Linear Equation Word Problems - Examples & Practice - Expii

Linear equation word problems - examples & practice, explanations (3), (video) slope-intercept word problems.

by cpfaffinator

how to solve linear models word problems

This video by cpfaffinator explains how to translate word problems into variables while working through a couple of examples.

(Note: the answer at 3:34 should be y=0.50x+3 instead of y=0.5x+b .)

When writing linear equations (also called linear models) from word problems, you need to know what the x and y variables refer to, as well as what the slope and y-intercept are. Here are some tricks to help translate :

  • The x variable. Usually, you'll be asked to find an equation in terms of some factor. This will usually be whatever affects the output of the equation.
  • The y variable. The y variable is usually whatever you're trying to find in the problem. It'll depend on the x variable.
  • The slope (m). The slope will be some sort of rate or other change over time.
  • The y-intercept (b). Look for a starting point or extra fees—something that will be added (or subtracted) on , no matter what the x variable is.

Using these parts, we can write the word problem in slope intercept form , y=mx+b, and solve.

A city parking garage charges a flat rate of $3.00 for parking 2 hours or less, and $0.50 per hour for each additional hour. Write a linear model that gives the total charge in terms of additional hours parked.

First up, we want the equation, or model, to give us the total charge. This will be what the y variable represents. y=total charge for parking The total charge depends on how many additional hours you've parked. This is the x variables. x=additional hours parked

Next, we need to find the slope, m. If we look at the slope-intercept equation , y=mx+b, the slope is multiplied with the input x. So, we want to find some rate in our problem that goes with the number of additional hours parked . In the problem, we have $0.50 per hour for each additional hour. The number of hours is being multiplied by the price of 50 cents every hour. m=0.50

The y-intercept is the starting point. In other words, if we parked our car in the garage for 0 hours, what would the price be? The problem says the garage *charges a flat rate of $3.00. A flat rate means that no matter what we're being charged 3 dollars. b=3

Plugging these into the slope-intercept equation will give us our model. y=mx+by=0.50x+3

Six 2 foot tall pine trees were planted during the school's observation of Earth Awareness Week in 1990. The trees have grown at an average rate of 34 foot per year. Write a linear model that gives the height of the trees in terms of the number of years since they were planted.

What does the y variable represent in our equation?

Years since the trees were planted

Number of trees planted

Radius of trees' roots

Height of the trees

Related Lessons

Translate to a linear equation.

Sometimes, the trick to solving a word problem will be to translate it into a linear equation. To clue you in, linear equation word problems usually involve some sort of rate of change , or steady increase (or decrease) based on a single variable. If you see the word rate , or even "per" or "each" , it's a safe bet that a word problem is calling for a linear equation.

There are a couple steps when translating from a word problem to a linear equation. Review them, then we'll work through an example.

  • Find the y variable, or output. What is the thing you're trying to find? This will often be a price, or an amount of time, or something else countable that depends on other things.
  • Find the x variable, or input. What is affecting the price, or amount of time, etc.?
  • Find the slope. What's the rate in the problem?
  • Find the y-intercept. Is there anything that's added or subtracted on top of the rate, no matter what what x is?
  • Plug all the numbers you know into y=mx+b !
Wally and Cobb are starting a catering business. They rent a kitchen for $350 a month, and charge $75 for each event they cater. If they cater 12 events in a month, how much do they profit?

First, let's find the y variable. What are we trying to find in this problem?

The name of the business

how to solve linear models word problems

Word Problems with Linear Relationships

When you're faced with a linear word problem, it can feel daunting to work out a solution. The good news is that there are few simple steps that you can take to tackle linear word problems.

how to solve linear models word problems

Image source: by Anusha Rahman

For future linear word problems, you can use these same steps to tackle the problem!

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Word Problems: Linear Models

Word problems are helpful for understanding why we are learning certain mathematical operations. They show real-life situations where we can put our math knowledge to work. Sometimes word problems ask us to write a linear function to model a particular situation.

Word problems using the slope-intercept form

Some word problems are phrased in such a way that we can easily find a linear function using the slope-intercept form of the equation for a line.

Remember, the slope-intercept form is in the form of:

y = m x + b

Elaine's electric company charges her $0.11 per kWh (kilowatt hour) of electricity, plus a basic connection charge of $15.00 per month. Write a linear function that models her monthly electricity bill as a function of electricity usage.

Here, when Elaine uses zero electricity, that is when x = 0 , the bill is $15.00. Therefore, the y-intercept is 15.

The rate of change is 0.11. That is, for each increase of x by 1 unit, in this case, kilowatt hours, there is an increase in y by $0.11.

Substitute the slope-intercept form y = m x + b

y = 0.11 x + 15

Word problems using the point-slope form

Some word problems are phrased in such a way that we can easily find a linear function using the point-slope form of the equation for a line.

Remember, the formula for the point-slope form of a line is:

y − y 1 = m ( x − x 1 )

Anthony lives in Telluride, Colorado, but his job is located in Denver, Colorado. Every Monday, he drives his car 332 miles from Telluride to Denver, spends the week in a company apartment, and then drives back to Telluride on Friday. He doesn't use his car for anything else. After 20 weeks, his odometer shows that he has traveled 240,218 miles since he bought the car.

Write a linear model that gives the odometer reading of the car as a function of the number of weeks since Anthony started his new job.

First, find the rate of change. Be sure to multiply the distance by 2 to find his round-trip distance since he has to go and come back.

2 ( 332 ) = 664 miles per week

This represents the slope of the line. But since we are not given the odometer reading of the car before he starts the job, we don't know the y-intercept yet.

That's okay, though. We have the coordinates of a point 20 240218 . So we can use the point-slope form y − y 1 = m ( x − x 1 ) .

y − 240218 = 664 ( x − 20 )

Note that you can use this equation to find the y-intercept if you want to.

y = 664 x + 226938

Topics related to the Word Problems: Linear Models

Solving Multi-Step Linear Equations

Word Problems

Graphing Linear Equations

Flashcards covering the Word Problems: Linear Models

Algebra 1 Flashcards

College Algebra Flashcards

Practice tests covering the Word Problems: Linear Models

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

Get help learning about word problems: Linear models

Solving word problems that involve linear models can be confusing for students. If your student needs help figuring out these types of word problems, having them work with a qualified tutor is an excellent idea. Their tutor can give them the boost they need to understand word problems thoroughly. To learn more about how tutoring can help your student with math concepts like solving word problems involving linear models, contact the Educational Directors at Varsity Tutors today.

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how to solve linear models word problems

Word Problems Linear Equations

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

how to solve linear models word problems

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Linear Programming Word Problems

Intro How-To Word Problems More Examples Four Variables

What are the constraints they "forget" to mention?

When you are doing a linear programming word problem, you are dealing with a (pretend) real-world situation. In this context, you will need to remember the understood (and thus usually omitted) constraints; namely, that you can't (generally) have negative amounts of inputs.

Content Continues Below

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This omitted fact requires that you remember to add non-negativity constraints for each of your graphing variables; namely, x  ≥ 0 and y  ≥ 0 .

  • A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.   If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits?

