcers informed the crowd that attendance for the game was 12,250. If this is two-thirds of the capacity, find the full seating capacity for the basketball arena.

We follow the Requirements for Word Problem Solutions .

1. Set up a Variable Dictionary . Let F represent the full seating capacity. Note: It is much better to use a variable that “sounds like” the quantity that it represents. In this case, letting F represent the full seating capacity is much more descriptive than using x to represent the full seating capacity.

2. Set up an Equation . Two-thirds of the full seating capacity is 12,250.

\[ \begin{aligned} \colorbox{cyan}{Two-thirds} & \text{ of } & \colorbox{cyan}{Full Seating Capacity} & \text{ is } & 12,250 \\ \frac{2}{3} & \cdot & F & = & 12,250 \end{aligned}\nonumber \]

Hence, the equation is

\[ \frac{2}{3} F = 12250.\nonumber \]

3. Solve the Equation . Multiply both sides by 3 to clear fractions, then solve.

\[ \begin{aligned} \frac{2}{3} F = 12250 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 3 \left( \frac{2}{3} F \right) = 3(12250) ~ & \textcolor{red}{ \text{ Multiply both sides by 3.}} \\ 2F = 36750 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \\ \frac{2F}{2} = \frac{36750}{2} ~ & \textcolor{red}{ \text{ Divide both sides by 2.}} \\ F = 18375 ~ & \textcolor{red}{ \text{ Simplify both sides.}} \end{aligned}\nonumber \]

4. Answer the Question . The full seating capacity is 18,375.

5. Look Back . The words of the problem state that 2/3 of the seating capacity is 12,250. Let’s take two-thirds of our answer and see what we get.

\[ \begin{aligned} \frac{2}{3} \cdot 18375 & = \frac{2}{3} \cdot \frac{18375}{1} \\ & = \frac{2}{3} \cdot \frac{3 \cdot 6125}{1} \\ & = \frac{2}{ \cancel{3}} \cdot \frac{ \cancel{3} \cdot 6125}{1} \\ & = 12250 \end{aligned}\nonumber \]

This is the correct attendance, so our solution is correct.

Attendance for the Celtics game was 9,510. If this is 3/4 of capacity, what is the capacity of the Celtics’ arena?

The area of a triangle is 20 square inches. If the length of the base is \(2 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

1. Set up a Variable Dictionary . Our variable dictionary will take the form of a well labeled diagram.

2. Set up an Equation . The area A of a triangle with base b and height h is

\[A = \frac{1}{2} bh.\nonumber \]

Substitute A = 20 and b = \(2 \frac{1}{2}\).

\[20 = \frac{1}{2} \left( 2 \frac{1}{2} \right) h.\nonumber \]

3. Solve the Equation . Change the mixed fraction to an improper fraction, then simplify.

\[ \begin{aligned} 20 = \frac{1}{2} \left( 2 \frac{1}{2} \right) h ~ & \textcolor{red}{ \text{ Original equation.}} \\ 20 = \frac{1}{2} \left( \frac{5}{2} \right) h ~ & \textcolor{red}{ \text{ Mixed to improper: } 2 \frac{1}{2} = \frac{5}{2}.} \\ 20 = \left( \frac{1}{2} \cdot \frac{5}{2} \right) h ~ & \textcolor{red}{ \text{ Associative property.}} \\ 20 = \frac{5}{4} h ~ & \textcolor{red}{ \text{ Multiply: } \frac{1}{2} \cdot \frac{5}{2} = \frac{5}{4}.} \end{aligned}\nonumber \]

Now, multiply both sides by 4/5 and solve.

\[ \begin{aligned} \frac{4}{5} (20) = \frac{4}{5} \left( \frac{5}{4} h \right) ~ & \textcolor{red}{ \text{ Multiply both sides by 4/5.}} \\ 16 = h ~ & \textcolor{red}{ \text{ Simplify: } \frac{4}{5} (20) = 16} \\ ~ & \textcolor{red}{ \text{ and } \frac{4}{5} \cdot \frac{5}{4} = 1.} \end{aligned}\nonumber \]

