This is the Mathematics 9 Quarter 1 self-learning module for Week 4. Module 10 is titled “Solving Word Problems on Rational Algebraic Expressions Transformable into Quadratic Equations.” It builds on Module 8, where learners solved rational algebraic equations that reduce to quadratic equations. Here they use that skill on real-life word problems about work, pipes and tanks, travel, and boats in a river current.
On this page 6 sections
What learners will learn
The module has one learning competency. Learners will be able to:
- solve problems involving quadratic equations and rational algebraic equations (M9AL-Ie-1).
What is inside the module
The module follows the usual DepEd self-learning module parts:
- What I Know: a 10-item multiple-choice pretest on work, pipe, and travel problems.
- What’s In: a review of adding and subtracting rational expressions, with a five-item “Try This!” exercise.
- What’s New: “Goods for All,” a conversation between Gabriel and Wynter about two barangays packing relief goods. Barangay Matipid takes 7 hours longer than Barangay Masipag, and together they finish in 12 hours. Learners are asked how long Barangay Matipid would take alone and what equation represents the problem.
- What Is It: two fully worked examples.
- What’s More: two practice problems.
- What I Have Learned: the step-by-step procedure for solving these problems.
- What I Can Do: four more problems. Answers are given to 2 decimal places where needed.
- Assessment: a 10-item multiple-choice post-test.
- Additional Activities: learners create and solve their own real-life problem.
- Problem-Based Worksheet: a billboard problem and a flag ceremony problem.
- Answer Key: answers for every part.
Key ideas: from rational equation to quadratic equation
Review: adding and subtracting rational expressions
To add or subtract rational expressions with unlike denominators:
- Find the LCM of the denominators.
- Write each expression using the LCD.
- Add or subtract the numerators.
- Simplify as needed.
For example, 1/x + 2x/5 uses the LCD 5x and becomes (5 + x²)/5x.
Work problems (rate, time, and work)
If a worker finishes a job alone in t hours, the worker’s rate is 1/t of the job per hour. Work done equals rate times time. The work done by each worker adds up to one whole job.
In Example 1, two factories produce a shipment of facemasks together in 6 hours. The slower factory would take 9 hours longer than the faster one.
- Let t be the faster factory’s time. The slower factory’s time is t + 9.
- The equation is 6/t + 6/(t + 9) = 1.
- Multiply both sides by the LCD t(t + 9) to get t² − 3t − 54 = 0.
- Factoring gives (t − 9)(t + 6) = 0, so t = 9 or t = −6.
- Time cannot be negative, so the faster factory takes 9 hours and the slower one takes 18 hours.
Motion problems (distance, rate, and time)
These problems use D = rt, so t = D/r. Example 2 has two rescue boats with the same speed in still water, in a river with a current of 5 km/h.
- The boat going downstream moves at r + 5. The boat going upstream moves at r − 5.
- The upstream boat left 1 hour earlier, has gone 44 km, and its time is 44/(r − 5).
- The downstream boat has gone 75 km, and its time is 75/(r + 5).
- The equation is 44/(r − 5) = 75/(r + 5) + 1.
- This leads to r² + 31r − 620 = 0. Solving it with the quadratic formula gives r ≈ 13.8 or r ≈ −44.8.
- The negative value is rejected, so each boat’s speed in still water is about 13.8 km/h.
The general procedure
The module’s “What I Have Learned” part lists these steps:
- Read and analyze the problem carefully.
- Collect the necessary data.
- Let the missing data be variables, and use algebraic expressions for the other data.
- Turn the remaining data into a mathematical sentence, the “working rational algebraic equation.”
- Solve the rational algebraic equation.
- Check whether the resulting equation is quadratic.
- If it is, find its roots using any method from earlier lessons.
- Substitute the roots into the working equation to check them, because you may get an extraneous root.
The module adds one important point. Because these are real-life problems, you must also decide whether both, one, or none of the solutions make sense. A negative time, speed, or length is not a reasonable answer.