The question asks for the optimal number of calculators, so my variables will stand for numbers of calculators:

x : number of scientific calculators produced

y : number of graphing calculators produced

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Since they can't produce negative numbers of calculators, I have my first two constraints; namely, x  ≥ 0 and y  ≥ 0 . However, as it turns out, I can ignore these constraints in this particular exercise, because they've already given me minimums:

The exercise also gives maximums:

The minimum shipping requirement gives me another constraint:

x + y ≥ 200

For graphing purposes, I'll restate this constraint as:

y ≥ − x + 200

The profit relation will be my optimization equation:

P = −2 x + 5 y

So my entire system is:

P = 2 x + 5 y , subject to:

100 ≤ x ≤ 200

80 ≤ y ≤ 170

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The feasibility region graphs as:

The corner points are at (100, 170) , (200, 170) , (200, 80) , (120, 80) , and (100, 100) . When you test these points in the optimization equation, you should obtain the maximum value of profit P  = 650 at ( x ,  y ) = (100, 170) . Interpreting this within the context of the original word problem, the solution will be " 100 scientific calculators and 170 graphing calculators".

(Yes, you can also find the answer just using logic: If the one kind loses you money and the other makes you money, obviously you'll want to maximize production of what makes you money.)

  • You need to buy some filing cabinets. You know that Cabinet X costs $10 per unit, requires six square feet of floor space, and holds eight cubic feet of files. Cabinet Y costs $20 per unit, requires eight square feet of floor space, and holds twelve cubic feet of files. You have been given $140 for this purchase, though you don't have to spend that much. The office has room for no more than 72 square feet of cabinets. How many of which model should you buy, in order to maximize storage volume?

The question ask for the number of cabinets I need to buy, so my variables will stand for that:

x : number of model X cabinets purchased y : number of model Y cabinets purchased

Naturally, x  ≥ 0 and y  ≥ 0 . I have to consider the costs and the floor space (that is, the footprint of each unit), while maximizing the storage volume, so costs and floor space will be my constraints, while volume will be my optimization equation.

cost: 10 x + 20 y ≤ 140 , or y ≤ −(½) x + 7

space: 6 x + 8 y ≤ 72 , or y ≤ −(¾) x + 9

volume: V = 8 x + 12 y

This system (along with the first two constraints) graphs as:

The corner points are at (0, 0) , (8, 3) , (0, 7) , and (12, 0) . When you plug them into the optimization equation, you should obtain a maximal volume of 100 cubic feet by buying eight of model X and three of model Y.

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how to solve linear models word problems

SOLVING WORD PROBLEMS USING LINEAR MODELS

Plan for Solving a Word Problem :

(i) Find out what numbers are asked for from the given information.

(ii) Choose a variable to represent the number(s) described in the problem. Sketch or a chart may be helpful.

(iii) Write an equation that represents relationships among the numbers in the problem.

(iv)  Solve the equation and find the required numbers.

(v) Answer the original question. Check that your answer is reasonable.

A linear function y = mx + b can be used as a model for many types of real life word problems which involve a constant rate of change.

Example 1 : 

A person travels home from work at a constant speed. Ten minutes after leaving work he is 20 miles from home, and 20 minutes after leaving work he is 12 miles from home. If he continues to travel at the same speed, how long will it take him to arrive home from work ?

The problem asks for the number of minutes it takes to travel from work to home.

Start with the linear equation y = mx + b, in which y is the distance in miles from home, and x is the time in minutes.

When x = 10, y = 20

When x = 20, y = 12

By subtracting the second equation from the first equation we get

Divide each side by -10.

- ⁴⁄₅  = m

Substitute m = -4/5 into the first equation. 

20 = 10( - ⁴⁄₅ ) + b

20 = -8 + b

Add 8 to each side.

In y = mx + b, replace m with - ⁴⁄₅ and b with 28.

y = ( - ⁴⁄₅ )x + 28

When he arrives home, y = 0. 

0 = (- ⁴⁄₅ )x + 28

Solve for x. 

( ⁴⁄₅ )x = 28

Multiply each side by 5.

Divide both sides by 4.

It takes 35 minutes from work to home.

Example 2 :

At the beginning of a trip, the tank of Chloe’s car was filled with 12 gallons of gas. When she travels constantly on the highway 60 miles per hour, the car consumes 1 gallon of gas per 35 miles. If she traveled 5 hours and 15 minutes on the highway with a constant speed of 60 miles per hour, how many gallons of gas are left in the tank?

(A)  3

(B)  4

(C)  5

(D)  6

Start with the linear equation y = mx + b , in which y is the number of gallons gas left in the tank after x miles of distance travelled.

It is given that the car consumes 1 gallon of gas for every 35 miles of distance. That is, amount of gas left in the tank is decreasing at the rate of 1 gallon per 35 miles.

y = (- ¹⁄₃₅ )x + b

Here, b is the amount of gas in the tank initially, that is, 12 gallons.

y = (- ¹⁄₃₅ )x + 12  ----(1)

Distance travelled in 5 hours 15 minutes :

= speed  ⋅ time

= 60  ⋅ 5 ¼

= 60 ⋅   ²¹⁄₄

= 15 ⋅  21

= 315 miles

Substitute x = 315 in (1).

y = (- ¹⁄₃₅ )(315) + 12 

y = -9 + 12

After Chloe traveled 5 hours and 15 minutes, amount of gas left in the tank is 3 gallons.

Example 3 :

A rock climber is climbing up a 450 feet high cliff. By 9:30 am. the climber reached 90 feet up the cliff and by 11:00 am, he has reached 210 feet up the cliff. If he climbs with a constant speed, by what time will he reach the top of the cliff?

(A)  1 : 45 pm

(B)  2 : 00 pm

(C)  2 : 15 pm

(D)  2 : 30 pm

Start with the linear equation y = mx + b , in which y is the height climbed in x hours.

We can assume the following values corresponding to thew 9:30 am and 11.00 am.

9.30 am ----> 0 hours

11.00 am ----> 1.5 hours

Average rate of climbing :

= 80 feet per hour

y = 80x + b

Here, b is the initial height. That is 90 feet at 9.30 am (or 0 hours).

y = 80x + 90

At the top of the cliff, y = 450.

450 = 80x + 90

x = 4.5 hours

4.5 hours later from 9 : 30 am is 2 pm.

Therefore, the correct answer is option (B).

Example 4 :

In 2005 a house was purchased for $280,000 and in 2013 it was sold at $334,000. Assuming that the value of the house increased at a constant annual rate what would be the price of the house in the year 2018?

(A)  $354,250

(B)  $361,000

(C)   $367,750

(D)   $374,500

We can assume the following values corresponding to thew 2005, 2013 and 2018.

2005 ----> 0

2013 ----> 8

2018 ----> 13

Average rate of change in the price of the house :

= $6,750 per year

y = 6,750x + b

Here, b is the initial height. That is  $280,000 in 2005.

y = 6,750x + 280,000

To find the price of the house in the year 2018, substitute x = 13.

y = 6,750(13) + 280,000

y = 87,750 + 280,000

y = 367,750

Therefore, the correct answer is option (C).

Example 5 :

To join Eastlake Country Club one must pay d  dollars for a one time membership fee and pay w dollars for a monthly fee. If the first month is free for the club, what is the total amount, y , x months after a person joined the club, in terms of d , w , and x ?