4. Answer the Question . The height of the triangle is 16 inches.

5. Look Back . If the height is 16 inches and the base is \(2 \frac{1}{2}\) inches, then the area is

\[ \begin{aligned} A & = \frac{1}{2} \left( 2 \frac{1}{2} \right) (16) \\ & = \frac{1}{2} \cdot \frac{5}{2} \cdot \frac{16}{1} \\ & = \frac{5 \cdot 16}{2 \cdot 2} \\ & = \frac{(5) \cdot (2 \cdot 2 \cdot 2 \cdot 2)}{(2) \cdot (2)} \\ & = \frac{5 \cdot \cancel{2} \cdot \cancel{2} \cdot 2 \cot 2}{ \cancel{2} \cdot \cancel{2}} & = 20 \end{aligned}\nonumber \]

This is the correct area (20 square inches), so our solution is correct.

The area of a triangle is 161 square feet. If the base of the triangle measures \(40 \frac{1}{4}\) feet, find the height of the triangle.

1. Is 1/4 a solution of the equation \(x + \frac{5}{8} = \frac{5}{8}\)?

2. Is 1/4 a solution of the equation \(x + \frac{1}{3} = \frac{5}{12}\)?

3. Is −8/15 a solution of the equation \(\frac{1}{4} x = − \frac{1}{15}\)?

4. Is −18/7 a solution of the equation \(− \frac{3}{8} x = \frac{25}{28}\)?

5. Is 1/2 a solution of the equation \(x + \frac{4}{9} = \frac{17}{18}\)?

6. Is 1/3 a solution of the equation \(x + \frac{3}{4} = \frac{13}{12}\)?

7. Is 3/8 a solution of the equation \(x − \frac{5}{9} = − \frac{13}{72}\)?

8. Is 1/2 a solution of the equation \(x − \frac{3}{5} = − \frac{1}{10}\)?

9. Is 2/7 a solution of the equation \(x − \frac{4}{9} = − \frac{8}{63}\)?

10. Is 1/9 a solution of the equation \(x − \frac{4}{7} = − \frac{31}{63}\)?

11. Is 8/5 a solution of the equation \( \frac{11}{14}x = \frac{44}{35}\)?

12. Is 16/9 a solution of the equation \(\frac{13}{18} x = \frac{104}{81}\)?

In Exercises 13-24, solve the equation and simplify your answer.

13. \(2x − 3=6x + 7\)

14. \(9x − 8 = −9x − 3\)

15. \(−7x +4=3x\)

16. \(6x +9= −6x\)

17. \(−2x = 9x − 4\)

18. \(−6x = −9x + 8\)

19. \(−8x = 7x − 7\)

20. \(−6x = 5x + 4\)

21. \(−7x +8=2x\)

22. \(−x − 7=3x\)

23. \(−9x +4=4x − 6\)

24. \(−2x +4= x − 7\)

In Exercises 25-48, solve the equation and simplify your answer.

25. \(x + \frac{3}{2 = \frac{1}{2}\)

26. \(x − \frac{3}{4} = \frac{1}{4}\)

27. \(− \frac{9}{5} x = \frac{1}{2}\)

28. \(\frac{7}{3} x = − \frac{7}{2}\)

29. \(\frac{3}{8} x = \frac{8}{7}\)

30. \(− \frac{1}{9} x = − \frac{3}{5}\)

31. \(\frac{2}{5} x = − \frac{1}{6}\)

32. \(\frac{1}{6} x = \frac{2}{9}\)

33. \(− \frac{3}{2} x = \frac{8}{7}\)

34. \(− \frac{3}{2} x = − \frac{7}{5}\)

35. \(x + \frac{3}{4} = \frac{5}{9}\)