How to use this module
- Learners: Start with the What I Know pretest and write your answers on a separate answer sheet. Check your work, and note the items you missed so you can find the right answers as you go. Then work through the parts in order. In the Assessment, if your answer is not among the choices, write E together with your final answer. The module also encourages trading solutions with classmates for a cooperative review of the Try This! items.
- Parents: Ask your child to explain the word equation behind each problem, such as “work done by the first + work done by the second = one whole job.” The answer key at the end lets you check the results together.
- Teachers: The pretest and the Assessment each have 10 items and can be compared to measure progress. The Additional Activities and the Problem-Based Worksheet work well as performance tasks. In those tasks, learners build, solve, and interpret their own real-life problems.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 9 Mathematics, Quarter 1, Module 10, intended for Week 4. Its learning code is M9AL-Ie-1.
What kinds of word problems does the module cover?
It covers work problems, such as factories, painters, pipes, and faucets working together, and motion problems, such as cars, boats moving against or with a river current, and airplanes flying into the wind. Each one leads to a rational equation that becomes a quadratic equation.
Does the module have an answer key?
Yes. The answer key covers What’s In, What’s More, What I Know, What I Can Do, the Assessment, and the Problem-Based Worksheet.
Why are some solutions rejected in these problems?
Some roots of the quadratic equation, such as a negative time or speed, do not make sense in a real-life situation. The module also reminds learners to substitute roots back into the working equation to check for extraneous roots.
What should I review before starting this module?
Review how to add and subtract rational expressions and how to solve rational equations that reduce to quadratic equations, the topic of Module 8. The module’s What’s In part gives a short review with practice items.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 1-Module 10: Solving Word Problems on Rational Algebraic Expressions Transformable into Quadratic Equations
Week 4 Learning Code – M9AL-Ie-1
9
[illustration: people learning online and at home, with a laptop, books, a video screen, and a tablet]
Copyright
Quarter 1 – Module 10 – New Normal Math for G9 First Edition 2020 Copyright © 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.
Borrowed materials (i.e. songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.
Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
Development Team of the Module
Writers: Junalisa A. Bartolome- TI, Rowena F. Reyes – TI, Marvin G. Sollera – MTII
Editor: Sally C. Caleja – Head Teacher VI, Luningning R. Tayamora- Head Teacher VI, Juvy G. Delos Santos – Head Teacher III
Validators: Remylinda T. Soriano, EPS, Math, Angelita Z. Modesto, PSDS, George B. Borromeo, PSDS
Illustrator: Writers
Layout Artist: Writers
Management Team: Malcolm S. Garma, Regional Director; Genia V. Santos, CLMD Chief; Dennis M. Mendoza, Regional EPS in Charge of LRMS and Regional ADM Coordinator; Maria Magdalena M. Lim, CESO V, Schools Division Superintendent; Aida H. Rondilla, Chief-CID; Lucky S. Carpio, Division EPS in Charge of LRMS and Division ADM Coordinator
MODULE 10
SOLVING WORD PROBLEMS ON RATIONAL ALGEBRAIC EXPRESSIONS TRANSFORMABLE INTO QUADRATIC EQUATIONS
In module 8, you have learned solving rational algebraic expressions reducible into quadratic equations. Mastery to this topic is a valuable tool to answer many real-life problems which you will learn from this module.
WHAT I NEED TO KNOW
LEARNING COMPETENCY
The learners will be able to:
- solve problems involving quadratic equations and rational algebraic equations. M9AL-Ie-1
WHAT I KNOW
[transcriber’s note: a decorative label reading “PPREPREVIER!” appears beside the “WHAT I KNOW” heading and is not transcribed]
Find out how much you already know about solving word problems on rational algebraic expression transformable into quadratic equation. Write the letter that you think is the best answer to each question on your answer sheet. Answer all items. After taking and checking this short test, take note of the items that you were not able to answer correctly and look for the right answer as you go through this module.