(A)  y = wx - 1 + d

(B)  y = w(x - 1) + d

(C)  y = d(x - 1) + w

(D)  y = dx - 1 + w

Since the first month is free for the club, the amount of monthly fee x months after a person joined the club is  w (x - 1), and the total amount including the one time membership is

y = w(x - 1) + d

Therefore, the corrrect answer is option (B).

Example 6 :

From 1990 to 2000 The population of city A rose from 12,000 to 28,000 and the population of city B  rose from 18,000 to 24,000. If the population of the two cities increased at a constant rate, in what year was the population of both cities the same?

Average rate of change in the population of city A :

= 1,600 per year

Average rate of change in the population of city B :

= 600 per year

Let x be the number of years from 1990 and y be the population after 1990.

The population of city A after 1990 would be

y = 1600x + 12000

The population of city B after 1990 would be

y = 600x + 18000 

To find the year the population of both cities were the same, let the two equations be equal.

1600x + 12000 = 600x + 18000

1000x = 6000

The year which is 6 years from 1990 is 1996.

In the year 1996, the  population of both cities was the same.

Example 7 :

An empty 1,200 gallon tank is filled with water at a rate of 6 gallons of water per minute. At the same time, another 1,200 gallon tank full of water is being drained at a rate of 9 gallons per minute. How many minutes will it take for the amount of water in both tanks to become the same?

Let y be the amount of water in the both the tanks after x minutes.

The empty 1,200 gallon tank is filled with water at a rate of 6 gallons of water per minute.

Amount of water in the emty tank after x minutes :

Another 1,200 gallon tank full of water is being drained at a rate of 9 gallons per minute.

Amount of water in the second tank full of water after x minutes :

y = -9x + 1200

When the amount of water in both tanks to become the same,

6x = -9x + 1200

It take 80 minutes for the amount of water in both tanks to become the same.

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Modeling Word Problems

Using models is a critical step in helping students transition from concrete manipulative work with word problems to the abstract step of generating an equation to solve contextual problems. By learning to use simple models to represent key mathematical relationships in a word problem, students can more easily make sense of word problems, recognize both the number relationships in a given problem and connections among types of problems, and successfully solve problems with the assurance that their solutions are reasonable.

Why is modeling word problems important?

Mr. Alexander and teachers from his grade level team were talking during their Professional Learning Community (PLC) meeting about how students struggle with word problems. Everyone felt only a few of their students seem to be able to quickly generate the correct equation to solve the problem. Many students just seem to look for some numbers and do something with them, hoping they solve the problem.

Mr. Alexander had recently learned about using modeling for word problems in a workshop he had attended.  He began to share the model diagrams with his teammates and they were excited to see how students might respond to this approach. They even practiced several model diagrams among themselves as no one had ever learned to use models with word problems. Since part of their PLC work freed them up to observe lessons in each others' rooms, they decided they would watch Mr. Alexander introduce modeling to his students. 

So, two days later they gathered in Mr. Alexander's room for the math lesson. Mr. Alexander presented the following problem:

Lily and her brother, Scotty, were collecting cans for the recycling drive. One weekend they collected 59 cans and the next weekend they collected 85 cans.  How many cans were collected in all?

Mr. Alexander went over the problem and drew a rectangular bar divided into two parts on the board, explaining that each part of the rectangle was for the cans collected on one of the weekends and the bracket indicated how many cans were collected in all. Reviewing the problem, Mr. Alexander asked students what was not known, and where the given numbers would go and why. This resulted in the following bar model:

how to solve linear models word problems

The class then discussed what equations made sense given the relationship of the numbers in the bar model. This time many students wrote the equation, 59 + 85 = ?, and solved the problem. In their discussion after the lesson, Mr. Alexander's teammates mentioned that they noticed a much higher degree of interest and confidence in problem solving when Mr. Alexander introduced the bar model. Everyone noticed that many more students were successful in solving problems once modeling was introduced and encouraged. As the class continued to do more word problems, the diagrams appeared to be a helpful step in scaffolding success with word problems.

Knowledge can be represented both linguistically and nonlinguistically. An increase in nonlinguistic representations allows students to better recall knowledge and has a strong impact on student achievement (Marzano, et. al., 2001, Section 5). In classic education research, Bruner (1961) identified three modes of learning:  enactive (manipulating concrete objects), iconic (pictures or diagrams), and symbolic (formal equation).  The iconic stage, using pictures and diagrams, is an important bridge to abstracting mathematical ideas using the symbols of an equation. Research has also validated that students need to see an idea in multiple representations in order to identify and represent the common core (Dienes, undated). For word problems involving the operation of addition, students need to experience several types of problems to generalize that when two parts are joined they result in a total or a quantity that represents the whole. Whether the items are bears, balloons, or cookies no longer matters as the students see the core idea of two subsets becoming one set. Dienes discovered that this abstraction is only an idea; therefore it is hard to represent. Diagrams can capture the similarity students notice in addition/joining problems where both addends are known and the total or whole is the unknown.  Diagrams will also be useful for missing addend situations. Like Bruner, Dienes saw diagrams as an important bridge to abstracting and formalizing mathematical ideas.  

Along with Bruner and Dienes, Skemp (1993) identified the critical middle step in moving from a real-life situation to the abstractness of an equation. While students need to experience many real-life situations they will get bogged down with the "noise" of the problem such as names, locations, kinds of objects, and other details. It is the teacher's role to help students sort through the noise to capture what matters most for solving the problem. A diagram can help students capture the numerical information in a problem, and as importantly, the relationship between the numbers, e.g. Do we know both the parts, or just one of the parts and the whole? Are the parts similar in size, or is one larger than the other? Once students are comfortable with one kind of diagram, they can think about how to relate it to a new situation. A student who has become proficient with using a part-part-whole bar model diagram when the total or whole is unknown, (as in the collecting cans problem in Mr. Alexander's class), cannot only use the model in other part-part-whole situations, but can use it in new situations, for example, a missing addend situation. Given several missing addend situations, students may eventually generalize that these will be subtractive situations, solvable by either a subtraction or adding on equation.

The work of Bruner, Dienes and Skemp informed the development of computation diagrams in some elementary mathematics curriculum materials in the United States. Interestingly, it also informed the development of curriculum in Singapore, as they developed the "Thinking Schools, Thinking Nation" era of reforming their educational model and instructional strategies (Singapore Ministry of Education, 1997). The bar model is a critical part of "Singapore Math."  It is used and extended across multiple grades to capture the relationships within mathematical problems. Singapore has typically scored near the top of the world on international assessments, a possible indicator of the strong impact of including the visual diagram step to represent and solve mathematical problems.

What is modeling word problems?

Models at any level can vary from simple to complex, realistic to representational. Young students often solve beginning word problems, acting them out, and modeling them with the real objects of the problem situation, e.g. teddy bears or toy cars. Over time they expand to using representational drawings, initially drawing pictures that realistically portray the items in a problem, and progressing to multi-purpose representations such as circles or tally marks. After many concrete experiences with real-life word problems involving joining and separating, or multiplying and dividing objects, teachers can transition students to inverted-V and bar model drawings which are multi-purpose graphic organizers tied to particular types of word problems.