36. \(x − \frac{1}{9} = − \frac{3}{2}\)

37. \(x − \frac{4}{7} = \frac{7}{8}\)

38. \(x + \frac{4}{9} = − \frac{3}{4}\)

39. \(x + \frac{8}{9} = \frac{2}{3}\)

40. \(x − \frac{5}{6} = \frac{1}{4}\)

41. \(x + \frac{5}{2} = − \frac{9}{8}\)

42. \(x + \frac{1}{2} = \frac{5}{3}\)

43. \(− \frac{8}{5} x = \frac{7}{9}\)

44. \(− \frac{3}{2} x = − \frac{5}{9}\)

45. \(x − \frac{1}{4} = − \frac{1}{8}\)

46. \(x − \frac{9}{2} = − \frac{7}{2}\)

47. \(− \frac{1}{4} x = \frac{1}{2}\)

48. \(− \frac{8}{9} x = − \frac{8}{3}\)

In Exercises 49-72, solve the equation and simplify your answer.

49. \(− \frac{7}{3} x − \frac{2}{3} = \frac{3}{4} x + \frac{2}{3}\)

50. \(\frac{1}{2} x − \frac{1}{2} = \frac{3}{2} x + \frac{3}{4}\)

51. \(− \frac{7}{2} x − \frac{5}{4} = \frac{4}{5}\)

52. \(− \frac{7}{6} x + \frac{5}{6} = − \frac{8}{9}\)

53. \(− \frac{9}{7} x + \frac{9}{2} = − \frac{5}{2}\)

54. \(\frac{5}{9} x − \frac{7}{2} = \frac{1}{4}\)

55. \(\frac{1}{4} x − \frac{4}{3} = − \frac{2}{3}\)

56. \(\frac{8}{7} x + \frac{3}{7} = \frac{5}{3}\)

57. \(\frac{5}{3} x + \frac{3}{2} = − \frac{1}{4}\)

58. \(\frac{1}{2} x − \frac{8}{3} = − \frac{2}{5}\)

59. \(− \frac{1}{3} x + \frac{4}{5} = − \frac{9}{5} x − \frac{5}{6}\)

60. \(− \frac{2}{9} x − \frac{3}{5} = \frac{4}{5} x − \frac{3}{2}\)

61. \(− \frac{4}{9} x − \frac{8}{9} = \frac{1}{2} x − \frac{1}{2}\)

62. \(− \frac{5}{4} x − \frac{5}{3} = \frac{8}{7} x + \frac{7}{3}\)

63. \(\frac{1}{2} x − \frac{1}{8} = − \frac{1}{8} x + \frac{5}{7}\)

64. \(− \frac{3}{2} x + \frac{8}{3} = \frac{7}{9} x − \frac{1}{2}\)

65. \(− \frac{3}{7} x − \frac{1}{3} = − \frac{1}{9}\)

66. \(\frac{2}{3} x + \frac{2}{9} = − \frac{9}{5}\)

67. \(− \frac{3}{4} x + \frac{2}{7} = \frac{8}{7} x − \frac{1}{3}\)

68. \(\frac{1}{2} x + \frac{1}{3} = − \frac{5}{2} x − \frac{1}{4}\)

69. \(− \frac{3}{4} x − \frac{2}{3} = − \frac{2}{3} x − \frac{1}{2}\)

70. \(\frac{1}{3} x − \frac{5}{7} = \frac{3}{2} x + \frac{4}{3}\)

71. \(− \frac{5}{2} x + \frac{9}{5} = \frac{5}{8}\)

72. \(\frac{9}{4} x + \frac{4}{3} = − \frac{1}{6}\)

73. At a local soccer game, announcers informed the crowd that attendance for the game was 4,302. If this is 2/9 of the capacity, find the full seating capacity for the soccer stadium.