Mary and Carol are doing a Math project. Carol can do the work 2 hours longer as Mary. If they work together, they can finish the project in 4 hours. How long does it take Mary working alone to do the same project?
- A. 6 h
- B. 4 h
- C. 8 h
- D. 10 h
Two pipes working together can fill a tank in 35 minutes. If the large pipe alone can fill the tank in 24 minutes less than the time taken by the smaller pipe then find the time taken by larger pipe working alone to fill the tank.
- A. 30 mins.
- B. 60 mins.
- C. 40 mins.
- D. 80 mins.
It takes Mary 3 hours more to do a job than it takes Jane. If they work together, they can finish the same job in 2 hours. How long would it take Mary to finish the job alone?
- A. 3 hours
- B. 5 hours
- C. 6 hours
- D. 8 hours
Two conveyor belts can unload a shipment of tomatoes in 6 hours. Working alone, the slower belt would take 9 hours longer than the faster belt to do the job. How many hours would it take the faster conveyor belt working alone to do the job?
- A. 9 hours
- B. 10 hours
- C. 8 hours
- D. 6 hours
A motorcyclist traveled at a constant speed for 60 km. If he had gone 10 km/h faster, he would have shortened his traveling time by one hour. Find the speed of the motorcyclist.
- A. 19 km/h
- B. 18 km/h
- C. 20 km/h
- D. 17 km/h
A motorboat heads upstream a distance of 48 km on a river whose current is running at 3 km per hour (km/h). Then the motorboat returns. The trip upstream and back takes 12 hours. Assuming that the motorboat maintained a constant speed relative to the water, what was its speed?
- A. 8 km/h
- B. 9 km/h
- C. 7 km/h
- D. 10 km/h
A swimming pool can be filled by two pipes together in six hours. If the larger pipe alone takes five hours less than the smaller pipe alone to fill the pool, find the time in which smaller pipes alone would fill the pool.
- A. 15 hours
- B. 12 hours
- C. 10 hours
- D. 8 hours
A car drove from a city A to a city B. The distance between the two cities is 350km. After 200 km, the speed of the car decreased by 20 km/h. The total trip took 5 hours. Find the speed of the car in the first part of the journey.
- A. 60 km/h
- B. 70 km/h
- C. 80 km/h
- D. 90 km/h
Alex can do a job in one hour less than Jane. If Alex and Jane work together the job takes 6/5 hours. How long would it take Alex working alone?
- A. 2 hours
- B. 4 hours
- C. 3 hours
- D. 6 hours
Two painters working together can paint the front of a house in 16 hours. One of the painters alone can finish this job in 24 hours less time than the other painter alone. How much time does the faster painter to do the job alone?
- A. 24 hours
- B. 32 hours
- C. 20 hours
- D. 18 hours
WHAT’S IN
[transcriber’s note: the word “PPREPREV IER!” appears in the page as a stylized graphic, transcribed as printed]
PPREPREV IER! Let us recall: ADD and SUBTRACT! Here are a few steps to follow when you add or subtract rational expressions with like/unlike denominators.
- To add or subtract rational expressions with unlike denominators, first find the LCM of the denominator.
- Write each expression using the LCD.
- Add or subtract the numerators.
- Simplify as needed.
Study the examples below.
a.) $\frac{1}{x} + \frac{2x}{5} = \frac{1(5) + 2x(x)}{5x} = \frac{5 + x^2}{5x}$ (LCD = $5x$)
b.) $\frac{x}{x+1} – \frac{2}{x+2} = \frac{x(x+2) – 2(x+1)}{(x+1)(x+2)} = \frac{x^2 + 2x – 2x – 2}{x^2 + 3x + 2} = \frac{x^2 – 2}{x^2 + 3x + 2}$ (LCD = $(x + 1)(x + 2)$)
Try This! You can trade your solutions and answers for cooperative review. Add or subtract rational algebraic expressions and express the results in simplest forms.