Modeling Basic Number Relationships

Simple diagrams, sometimes known as fact triangles, math mountains, situation diagrams, or representational diagrams have appeared sporadically in some curriculum materials. But students' problem solving and relational thinking abilities would benefit by making more routine use of these diagrams and models.

Young children can begin to see number relationships that exist within a fact family through the use of a model from which they derive equations. An inverted-V is one simple model that helps students see the addition/subtraction relationships in a fact family, and can be used with word problems requiring simple joining and separating. The inverted-V model can be adapted for multiplication and division fact families. For addition, students might think about the relationships among the numbers in the inverted-V in formal terms, addend and sum , or in simpler terms, part and total , as indicated in the diagrams below.

how to solve linear models word problems

A specific example for a given sum of 10 would be the following, depending on which element of the problem is unknown.

how to solve linear models word problems

      6 + 4 = ?                   6 + ? = 10                       ? + 4 = 1

         4 + 6 = ?                 10 - 6 = ?                       10 - 4 = ?

While often used with fact families, and the learning of basic facts, inverted-V diagrams can also work well with solving word problems. Students need to think about what they know and don't know in a word problem - are both the parts known, or just one of them?  By placing the known quantities correctly into the inverted-V diagram, students are more likely to determine a useful equation for solving the problem, and see the result as reasonable for the situation. For example, consider the following problem:

Zachary had 10 train cars. Zachary gave 3 train cars to his brother. How many train cars does Zachary have now? 

Students should determine they know how many Zachary started with ( total or whole ), and how many he gave away ( part of the total ). So, they need to find out how many are left ( other part of the total ). The following inverted-V diagram represents the relationships among the numbers of this problem:

how to solve linear models word problems

3 + ? = 10 or 10 - 3 = ?, so Zachary had 7 train cars left.

As students move on to multiplication and division, the inverted-V model can still be utilized in either the repeated addition or multiplicative mode. Division situations do not require a new model; division is approached as the inverse of multiplication or a situation when one of the factors is unknown.

how to solve linear models word problems

Again, the inverted-V diagram can be useful in solving multiplication and division word problems. For example, consider the following problem:

Phong planted 18 tomato plants in 3 rows. If each row had the same number of plants, how many plants were in each row?   

Students can see that they know the product and the number OF rows. The number IN A row is unknown. Either diagram below may help solve this problem, convincing students that 6 in a row is a reasonable answer.

how to solve linear models word problems

While the inverted-V diagram can be extended to multi-digit numbers, it has typically been used with problems involving basic fact families. Increasing the use of the inverted-V model diagram should heighten the relationship among numbers in a fact family making it a useful, quick visual for solving simple word problems with the added benefit of using and increasing the retention of basic facts.

Models and Problem Types for Computation

As children move to multi-digit work, teachers can transition students to bar model drawings, quick sketches that help students see the relationships among the important numbers in a word problem and identify what is known and unknown in a situation. 

With bar models the relationships among numbers in all these types of problems becomes more transparent, and helps bridge student thinking from work with manipulatives and drawing pictures to the symbolic stage of writing an equation for a situation. With routine use of diagrams and well-facilitated discussions by teachers, student will begin to make sense of the parts of a word problem and how the parts relate to each other.

Part-Part-Whole Problems. Part-Part-Whole problems are useful with word problems that are about sets of things, e.g. collections. They are typically more static situations involving two or more subsets of a whole set. Consider the problem,

Cole has 11 red blocks and 16 blue blocks. How many blocks does Cole have in all?

Students may construct a simple rectangle with two parts to indicate the two sets of blocks that are known (parts/addends). It is not important to have the parts of the rectangle precisely proportional to the numbers in the problem, but some attention to their relative size can aid in solving the problem. The unknown in this problem is how many there are altogether (whole/total/sum), indicated by a bracket (or an inverted-V) above the bar, indicating the total of the 2 sets of blocks. The first bar model below reflects the information in the problem about Cole's blocks. 

how to solve linear models word problems

11 + 16 = ?  so Cole has 27 blocks in all.

A similar model would work for a problem where the whole amount is known, but one of the parts (a missing addend) is the unknown. For example:

Cole had 238 blocks. 100 of them were yellow. If all Cole's blocks are either blue or yellow, how many were blue? 

The following bar model would be useful in solving this problem.

how to solve linear models word problems

100 + ? = 238 or 238 - 100 = ? so Cole has 138 blue blocks.

The answer has to be a bit more than 100 because 100 + 100 is 200 but the total here is 238 so the blue blocks have to be a bit more than 100.

The part-part-whole bar model can easily be expanded to large numbers, and other number types such as fractions and decimals. Consider the problem:

Leticia read 7 ½ books for the read-a-thon.She wants to read 12 books in all. How many more books does she have to read?   

The first diagram below reflects this problem. Any word problem that can be thought of as parts and wholes is responsive to bar modeling diagrams. If a problem has multiple addends, students just draw enough parts in the bar to reflect the number of addends or parts, and indicate whether one of the parts, or the whole/sum, is the unknown, as shown in the second figure below.

how to solve linear models word problems

     12 - 7 ½ = ? or  7 ½ + ? = 12 so Leticia needs to read 4 ½ more books.

Join (Addition) and Separate (Subtraction) Problems.

Consider this joining problem:

Maria had $20.  She got $11 more dollars for babysitting.  How much money does she have now?

Students can identify that the starting amount of $20 is one of the parts, $11 is another part (the additive amount), and the unknown is the sum/whole amount, or how much money she has now. The first diagram below helps represent this problem.   

how to solve linear models word problems

Consider the related subtractive situation:

Maria had $31.  She spent some of her money on a new CD.  Maria now has $16 left.

The second diagram above represents this situation. Students could use the model to help them identify that the total or sum is now $31, one of the parts (the subtractive change) is unknown, so the other part is the $16 she has left. 

Comparison Problems. Comparison problems have typically been seen as difficult for children. This may partially be due to an emphasis on subtraction as developed in word problems that involve "take away" situations rather than finding the "difference" between two numbers. Interestingly, studies in countries that frequently use bar models have determined that students do not find comparison problems to be much more difficult than part-part-whole problems (Yeap, 2010, pp. 88-89). 

A double bar model can help make comparison problems less mysterious. Basically, comparison problems involve two quantities (either one quantity is greater than the other one, or they are equal), and a difference between the quantities. Two bars, one representing each quantity, can be drawn with the difference being represented by the dotted area added onto the lesser amount. For example, given the problem:

Tameka rode on 26 county fair rides. Her friend, Jackson, rode on 19 rides. How many more rides did Tameka ride on than Jackson?

Students might generate the comparison bars diagram shown below, where the greater quantity, 26, is the longer bar. The dotted section indicates the difference between Jackson's and Tameka's quantities, or how much more Tameka had than Jackson, or how many more rides Jackson would have had to have ridden to have the same number of rides as Tameka.

how to solve linear models word problems

26 - 19 = ?  or  19 + ? = 26; the difference is 7 so Tameka rode 7 more rides.