74. At a local basketball game, announcers informed the crowd that attendance for the game was 5,394. If this is 2/7 of the capacity, find the full seating capacity for the basketball stadium.

75. The area of a triangle is 51 square inches. If the length of the base is \(8 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

76. The area of a triangle is 20 square inches. If the length of the base is \(2 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

77. The area of a triangle is 18 square inches. If the length of the base is \(4 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

78. The area of a triangle is 44 square inches. If the length of the base is \(5 \frac{1}{2}\) inches, find the height (altitude) of the triangle.

79. At a local hockey game, announcers informed the crowd that attendance for the game was 4,536. If this is 2/11 of the capacity, find the full seating capacity for the hockey stadium.

80. At a local soccer game, announcers informed the crowd that attendance for the game was 6,970. If this is 2/7 of the capacity, find the full seating capacity for the soccer stadium.

81. Pirates . About one-third of the world’s pirate attacks in 2008 occurred off the Somali coast. If there were 111 pirate attacks off the Somali coast, estimate the number of pirate attacks worldwide in 2008.

82. Nuclear arsenal . The U.S. and Russia agreed to cut nuclear arsenals of long-range nuclear weapons by about a third, down to 1, 550. How many long-range nuclear weapons are there now? Associated Press-Times-Standard 04/04/10 Nuclear heartland anxious about missile cuts.

83. Seed vault . The Svalbard Global Seed Vault has amassed half a million seed samples, and now houses at least one-third of the world’s crop seeds. Estimate the total number of world’s crop seeds. Associated Press-Times-Standard 03/15/10 Norway doomsday seed vault hits half-million mark.

84. Freight train . The three and one-half mile long Union Pacific train is about 2 1 2 times the length of a typical freight train. How long is a typical freight train? Associated Press-Times-Standard 01/13/10 Unusally long train raises safety concerns.

13. \(− \frac{5}{2}\)

15. \(\frac{2}{5}\)

17. \(\frac{4}{11}\)

19. \(\frac{7}{15}\)

21. \(\frac{8}{9}\)

23. \(\frac{10}{13}\)

25. \(−1\)

27. \(− \frac{5}{18}\)

29. \(\frac{64}{21}\)

31. \(− \frac{5}{12}\)

33. \(− \frac{16}{21}\)

35. \(− \frac{7}{36}\)

37. \(\frac{81}{56}\)

39. \(− \frac{2}{9}\)

41. \(− \frac{29}{8}\)

43. \(− \frac{35}{72}\)

45. \(\frac{1}{8}\)

47. \(−2\)

49. \(− \frac{16}{37}\)

51. \(− \frac{41}{70}\)

53. \(\frac{49}{9}\)

55. \(\frac{8}{3}\)

57. \(− \frac{21}{20}\)

59. \(− \frac{49}{44}\)

61. \(− \frac{7}{17}\)

63. \(\frac{47}{35}\)

65. \(− \frac{14}{27}\)

67. \(\frac{52}{159}\)

69. \(− 2\)

71. \(\frac{47}{100}\)

81. There were about 333 pirate attacks worldwide.

83. 1,500,000

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2.6: How to solve problems

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In general you can use the following flowchart [9] to solve problems in a structured way. The steps will be outlined in more detail in Ch. 14.

Follow these steps to solve a problem:

• Make sure you read the theory, at least up to the point needed to solve the problem.
• Read the entire problem description carefully, from beginning to end.
• Write down a list of variables given and a second list of variables to be calculated.
• Determine the type of problem.
• Determine for yourself whether this is a problem of kinematics, kinetics, or other.
• Break the problem down into sub-problems as appropriate and search for the required theory subsection in the book. Read this part again if needed.
• Extract the general methods and formulae that need to be applied. Write down the equation number(s) from the book. Make sure you use the original, generic form of the formulae (not a similar-looking version from an example or a problem solution).
• Make drawings, such as FBDs, or kinematic diagrams of your system. Use different views where needed. Clearly define at least one coordinate system including unit vector directions and a coordinate origin. Specify it if the coordinate system is moving or rotating.
• Optionally create a table of variables where you clearly associate the variables in the problem and in your drawings with the variables given in the theoretic formulae.
• Check all assumptions that need to hold for the formulae to be applicable
• Write down all assumptions you made for your specific problem, and highlight those that can only be checked later.
• In many problems you have to solve for one or more scalar variables. In general, you have \(E\) equations, and need to solve a system with \(U\) unknowns. To check if you have sufficient information \((E)\) to solve the problem apply the following steps:

a) Count the number \(U\) of scalar unknowns. Be careful to include all unique components of a vector or a matrix individually.

b) Count the number \(E\) of scalar equations given by the equations you found.

c) Compare \(E\) and \(U\).

• If \(E=U\), continue.
• If \(E<U\), do not start solving yet. Instead, find more equations, either in the theory or in the problem itself. Are there kinematic relationships between variables that can reduce \(U\) or increase E?
• If \(E>U\), continue, but make a note to come back to check if some of the equations do not contradict each other and if all of them are correct.

d) Solve the system of equations. Do this symbolically as much as possible, do not insert any numbers.

e) Check the original assumptions. Do they indeed hold? If not, go back to step 11, make other assumptions and solve the problem again.

f) Once you have found a result, conduct multiple plausibility checks. Also re-insert your solution into your original equations, especially if \(E>U\). If in doubt about your result, start again at step 1.

g) If requested compute the numerical values of the \(U\) unknowns using the \(E\) obtained equations and the given numerical quantities.

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Unit 4: Understand fractions

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From cooking measurements to geometry, fractions are all around us. By understanding how the numerator and denominator work together, you'll be able to break down numbers into smaller parts, compare different fractions, and get a grasp on concepts like equivalent fractions.

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Fractions and whole numbers

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Equivalent fractions

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How to Solve Maths Problems with Google Circle for Students

Math can be a challenging subject for many students. Between understanding word problems and applying the right formulas, getting stuck is a common experience. Google has introduced a powerful tool called Google Circle, designed to help students solve math problems with confidence.

Read In Short: What Google Circle is and how it can help students solve math problems. How to use Google Circle to break down and solve different types of math problems. Additional tips and tricks to maximize your learning experience with Google Circle

What is Google Circle?

Google Circle is a feature within the Google Search app for Android devices . It allows you to solve math problems directly from your phone or tablet. Instead of simply providing answers, Google Circle focuses on math problem-solving by offering step-by-step guidance. This makes it a valuable tool for understanding the concepts behind the calculations, not just getting the final answer.

How Does Google Circle Work

Using Google Circle is incredibly simple. Here’s how it goes:

Step 1: Open the Google Search app on your Android device.

Step 2: find the math problem you’re working on., step 3: circle the specific problem you want help with using your finger., step 4: google circle will recognize it and use its artificial intelligence to analyze it., how to enable circle to search, step 1: open settings on your phone, step 2: scroll down and select display, step 3: scroll down and select navigation bar, step 4: select the toggle and turn on circle to search, which devices are getting the circle to search update.

Below is the list of android which is getting Google Circle App

• Samsung Galaxy S24, S24 Plus, and S24 Ultra
• Samsung Galaxy S23, S23 Plus, and S23 Ultra
• Samsung Galaxy S23 FE
• Samsung Galaxy Z Fold 5
• Samsung Galaxy Z Flip 5
• Samsung Galaxy Tab S9, S9 Plus, and S9 Ultra
• Google Pixel 8 and Pixel 8 Pro
• Google Pixel 7 and Pixel 7 Pro
• Google Pixel 6 and Pixel 6 Pro
• Google Pixel 7a
• Google Pixel 6a
• Google Pixel Fold
• Google Pixel Tablet

What Types of Math Problems Can Google Circle Solve?