- $\frac{3}{x} + \frac{x}{6}$
- $\frac{x}{2} + \frac{2x+1}{x}$
- $\frac{5}{y} – \frac{3y+1}{2}$
- $\frac{a+1}{a} – \frac{3}{2a+1}$
- $\frac{m+5}{2m} + \frac{m-2}{m+1}$
WHAT’S NEW
Communication, character building, critical thinking and collaboration
Read and understand the conversation below and answer the given questions
GOODS FOR ALL
To address the needs of the people in surviving the present pandemic, the government utilized the reserved funds in providing relief goods (mostly food consisting of rice and canned goods) to households and mobilized disinfection efforts.
Gabriel and Wynter came from two different barangays. They talked about the distribution of relief goods in their area during the quarantine period through phone.
I just…
[illustration: two children, Gabriel and Wynter, greeting each other; speech bubbles read “Hi wynter! How are you?” and “I’m fine. I’m just finished my paper works at home”]
Gabriel: Hello Wynter. How are you? I just heard the news that ECQ is extended till the end of the month.
Wynter: I’m fine. I just finished my paper works at home. How’s the distribution of relief goods in Barangay Masipag?
Gabriel: According to our barangay officials, they will distribute the goods this afternoon. How about in Barangay Matipid?
Wynter: They are still packing the goods. The chairman said that our barangay will take 7 hours longer to pack all the goods compared to your barangay.
Gabriel: Just have more patience Wynter. Last week, Barangay Masipag and Barangay Matipid worked together in packing the goods and it took 12 hours to finish it.
Wynter: Oh that’s a very long and tiring day for them.
Gabriel: Yes I agree. Now I am just wondering that since our barangays decided to work separately, how long it will take to pack all the relief goods in each barangay?
Wynter: I am also curious about that!
a. How long would it take Barangay Matipid to pack the relief goods by itself? b. What equation would represent the problem? c. How will you use the equation to solve what is asked?
DID YOU KNOW
The ancient greek mathematician Pythagoras believed that all numbers were rational, but one of his students Hippasus proved (using geometry, it is thought) that you could not write the square root of 2 as a fraction, and so it was irrational.
It is important to be able to relate the language of mathematics to your everyday language. The problems you encounter at work, for example, will likely be expressed orally or in written form. It is important to practice expressing your ideas in words, or what we call word equations, and practice writing your answers in complete sentences.
WHAT IS IT
Communication, Critical Thinking, and Collaboration
Example 1: Two factories can produce a shipment of facemasks in 6 hours. Working alone, the slower factory would take 9 hours longer than the faster factory to do the job. How many hours would it take each factory working alone to do the job?
Solution:
Let $t$ = time in hours for the faster factory $\frac{1}{t}$ = rate of the faster factory $t + 9$ = time in hours for slower factory $\frac{1}{t+9}$ = rate of the slower factory
| Rate | Time | Work | |
|---|---|---|---|
| Faster factory | $\frac{1}{t}$ | 6 | $\frac{6}{t}$ |
| Slower factory | $\frac{1}{t + 9}$ | 6 | $\frac{6}{t + 9}$ |
[chart: word equation diagram; Work done by faster factory + Work done by slower factory = One whole job]
$\frac{6}{t} + \frac{6}{t+9} = 1$ — Substitute the values from the table into the word equation.
$t(t + 9)\left(\frac{6}{t} + \frac{6}{t+9}\right) = t(t + 9)(1)$ — Multiply both members by the LCD $t(t+9)$.
$6(t + 9) + 6t = t(t + 9)$
$6t + 54 + 6t = t^2 + 9t$ — Simplify and write the quadratic equation so that one member is 0.
$t^2 – 3t – 54 = 0$
$(t – 9)(t + 6) = 0$ — Factor and solve for t.