Comparison problems express several differently worded relationships. If Tameka rode 7 more rides than Jackson, Jackson rode 7 fewer rides than Tameka.  Variations of the double bar model diagram can make differently worded relationships more visual for students. It is often helpful for students to recognize that at some point both quantities have the same amount, as shown in the model below by the dotted line draw up from the end of the rectangle representing the lesser quantity. But one of the quantities has more than that, as indicated by the area to the right of the dotted line in the longer bar. The difference between the quantities can be determined by subtracting 19 from 26, or adding up from 19 to 26 and getting 7, meaning 26 is 7 more than 19 or 19 is 7 less than 26.

how to solve linear models word problems

Comparison word problems are especially problematic for English Learners as the question can be asked several ways. Modifying the comparison bars may make the questions more transparent. Some variations in asking questions about the two quantities of rides that Tameka and Jackson rode might be:  

  • How many more rides did Tameka ride than Jackson?
  • How many fewer rides did Jackson go on than Tameka?
  • How many more rides would Jackson have had to ride to have ridden the same number of rides as Tameka?
  • How many fewer rides would Tameka have had to ride to have ridden the same number of rides as Jackson?

Comparisons may also be multiplicative. Consider the problem:

Juan has 36 CDs in his collection. This is 3 times the amount of CDs that his brother, Marcos, has. How many CDs does Marcos have?  

In this situation, students would construct a bar model, shown below on the left, with 3 parts. Students could divide the 36 into 3 equal groups to show the amount that is to be taken 3 times to create 3 times as many CDs for Juan.

how to solve linear models word problems

  36 ¸ 3 = ? or  3 x ? = 36             12 + 12 + 12 = ? (or 3 x 12 = ?)

so Marcos has 12 CDs.                    so Juan has 36 CDs.

A similar model can be used if the greater quantity is unknown, but the lesser quantity, and the multiplicative relationship are both known. If the problem was:

Juan has some CDs. He has 3 times as many CDs as Marcos who has 12 CDs. How many CDs does Juan have?

As seen in the diagram above on the right, students could put 12 in a box to show the number of CDs Marcos has; then duplicate that 3 times to sow that Juan has 3 times as many CDs. Then the total number that Juan has would be the sum of those 3 parts. 

Multiplication and Division Problems. The same model used for multiplicative comparisons will also work for basic multiplication word problems, beginning with single digit multipliers. Consider the problem:

Alana had 6 packages of gum. Each package holds 12 pieces of gum. How many pieces of gum does Alana have in all?

The following bar model uses a repeated addition view of multiplication to visualize the problem.

how to solve linear models word problems

12 + 12 + 12 + 12 + 12 + 12 = 72 (or 6 x 12 = 72)

so Alana has 72 pieces of gum.

As students move into multi-digit multipliers, they can use a model that incorporates an ellipsis to streamline the bar model. For example:

Sam runs 32 km a day during April to get ready for a race. If Sam runs every day of the month, how many total kilometers did he run in April?

how to solve linear models word problems

30 x 32 km = 30 x 30 km + 30 x 2 km = 960 km

Sam ran 960 km during the 30 days of April.

Since division is the inverse of multiplication, division word problems will utilize the multiplicative bar model where the product (dividend) is known, but one of the factors (divisor or quotient) is the unknown. 

Problems Involving Rates, Fractions, Percent & Multiple Steps. As students progress through the upper grades, they can apply new concepts and multi-step word problems to bar model drawings. Skemp (1993) identified the usefulness of relational thinking as critical to mathematical development. A student should be able to extend their thinking based on models they used earlier, by relating and adapting what they know to new situations. 

Consider this rate and distance problem:

Phong traveled 261 miles to see her grandmother. She averaged 58 mph. How long did it take her to get to her grandmother's house?

The following model builds off of the part-part-whole model using a repeated addition format for multiplication and division. It assumes that students have experience with using the model for division problems whose quotients are not just whole numbers. As they build up to (or divide) the total of 261 miles, they calculate that five 58's will represent 5 hours of travel, and the remaining 29 miles would be represented by a half box, so the solution is it would take Phong 5½ hours of driving time to get to her grandmother's house.

how to solve linear models word problems

Even a more complex rate problem can be captured with a combination of similar models.  Consider this problem: 

Sue and her friend Anne took a trip together.  Sue drove the first 2/5 of the trip and Anne drove 210 miles for the last 3/5 of the trip.  Sue averaged 60 mph and Anne averaged 70 mph.  How long did the trip take them?

There are several ways students might combine or modify a basic bar model. One solution might be the following, where the first unknown is how many miles Sue drove. A bar divided into fifths represents how to calculate the miles Sue drove. Since we know that the 210 miles Anne drove is 3/5 of the total trip, each one of Anne's boxes, each representing 1/5 of the trip, is 70 miles. Therefore, Sue drove two 70 mile parts, or 140 miles, to equal 2/5 of the total trip.

how to solve linear models word problems

The diagram now needs to be extended to show how to calculate the number of hours. Anne's 210 mile segment, divided by her 70 mph rate will take 3 hours, as recorded on the following extension of the diagram. Sue's distance of 140 miles now needs to be divided into 60 mph segments to determine her driving time of 2 1/3 hours.  So, the total trip of 350 miles would take 5 1/3 hours of driving time, considering the two driving rates. 

how to solve linear models word problems

Certainly, a foundation of using simple bar model drawings needs to be well developed in early grades for students to extend diagrams with understanding in later grades. The Sue and Anne rate-time-distance problem would not be the place to begin using bar models!  But, by building on work in earlier grades with models, this extended model makes the mathematics of this complex problem more transparent, and helps students think through the steps. 

Consider a simpler multi-step problem:

Roberto purchased 5 sports drinks at $1.25 each. Roberto gave the cashier $20. How much change did he get back?

Again, there may be student variations when they begin to extend the use of diagrams in multi-step or more complex problems.  Some students might use two diagrams at once, as show below on the left.  Others may indicate computation within one diagram, as shown in the diagram on the right.

how to solve linear models word problems

With routine experience with bar modeling, students can extend the use of the models to problems involving relationships that can be expressed with variables.  Consider this simple problem that could be represented algebraically:

Callan and Avrielle collected a total of 190 bugs for a science project.  Callan collected 10 more bugs than Avrielle.  How many bugs did Callan collect?

Let n equal the number of bugs Avrielle collected, and n + 10 equal the number of bugs Callan collected.  The following model might be created by students:

how to solve linear models word problems

Since n + n = 180 (or 2 n = 180), n = 90.  Therefore, Callan collected 90 + 10 or 100 bugs and Avrielle collected 90 bugs for a total of 190 bugs collected together.

Planning and Instruction

How do I intentionally plan for and use modeling?

If modeling is not a way you learned to identify the important information and numerical relationships in word problems, you may want to review some of the resources on problem types (see Carpenter's book in References and Resources section below), or bar modeling (see books by Forsten, Walker, or Yeap in the References and Resources section below).  You may also want to practice the different types of models.  Decide which are most accessible for your students, and start with introducing one model at a time, helping students determine what is unknown in the problem, and where that unknown and the other numerical information should be placed in the bar model.  A question mark, box, or a variable can be used for the unknown.  As students become comfortable with that model, introduce, and compare and contrast a second model with the known model. 