Currently, Google Circle is particularly helpful with math word problems. These can be some of the trickiest problems to tackle, as they require translating real-world situations into mathematical terms.

Google Circle excels at helping students understand the key information in a word problem, identify the relevant mathematical concepts, and apply the appropriate formulas to reach the solution.

Here are some specific areas where Google Circle can be helpful:

• Algebra: Solve for variables, manipulate equations, and understand linear relationships.
• Geometry: Calculate areas, and volumes, and understand geometric shapes and their properties.
• Statistics: Analyze data sets, calculate measures of central tendency, and interpret graphs.

How to Solve Math Probelm Using Google Search

Step 1: activate circle to search.

On your Android smartphone or tablet, activate the Circle to Search feature.

Step 2: Circle the Problem

Find the math problem you’re stuck on and circle it on your screen. This prompts the feature to focus on that particular problem.

Step 3: Get Step-by-Step Instructions

The feature breaks down the problem and provides step-by-step instructions to solve it

Google Circle Vs Photomath Vs Socratic by Google

Tips for getting the most out of google circle.

• Practice makes perfect: Don’t just rely on Google Circle for the answer. Use it as a guide to understand the steps and then try solving similar problems on your own. This will solidify your learning and build your confidence.
• Focus on understanding, not just answers: While getting the right answer is important, the true power of Google Circle lies in its explanations. Pay attention to why each step is taken and how the concepts are applied.
• Don’t be afraid to experiment: Google Circle can handle various math problems. Try circling different parts of a word problem to see how it breaks down the information and guides you toward the solution.
• Take notes: As Google Circle explains the steps, jot down important points or formulas for future reference. This will create a personal study guide tailored to your learning needs.
• Combine it with other resources: Google Circle is a powerful tool, but it shouldn’t be your only resource. Use it alongside your textbook, class notes, or online tutorials for a well-rounded understanding.

Google Circle is a new tool designed to help students to become independent math problem solvers. By using its step-by-step guidance and clear explanations, you can transform your approach to math, gaining a deeper understanding of concepts and building the confidence to tackle even the most challenging problems. So, next time you’re stuck on a math question, don’t hesitate to circle it with Google Circle and watch your problem-solving skills soar!

Google Circle for Students – FAQs

Is google circle available on iphones.

Unfortunately, as of now, Google Circle is only available on Android devices within the Google Search app. However, there might be plans for future expansion to other platforms.

Is Google Circle safe to use?

Google Circle is a feature within the official Google Search app, making it a safe and reliable tool. However, it’s always a good practice to be mindful of the information you share online and only use it for educational purposes.

What if Google Circle can’t solve my math problem?

While Google Circle is constantly learning and expanding its capabilities, it might encounter problems it can’t handle yet. In such cases, don’t hesitate to consult your teacher, a tutor, or online resources for alternative solutions.

Will Google Circle do my math homework for me?

Google Circle is not a magic shortcut. It’s designed to be a learning tool. While it can guide you through the problem-solving process, it’s still important for you to understand the steps and concepts involved.

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Easy Finger Math Tricks to Help Kids Solve Problems

While using your fingers isn't the fastest way to recall a multiplication fact while doing a problem, finger math tricks can help kids figure out how to answer the problem at hand — and as they work on their math, they will eventually learn all the facts by repetition.

Note that before your child can understand other finger tricks, they must be able to count by 2s, 5s, and 10s and multiply by 2s, 3s, and 4s.

Quick Finger Math Tricks for Threes and Fours

The tricks for multiplying by threes and fours are really a matter of counting out the answer on your fingers. As your children count out the answer repeatedly, they'll memorize it and then be able to move on to larger numbers.

Multiplying by Three

Did you realize that all of your fingers have three segments? Therefore, you can figure out anything from 3 x 1 to 3 x 10 by counting the segments on each finger. To start:

• Hold up the number of fingers you're going to multiply by 3. For example, if the problem is 3 x 4 — hold up four fingers.
• Count each segment on each finger you're holding up, and you should come up with 12 — which is the correct answer.