$t – 9 = 0$ or $t + 6 = 0$
$t = 9$ or $t = -6$ — Since the time can’t be negative, t=-6 is not an answer.
Solving for the number of hours the slower factory can do the job, $t + 9 = 18$.
Answer: The faster factory could do the job in 9 h, while the slower factory could do the job in 18 h.
Example 2: Two rescue boats that have the same speed in still water travel in opposite directions in a river with a current of 5 km/h. The rescue boat going upstream departed 1h before the boat going downstream. A period after the boat going downstream has departed, a radio conversation between the boats indicated that one boat is 44 km upstream and the other boat is 75 km downstream. Approximate, to the nearest tenths of kilometer per hour, the speed of each boat in still water.
Solution:
Let $r$ = rate of each rescue boat in still water in km/h $r + 5$ = rate of the rescue boat going downstream in km/h $r – 5$ = rate of the rescue boat going upstream in km/h
Since the distance D is the product of the rate r and time t, or D = rt, then $t = \frac{D}{r}$. Use this equation and a table to organize the given information.
| Distance | rate | time | |
|---|---|---|---|
| Boat going upstream | 44 | $(r – 5)$ | $\frac{44}{r – 5}$ |
| Boat going downstream | 75 | $(r + 5)$ | $\frac{75}{r + 5}$ |
If the boat traveling upstream departed 1hr before the boat going downstream, then we have
(time for boat going upstream) = (time for boat going downstream) + 1
$\frac{44}{r – 5} = \frac{75}{r + 5} + 1$
$(r – 5)(r + 5)\left(\frac{44}{r – 5}\right) = (r – 5)(r + 5)\left[\frac{75}{r + 5} + 1\right]$
$44(r + 5) = 75(r – 5) + (r – 5)(r + 5)$
$44r + 220 = 75r – 375 + r^2 – 25$
$0 = r^2 + 31r – 620$
$r = \frac{-31 \pm \sqrt{(31)^2 – 4(1)(-620)}}{2(1)}$
[transcriber’s note: the title of the module is not printed on these pages; the earlier pages carry it]
[transcriber’s note: the solution steps below are printed above the section heading and are not labeled]
𝑟 ≈ (−31 ± √3441) / 2 → 𝑟 ≈ (−31 ± 58.660) / 2
𝑟 ≈ 13.8 or 𝑟 ≈ −44.8 [callout: Not an appropriate solution]
Answer: The speed of each boat in still water is approximately 13.8 km/h.
WHAT’S MORE
Critical Thinking
Solve the following problems:
- A positive integer is six less than the other integer. The sum of the reciprocal of the two numbers is 14/40. What are the two integers?
- Two groups of volunteers can repack relief goods in 4 hours. One group would take 2 hours longer than the faster factory to the repacking. How many hours would it take each group working alone?
WHAT I HAVE LEARNED
To solve problems involving rational algebraic equation transformable to quadratic equations you may use the following procedures: (this is just almost the same with the previous module)
In solving problems involving the quadratic equation, you may use the following procedures:
- Read and analyze the problem carefully.
- Collect the necessary data.
- Let the missing data be any variables and use algebraic expression to represent some data.
- Convert the remaining data into mathematical sentence. This will serve as the “working rational algebraic equation”.
- Solve the rational algebraic equations.
- Upon solving, observe if the resulted equation is quadratic.
- If yes, find the solution set or roots of the quadratic equation using any method discussed on the previous lessons.
- For assurance, just substitute the obtained roots to the “working rational algebraic equation”. [you may obtain *extraneous root]
NOTE: Solving problems involving equations transformable to quadratic equations does not end in finding and checking the roots. Since, it is a real-life problem you must decide whether both or one or none of the solutions is reasonable.
WHAT I CAN DO
Critical Thinking
MORE PRACTICE MORE FUN!
Solve the following problems. Give your answers in 2 decimal places if necessary.