You might introduce bar model drawings, or inverted-V diagrams, when there is a unit in your curriculum that contains several word problems.  If word problems are sporadic in your curriculum, you might introduce a "Word-Problem-of-the-Day" format where students solve a problem, or cluster of related problems, each day. 

To emphasize model drawings, you might have students take a set of problems, and classify them as to which model would help them solve the problem, or do a matching activity between word problems and model drawings.  Ask students to explain why a particular equation matches a model and would be useful in finding the solution.  Another activity is to present a bar model with some numerical information and an unknown.  Then ask students to write a word problem that could logically be solved using that model.  Ask students to explain why the word problem created matches a diagram well.  As students use models for solving word problems, they may generate different equations to solve a problem even though their models are the same.  Plan for class discussions where students may discuss why there can be different equations from the same bar model. 

Several studies have shown that students who can visualize a word problem through modeling increase their problem solving ability and accuracy.  This has been particularly documented in Singapore and other high performing countries where bar modeling is used extensively across grades.  Students are more likely to solve problems correctly when they incorporate bar model drawings.  On difficult problems, students who have been able to easily generate equations with simple problems often find that bar model drawings are especially helpful in increasing accuracy as problems increase in difficulty or involve new concepts (Yeap, 2010, pp. 87-89).

TALK:  Reflection and Discussion

  • Are there particular types of word problems that your students solve more easily than others?  What characterizes these problems?
  • Identify some basic facts with which your students struggle.  How could you incorporate those facts into word problems, and how might the use of the inverted-V model help?
  • How do bar model drawings help extract and represent the mathematical components and numerical relationships of a word problem?
  • With which type of word problems would you begin to show your students the use of bar model drawings?

DO:  Action Plans

  • Select several story problems from your curriculum, MCA sample test items, or the Forsten, Walker, or Yeap resources on bar model drawing. Practice creating a bar model for several problems.  Compare your models with others in your grade level, team, or PLC group.  Practice until you feel comfortable with various model drawings.
  • Investigate the types of multiplication and division problems, and how bar models can be used with different types such as measurement and partitive division, arrays, equal groups, rates.  The Carpenter resource may be helpful.
  • Select some problems from your curriculum that are of a similar type.  Which bar model would be helpful in solving this type of problem?  Practice using the model yourself with several problems of this type.  How will you introduce the model to your students?
  • Identify some basic facts with which your students struggle.  Craft some rich word problems utilizing these fact families.  Introduce the inverted-V diagrams with the word problems to make sense of the information in the word problem, and discuss strategies for solving the problems.
  • Initiate a "Word-Problem-of-the-Day".  Students might want to keep problem solving notebooks.  Begin with problems of a particular type, and show students how to use a bar model to represent the information in a problem.  Cluster several problems of a given type during the week.  What improvements do you see in student selection of appropriate equations, accuracy of solutions, and ability to estimate or justify their answers as they increase the use of bar models to solve the word problems?  A quick way to disseminate the "Word-Problem-of-the-Day" is to duplicate the problem on each label on a sheet of address labels.  Students can just peel off the daily problem, add it to their problem solving notebook or a sheet of paper and solve away.
  • When your district is doing a curriculum materials review, advocate to include a criteria that requires the use of visual models in helping students make sense of mathematical problems. 
  • Watch some of the videos of students using models on the Powerful Practices CD (see Carpenter and Romberg in References and Resources Section). 

References and Resources

Bruner, J. S. (1961). The act of discovery.  Harvard Educational Review, 31, pp. 21-32, in Yeap, Ban Har. (2010). Bar modeling:  A problem solving tool.  Singapore: Marshall Cavendish Education. 

Carpenter, T. P., Fennema, E., Franke, M. L., Levi, L. & Empson, S. B. (1999). Children's mathematics: Cognitively guided instruction.  Portsmouth, NH:  Heinemann.  (Book and CD)

Carpenter, T. P. & Romberg. T. A. (2004). Modeling, generalization, and justification in mathematics cases, in Powerful practices in mathematics & science.  Madison, WI:  National Center for Improving Student Learning and Achievement in Mathematics and Science.  www.wcer.wisc.edu/ncisla   (Booklet and CD)

Dienes, Z. (undated). Zoltan Dienes' six-state theory of learning mathematics. Retrieved from http://www.zoltandienes.com

Forsten, C. (2009). Step-by-step model drawing:  Solving math problems the Singapore way.  Peterborough, NH: SDE:  Crystal Spring Books.  http:// www.crystalspringsbooks.com

Hoven, J. & Garelick, B. (2007). Singapore math: Simple or complex? Educational Leadership, 65 (3), 28-31.

Leinwand, S. (2009). Accessible mathematics: 10 instructional shifts that raise student achievement.  Portsmouth, NH:  Heinemann.

Marzano, R. J., Norford, J. S., Paynter, D. E., Pickering, D. J., & Gaddy, B. B. (2001).   A handbook for classroom instruction that works.  Alexandria, VA:  Association for Supervision and Curriculum Development.

Singapore Ministry of Education. (1997). Retrieved http://moe.gov.sg

Skemp, R. R. (January, 1993). "Theoretical foundations of problem solving: A position paper."  University of Warwick. Retrieved from http://www.grahamtall.co.uk/skemp/sail/theops.html

Walker, L. (2010). Model drawing for challenging word problems:  Finding solutions the Singapore way.   Peterborough, NH: SDE:  Crystal Spring Books.  http:// www.crystalspringsbooks.com

Yeap, B. H. (2010). Bar modeling:  A problem solving tool.  Singapore: Marshall Cavendish Education.  http:// www.singaporemath.com

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Solving Linear Equation Word Problems 9 Terrific Examples!

// Last Updated: January 20, 2020 - Watch Video //

What’s the first thing that comes to mind when you hear the phrase Word problems?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching linear word problems

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

For some, it’s a chance to solve a real-world example, so there’s a level of excitement and sense of wonder. For others, it’s groaning, and frustration on where to even begin.

Well, in this lesson we’re going to make Solving Linear Equation Word Problems manageable with easy to follow tricks and steps.

We already know how to solve all different types of equations. Yay!

And we also know how to translate algebraic expressions and equations. Double Yay!

Now it’s time to bring both of these together.

Finding the length of the missing side of a triangle

Solving equations and word problems Example

But, what about the tricks and steps?

Yes, there are some easy to follow steps that we are going to use to solve linear word problems.

  • Read the problem carefully and determine what is being asked.
  • Create a sidebar! Using different colors, symbols and diagrams and write an equation the relates all the information given.
  • Solve your equation and check your answer(s).

Now, these steps might not seem all that remarkable, but once you see them in action I guarantee that writing equations from word problems and solving them will become like second nature!

Again, the secret to success is your Sidebar. This is where you will write down all the information you’ve gleaned from the problem, and formulate a solution by writing an equation to model the situation, as Khan Academy accurately states.

Together we will walk through 9 examples in detail ranging from finding consecutive integers to finding hourly wages, profit and cost, distances for rectangles and triangles, and people’s ages.