Multiplying by Four

Multiplying by four is the same as multiplying by two — twice. To start:

• Hold up the number of fingers to correspond with the number you are multiplying by four. For example, if you are multiplying 4 x 6 — hold up six fingers.
• Count each finger by two, moving from left to right. Then count each finger again, continuing to count by twos, until you've counted every finger twice.

Helpful Hack To keep track of the fingers you've counted twice, sometimes it's easier to put your finger down as you count the first time, and back up as you count the second time.

Finger Math Tricks for Multiplying by 6, 7, 8, and 9

While numbers one through five are easy for most kids to remember, six and up often pose a problem. This handy trick will make it a little easier to work those problems out.

Multiplying 6, 7, 8, and 9 by Hand

To begin, assign each finger a number. For example, your thumbs represent 6, your index fingers each represent 7, etc. This will remain the same throughout the finger math hack.

Your left hand will represent the first number that you are multiplying and your right hand will represent the second number you are multiplying. In this example, we are multiplying 7 x 8. 

To Determine the Part of Your Answer:

• On your left hand, put down the finger that represents the number you are multiplying as well as any fingers whose number value is less than this figure. In this example, you are multiplying 7 x 8, so the left hand will represent 7. You will drop your index finger (number 7) and your thumb (number 6).
• Similarly, the right hand will represent eight, so you will drop down your middle finger (number 8), your index finger (number 7), and your thumb (number 6).
• Now, just multiply the fingers that are still pointed upwards. In this case, you will have three fingers on your left hand and two on your right, so you will multiply 3 x 2 to get 6. This is the first part of your answer!

To Determine the Second Part of Your Answer:

• Keeping your fingers in the same positions, count how many fingers are folded down. In the 7 x 8 example, you should have five fingers folded. 
• You will count each of these in quantities of ten. So, 10, 20, 30, 40, 50.
• 50 is your answer.

To Determine Your Final Answer:

• Add your two numbers together. In this example, you would add 6 + 50, which gives you 56!

Another Finger Math Trick Just for Nine

There is a trick that works separately, just for multiplying by the number nine.

• To start, hold up all ten fingers, with your palms facing you.
• Assign each finger a number, starting with your left-hand thumb and ending with your right-hand thumb. The left-hand thumb will be one, the left-hand index finger will be two, and so on until you reach the number 10 for your right-hand thumb.
• To tackle a problem, put down the corresponding finger of the number you're multiplying by nine. For example, if you are multiplying 9 x 8, you'd put down the eighth finger (which will be on your right hand).
• Count all the fingers to the left of the finger you have folded down. This will give you 7. This is the first digit of your answer.
• Count all the fingers to the right of the finger you have folded down. This will give you 2. This is the second digit of your answer.
• Put the numbers together! Your answer is 72.

Finger Multiplication Tricks Can Make Math Easy and Fun

While the hope is that your kids will eventually memorize their multiplication charts , using some quick hand tricks for multiplication and letting them count things out on their fingers is not a bad way to learn. It keeps frustration at bay since the answer is always a fingertip away, and the repetition of having to figure it out will help cement those facts into their brains.

Experience Google AI in even more ways on Android

May 14, 2024

[[read-time]] min read

By building AI right into the Android operating system, we're reimagining how you can interact with your phone.

• Bullet points
• Circle to Search gets smarter, helping students solve physics and math problems directly from their phones and tablets.
• Gemini on Android improves context understanding, allowing users to drag and drop generated images and ask questions about videos and PDFs.
• Gemini Nano with Multimodality coming to Pixel, bringing multimodal capabilities for richer image descriptions and scam alerts during phone calls.
• Android 15 and ecosystem updates coming tomorrow.
• Basic explainer

Google is making Android phones smarter with AI.