- A boat can travel at a speed of 12km/h in still water. If it takes three hours to go 15 km upstream and back, find the speed of the current, giving your answer to 2 decimal places.
- Gabo can do a job in one hour less than Wynter. If Gabo and Wynter work together the job takes 6/5 hours. How long would it take each person working alone?
- The time for the bus trip between two towns is 45 minutes shorter than the time for the train trip between the same two towns. If the towns are 350 km apart and the train speed is 15 km/h slower than the bus, find the speed of the bus in km/h.
- A cistern can be filled by two pipes. The small pipe alone will take 24 minutes longer than the larger pipe to fill the cistern alone. The small pipe alone will take 32 minutes longer to fill the cistern alone than when the two pipes are operating together. How long will it take the larger pipe to fill the cistern alone?
ASSESSMENT
Write the letter of the correct answer on your answer sheet. If your answer is not among the choices, write E together with your final answer.
A worker can clean a pool in four hours less time than it takes another worker. If the men work together the job takes 8/3 hours. How long would it take the slower man working alone?
A. 6 hours
B. 8 hours
C. 7 hours
D. 4 hoursJessie and Mark are planning to paint a house together. Jessie thinks that if he works alone, it will take him 5 hours more than the time Mark takes to paint the entire house. Working together, they can complete the job in 6 hours. How long will it take Jessie to finish the job?
A. 5 hours
B. 15 hours
C. 10 hours
D. 20 hoursThe distance between two cities A and B is 140 km. A car driving from A to B left at the same time as a car driving from B to A. The cars met after one hour, then the first car reached city B 35 minutes later than the second car reached city A. Find the speed of the car from city A to B.
A. 60 km/h
B. 50 km/h
C. 80 km/h
D. 70 km/hA river is flowing at a rate of 5 km/h. A boat travels 12 km upstream and 10 km downstream in a total of 5 hours. What is the speed of the boat in still water?
A. 17.6 km/h
B. 27.6 km/h
C. 43.5 km/h
D. 54.6 km/hGabriel drove 25 km and then biked 10 km to the campsite. The trip took 2 hours. If he drove 20 km/h faster than he biked, how fast did he drive?
A. 12.6 km/h
B. 20.4 km/h
C. 27.8 km/h
D. 45.8 km/hTwo taps A and B fill a swimming pool together in two hours. Alone, it takes tap A three hours less than B to fill the same pool. How many hours does it take tap A to fill the pool separately?
A. 3 hrs.
B. 6 hrs.
C. 9 hrs.
D. 12 hrs.Two faucets can fill a tank in 1 hour and 20 minutes. The first faucet takes more than two hours longer to fill the same tank when functioning without the second tap. How long does it take to fill the tank at first?
A. 2 hours
B. 6 hours
C. 7 hours
D. 4 hoursA motorboat makes a round trip on a river of 45 miles upstream and 45 miles downstream, maintaining the constant speed 12 miles per hour relative to the water. The entire round trip takes 8 hours. What is the speed of the current?
A. 3 mi/h
B. 2 mi/h
C. 9 mi/h
D. 6 mi/hAn airplane makes a trip of 2400 miles long into the wind and 2400 miles back with the same tail wind, maintaining the constant speed of 440 miles per hour relative to the air. The entire round trip takes 11hour. Find the speed of the wind.
A. 30 mi/h
B. 40 mi/h
C. 20 mi/h
D. 60 mi/hA boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/h, find the speed of the boat in still water.
A. 3.5 km/h
B. 2.5 km/h
C. 2 km/h
D. 3 km/h
ADDITIONAL ACTIVITIES
Communication, Creativity and Collaboration
A REALITY OF RATIONAL ALGEBRAIC EXPRESSIONS:
Cite and create a real-life situation where the concept of a rational algebraic equation transformable into a quadratic equation is being applied. Use the situation to answer the following questions.
- How are the concept of a rational algebraic equations transformable into a quadratic equation applied in the situation?
- What quantities are involved in the situation?