Linear Word Problems (How-To) – Video

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Mathematics LibreTexts

1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

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

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

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

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by \(x .8\) less than \(x\) mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

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

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

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

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

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

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

Solution: Each dime is worth 10 cents, so that this gives a total of \(10 x\) cents. Each quarter is worth 25 cents, so that this gives a total of \(25 y\) cents. Adding the two amounts gives a total of

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

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

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

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

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

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

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

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

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

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

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

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

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

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

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

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

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

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as \(x+9,\) while subtracting 7 from three times the number is written as \(3 x-7\). We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

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

\[x + 16 = 3x.\nonumber\]

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

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

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by \(5 \%\). How much does the loaf of bread cost now, when its price was \(\$ 2.40\) last year?

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

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

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

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

b) To complete a job, three workers get paid at a rate of \(\$ 12\) per hour. If the total pay for the job was \(\$ 180,\) then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by \(x\). Then the total price is calculated as the price per hour \((\$ 12)\) times the number of workers times the number of hours \((3) .\) We obtain the equation

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

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

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

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

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

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

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is \(4 \times 60\) feet \(=240\) feet.

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

Solution: We denote the weight of a block by \(x .\) If 4 blocks weigh \(28,\) then a block weighs \(x=\frac{28}{4}=7\)

How many blocks weigh \(70 ?\) Well, we only need to find \(\frac{70}{7}=10 .\) So, the answer is \(10 .\)

Note You can solve this problem by setting up and solving the fractional equation \(\frac{28}{4}=\frac{70}{x}\). Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by \(x\). Then the length is \(2 x+3\). The perimeter is 20 in on one hand and \(2(\)length\()+2(\)width\()\) on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

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

Solution: We know that \(C=2 \pi r\) where \(C\) is the circumference and \(r\) is the radius. So in this case

\[4=2 \pi r\nonumber\]

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

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

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

Solution: Let \(x\) equal the side of the triangle. Then the perimeter is, on the one hand, \(60,\) and on other hand \(3 x .\) So \(3 x=60\) and dividing both sides of the equation by 3 gives \(x=20\) meters.

h) If a gardener has \(\$ 600\) to spend on a fence which costs \(\$ 10\) per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is \(P=2 L+2 W\). Let \(x\) be the width of the rectangle. Then the length is \(2 x .\) The perimeter is \(P=2(2 x)+2 x=6 x\). The largest perimeter is \(\$ 600 /(\$ 10 / f t)=60\) ft. So \(60=6 x\) and dividing both sides by 6 gives \(x=60 / 6=10\). So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let \(b\) be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is \(\frac{1}{2}(3.2+b) \cdot 4=\) \(6.4+2 b .\) So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

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

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

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how to solve linear models word problems

Linear Programming: Word Problems and Applications

Several word problems and applications related to linear programming are presented along with their solutions and detailed explanations. Methods of solving inequalities with two variables , system of linear inequalities with two variables along with linear programming and optimization are used to solve word and application problems where functions such as return, profit, costs, etc., are to be optimized.

A store sells two types of toys, A and B. The store owner pays $8 and $14 for each one unit of toy A and B respectively. One unit of toys A yields a profit of $2 while a unit of toys B yields a profit of $3. The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than $20,000 in inventory of these toys. How many units of each type of toys should be stocked in order to maximize his monthly total profit profit?

Solution to Example 1 Let x be the total number of toys A and y the number of toys B; x and y cannot be negative, hence x ≥ 0 and y ≥ 0 The store owner estimates that no more than 2000 toys will be sold every month x + y ≤ 2000 One unit of toys A yields a profit of $2 while a unit of toys B yields a profit of $3, hence the total profit P is given by P = 2 x + 3 y The store owner pays $8 and $14 for each one unit of toy A and B respectively and he does not plan to invest more than $20,000 in inventory of these toys 8 x + 14 y ≤ 20,000 What do we have to solve? Find x and y so that P = 2 x + 3 y is maximum under the conditions \[ \begin{cases} \ x \ge 0 \\ \ x \ge 0 \\ \ x + y \le 2000 \\ \ 8 x + 14 y \le 20,000 \\ \end{cases} \]

Pin it!

Vertices of the solution set: A at (0 , 0) B at (0 , 1429) C at (1333 , 667) D at (2000 , 0) Calculate the total profit P at each vertex P(A) = 2 (0) + 3 ()) = 0 P(B) = 2 (0) + 3 (1429) = 4287 P(C) = 2 (1333) + 3 (667) = 4667 P(D) = 2(2000) + 3(0) = 4000 The maximum profit is at vertex C with x = 1333 and y = 667. Hence the store owner has to have 1333 toys of type A and 667 toys of type B in order to maximize his profit.

A company produces two types of tables, T1 and T2. It takes 2 hours to produce the parts of one unit of T1, 1 hour to assemble and 2 hours to polish.It takes 4 hours to produce the parts of one unit of T2, 2.5 hour to assemble and 1.5 hours to polish. Per month, 7000 hours are available for producing the parts, 4000 hours for assembling the parts and 5500 hours for polishing the tables. The profit per unit of T1 is $90 and per unit of T2 is $110. How many of each type of tables should be produced in order to maximize the total monthly profit?

Solution to Example 2 Let x be the number of tables of type T1 and y the number of tables of type T2. Profit P(x , y) = 90 x + 110 y \[ \begin{cases} \ x \ge 0 \\ \ x \ge 0 \\ \ 2x + 4y \le 7000 \\ \ x + 2.5y \le 4000 \\ \ 2x + 1.5y \le 5500 \\ \end{cases} \]

A at (0,0) B at (0,1600) C at (1500,1000) D at (2300,600) E at (2750,0)

Evaluate profit P(x,y) at each vertex A at (0,0) : P(0 , 0) = 0 B at (0,1600) : P(0 , 1600) = 90 (0) + 110 (1600) = 176000 C at (1500,1000) : P(1500,1000) = 90 (1500) + 110 (1000) = 245000 D at (2300,600): P(2300,600) = 90 (2300) + 110 (600) = 273000 E at (2750,0) : P(2750,0) = 90 (2750) + 110 (0) = 247500

The maximum profit of $273000 is at vertex D. Hence the company needs to produce 2300 tables of type T1 and 600 tables of type T2 in order to maximize its profit.

A farmer plans to mix two types of food to make a mix of low cost feed for the animals in his farm. A bag of food A costs $10 and contains 40 units of proteins, 20 units of minerals and 10 units of vitamins. A bag of food B costs $12 and contains 30 units of proteins, 20 units of minerals and 30 units of vitamins. How many bags of food A and B should the consumed by the animals each day in order to meet the minimum daily requirements of 150 units of proteins, 90 units of minerals and 60 units of vitamins at a minimum cost?