Circle to Search can now help students with homework.

Gemini, a new AI assistant, can understand what's on your screen and help you do things.

Android phones will soon be able to alert you to suspected scams during phone calls.

Explore other styles:

We’re at a once-in-a-generation moment where the latest advancements in AI are reinventing what phones can do. With Google AI at the core of Android’s operating system, the billions of people who use Android can now interact with their devices in entirely new ways.

Today, we’re sharing updates that let you experience Google AI on Android.

Circle to Search can now help students with homework

With Circle to Search built directly into the user experience, you can search anything you see on your phone using a simple gesture — without having to stop what you’re doing or switch to a different app. Since launching at Samsung Unpacked , we’ve added new capabilities to Circle to Search, like full-screen translation , and we’ve expanded availability to more Pixel and Samsung devices.

Starting today, Circle to Search can now help students with homework, giving them a deeper understanding, not just an answer — directly from their phones and tablets. When students circle a prompt they’re stuck on, they’ll get step-by-step instructions to solve a range of physics and math 1 word problems without leaving their digital info sheet or syllabus. Later this year, Circle to Search will be able to help solve even more complex problems involving symbolic formulas, diagrams, graphs and more. This is all possible due to our LearnLM effort to enhance our models and products for learning.

Circle to Search is already available on more than 100 million devices today. With plans to bring the experience to more devices, we’re on track to double that by the end of the year.

Gemini will get even better at understanding context to assist you in getting things done

Gemini on Android is a new kind of assistant that uses generative AI to help you be more creative and productive. This experience, which is integrated into Android, is getting even better at understanding the context of what’s on your screen and what app you’re using.

Soon, you’ll be able to bring up Gemini's overlay on top of the app you're in to easily use Gemini in more ways. For example, you can drag and drop generated images into Gmail, Google Messages and other places, or tap “Ask this video” to find specific information in a YouTube video. If you have Gemini Advanced, you’ll also have the option to “Ask this PDF” to quickly get answers without having to scroll through multiple pages. This update will roll out to hundreds of millions of devices over the next few months.

And we’ll continue to improve Gemini to give you more dynamic suggestions related to what’s on your screen.

Full multimodal capabilities coming to Gemini Nano

Android is the first mobile operating system that includes a built-in, on-device foundation model. With Gemini Nano, we’re able to bring experiences to you quickly and keep your information completely private to you. Starting with Pixel later this year, we’ll be introducing our latest model, Gemini Nano with Multimodality. This means your phone will not just be able to process text input but also understand more information in context like sights, sounds and spoken language.

Clearer descriptions with TalkBack

Later this year, Gemini Nano’s multimodal capabilities are coming to TalkBack, helping people who experience blindness or low vision get richer and clearer descriptions of what’s happening in an image. On average, TalkBack users come across 90 unlabeled images per day. This update will help fill in missing information — whether it’s more details about what’s in a photo that family or friends sent or the style and cut of clothes when shopping online. Since Gemini Nano is on-device, these descriptions happen quickly and even work when there's no network connection.

Receive alerts for suspected scams during phone calls

According to a recent report , in a 12-month period, people lost more than $1 trillion to fraud. We’re testing a new feature that uses Gemini Nano to provide real-time alerts during a call if it detects conversation patterns commonly associated with scams. For example, you would receive an alert if a “bank representative” asks you to urgently transfer funds, make a payment with a gift card or requests personal information like card PINs or passwords, which are uncommon bank requests . This protection all happens on-device, so your conversation stays private to you. We’ll share more about this opt-in feature later this year.

More to come on Android

We’re just getting started with how on-device AI can change what your phone can do, and we’ll continue building Google AI into every part of the smartphone experience with Pixel, Samsung and more. If you’re a developer, check out the Android Developers blog to learn how you can build with our latest AI models and tools, like Gemini Nano and Gemini in Android Studio.

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Available for some math word problems when opted into Search Labs.

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