- Formulate then solve a problem out of the given situation.
- What do solutions obtained represent? Explain your answer.
PROBLEM – BASED WORKSHEET
A. Advertising
An amusement park wants to place a new rectangular billboard to inform visitors of their new attractions. Suppose the length of the billboard to be placed is 4 meters longer than its width and the area is 96 square meters. What will the length and width of the billboard?”
[illustration: a rectangular billboard on a steel frame among trees]
- How will you represent the width and the length of the billboard?
- What equation describes the area of the billboard?
- What is the length and the width of the billboard?
B. Flag Ceremony
In one of the Flag raising ceremony, the school counsellor noticed that 780 students stand in rows and columns. Each row has equal number of students and each column has equal number of students. If the number of students in each row is 4 more than the number of rows, find the number of students in each row.
[transcriber’s note: the printed page numbers on this scan read 11; the page marker follows the requested numbering]
E-Search
You may also check the following links for your reference and further learnings on solving problems involving quadratic equations and rational algebraic equations.
- https://www.youtube.com/watch?v=ylgYZkzp4CA
- https://www.analyzemath.com/Algebra2/solutions-to-equations.html
- https://study.com/academy/topic/rational-expressions-quadratic-equations.html
- http://mathvids.com/lesson/mathhelp/1437-solving-rational-equation-to-quadratic-1
REFERENCES
Fadrigal, Catalina M, et.al. College Algebra .
Mathematics Learner’s Material 9
The Keng Seng, Loh Cheng Yee et. al. Mathematics Workbook 3, Shinglee Publishers PTE LTD
Sinom L. Chua, Benson Tan, et. al. Mastering Intermediate Algebra II, SIBS Publishing House, Inc.
https://www.freepik.com/free-vector/woman-with-long-hair-teaching-online_7707557.htm https://www.freepik.com/free-vector/kids-having-online-lessons_7560046.htm https://www.freepik.com/free-vector/illustration-with-kids-taking-lessons-online-design_7574030.htm
[transcriber’s note: this page is printed upside down in the scan; the content below is transcribed in upright reading order, and the printed page number reads 12]
ANSWER KEY
WHAT’S IN
- (x+18)/(6x)
- (x²+4x+2)/(2x)
- (−3y²−y+10)/(2y)
- (2a²+1)/(a(2a+1))
- (3m²+2m+5)/(2m(m+1))
WHAT I CAN DO
- 4.90 km/h
- 3 hours, 2 hours
- 91.5 km/h
- 24 minutes
WHAT’S MORE
- 4,10
- 3+√17 hours ≈ 7.12 hours, 5+√17 hours ≈ 9.12 hours
WHAT I KNOW
- C
- B
- C
- A
- C
- B
- A
- C
- A
- A
ASSESSMENT
- B
- B
- A
- B
- C
- A
- D
- A
- B
- A
PROBLEM – BASED WORKSHEET
A.
- If we represent the width, in meters, of the billboard by x, then its length is x + 4.
- Since the area of the billboard is 96 m², then x(x + 4) = 96.
- Solving the equation, we have:
x(x + 4) = 96 x² + 4x − 96 = 0 (x + 12)(x − 8) = 0 x + 12 = 0 or x − 8 = 0 x = −12 or x = 8
We only consider the positive value of x since the situation involves measures of length. Thus, the dimensions of the billboard are 8 meters (width) and 12 meters (length).
B. Let r be the number of rows, and r + 4 be the number of students in each row. The sum of the product of the number of rows and the number of students in each row add up to 780. In symbols, (r)(r + 4) = 780. Solving the resulting quadratic equation,
r(r + 4) = 780 r² + 4r − 780 = 0 (r + 30)(r − 26) = 0 r + 30 = 0 or r − 26 = 0 r = −30 or r = 26
Since the number of rows must be a positive number, we conclude that r = 26. And the number of students in each row r + 4 = 26 + 4 = 30 students.
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