Solution to Example 3 Let x be the number of bags of food A and y the number of bags of food B. Cost C(x,y) = 10 x + 12 y \[ \begin{cases} \ x \ge 0 \\ \ y \ge 0 \\ \ 40x + 30y \ge 150 \\ \ 20x + 20y \ge 90 \\ \ 10x + 30y \ge 60 \\ \end{cases} \]

Vertices: A at intersection of \( 10x + 30y = 60 \) and \( y = 0 \) (x-axis) coordinates of A: (6 , 0) B at intersection of \( 20x + 20y = 90 \) and \( 10x + 30y = 60 \) coordinates of B: (15/4 , 3/4) C at intersection of \( 40x + 30y = 150 \) and \( 20x + 20y = 90 \) coordinates of C : (3/2 , 3) D at at intersection of \( 40x + 30y = 150 \) and \( x = 0 \) (y-axis) coordinates of D: (0 , 5)

Evaluate the cost c(x,y) = 10 x + 12 y at each one of the vertices A(x,y), B(x,y), C(x,y) and D(x,y). At A(6 , 0) : c(6 , 0) = 10 (6) + 12 (0) = 60 At B(15/4 , 3/4) : c(15/4 , 3/4) = 10 (15/4) + 12 (3/4) = 46.5 At C(3/2 , 3) : c(3/2 , 3) = 10 (3/2) + 12 (3) = 51 At D(0 , 5) : c(0 , 5) = 10 (0) + 12 (5) = 60 The cost c(x , y) is minimum at the vertex B(15/4 , 3/4) where x = 15/4 = 3.75 and y = 3/4 = 0.75. Hence 3.75 bags of food A and 0.75 bags of food B are needed to satisfy the minimum daily requirements in terms of proteins, minerals and vitamins at the lowest possible cost.

John has $20,000 to invest in three funds F1, F2 and F3. Fund F1 is offers a return of 2% and has a low risk. Fund F2 offers a return of 4% and has a medium risk. Fund F3 offers a return of 5% but has a high risk. To be on the safe side, John invests no more than $3000 in F3 and at least twice as much as in F1 than in F2. Assuming that the rates hold till the end of the year, what amounts should he invest in each fund in order to maximize the year end return?

Solution to Example 4 Let x be the amount invested in F1, y the amount invested in F2 and z the amount invested in F1. x + y + z = 20,000 z = 20,000 - (x + y) Total return R of all three funds is given by R = 2% x + 4% y + 5% z = 0.02 x + 0.04 y + 0.05 (20,000 - (x + y)) Simplifies to R(x ,y) = 1000 - 0.03 x - 0.01 y : This is the return to maximize Constraints: x, y and z are amounts of money and they must satisfy x ≥ 0 y ≥ 0 z ≥ 0 Substitute z by 20,000 - (x + y) in the above inequality to obtain 20,000 - (x + y) ≥ 0 which may be written as x + y ≤ 20,000 John invests no more than $3000 in F3, hence z ≤ 3000 Substitute z by 20,000 - (x + y) in the above inequality to obtain 20,000 - (x + y) ≤ 3000 which may be written as x + y ≥ 17,000 Let us put all the inequalities together to obtain the following system \[ \begin{cases} \ x \ge 0 \\ \ y \ge 0 \\ \ (x + y) \ge 17,000 \\ \ (x + y) \le 20,000 \\ \ x \ge 2 y \\ \end{cases} \]

Vertices: A at intersection of \( x + y = 20000 \) and \( y = 0 \) , coordinates of A: (20000 , 0) B at intersection of \( x+y = 17000 \) and \( y=0 \) , coordinates of B: (17000 , 0) C at intersection of \( x+y = 17000 \) and \( x = 2y \) , coordinates of C : (11333 , 5667) D at at intersection of \( x = 2y \) and \( x + y = 20000 \) , coordinates of D: (13333 , 6667)

Evaluate the return R(x,y) = 1000 - 0.03 x - 0.01 y at each one of the vertices A(x,y), B(x,y), C(x,y) and D(x,y). At A(20000 , 0) : R(20000 , 0) = 1000 - 0.03 (20000) - 0.01 (0) = 400 At B(17000 , 0) : R(17000 , 0) = 1000 - 0.03 (17000) - 0.01 (0) = 490 At C(11333 , 5667) : R(11333 , 5667) = 1000 - 0.03 (11333) - 0.01 (5667) = 603 At D(13333 , 6667) : R(13333 , 6667) = 1000 - 0.03 (13333) - 0.01 (6667) = 533 The return R is maximum at the vertex At C(11333 , 5667) where x = 11333 and y = 5667 and z = 20,000 - (x+y) = 3000 For maximum return, John has to invest $11333 in fund F1, $5667 in fund F2 and $3000 in fund F3.

Each month a store owner can spend at most $100,000 on PC's and laptops. A PC costs the store owner $1000 and a laptop costs him $1500. Each PC is sold for a profit of $400 while laptop is sold for a profit of $700. The sore owner estimates that at least 15 PC's but no more than 80 are sold each month. He also estimates that the number of laptops sold is at most half the PC's. How many PC's and how many laptops should be sold in order to maximize the profit?

Solution to Example 5 Let x and y be the numbers of PC's and laptops respectively that should be sold. Profit = 400 x + 700 y to maximize Constraints 15 ≤ x ≤ 80 "least 15 PC's but no more than 80 are sold each month" y ≤ (1/2) x 1000 x + 1500 y ≤ 100,000 "store owner can spend at most $100,000 on PC's and laptops" \[ \begin{cases} \ 15 \le x \le 80 \\ \ y \ge 0 \\ \ y \le (1/2) x \\ \ 1000 x + 1500 y \le 100,000 \\ \end{cases} \]

Vertices: A at intersection of \( x = 15 \) and \( y = 0 \) (x-axis) coordinates of A: (15 , 0) B at intersection of \( x = 15 \) and \( y = (1/2) x \) coordinates of B: (15 , 7.5) C at intersection of \( y = (1/2) x \) and \( 1000 x + 1500 y = 100000 \) coordinates of C : (57.14 , 28.57) D at at intersection of \( 1000 x + 1500 y = 100000 \) and \( x = 80 \) (y-axis) coordinates of D: (80 , 13.3)

Evaluate the profit at each vertex A(15 , 0), P = 400 × 15 + 700 × 0 = 6000 B(15 , 7.5) , P = 400 × 15 + 700 × 7.5 = 11250 C(57.14 , 28.57) , P = 400 × 57.14 + 700 × 28.57 = 42855 D (80 , 13.3) , P = = 400 × 80 + 700 × 13.3 = 41310

The profit is maximum for x = 57.14 and y = 28.57 but these cannot be accepted as solutions because x and y are numbers of PC's and laptops and must be integers. We need to select the nearest integers to x = 57.14 and y = 28.57 that are satisfy all constraints and give a maximum profit x = 57 and y = 29 do not satisfy all constraints x = 57 and y = 28 satisfy all constraints Profit = 400 × 57 + 700 × 28 = 42400 , which is maximum.

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Course: Algebra 1   >   Unit 5

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Linear equations word problems

  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
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Mathx

  • Linear systems – word problems

When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems . The same rules apply. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you accustomed to finding elements of linear systems inside of word problems. Example 1 : A farmhouse shelters 16 animals. Some of them are chickens and the others are cows. Altogether these animals have 60 legs. How many chickens and how many cows are in the farmhouse?

First, to make the calculations clearer, we will choose symbols to represent the number of cows and the number of chickens. Let us say that the chickens will be represented with x and the cows with y. Now, this task gave us enough information to make two equations. The first one is that the sum of the number of chickens (x) and the number of cows (y) is 16, since there are only 16 animals in the farmhouse. That equation should look like this: x + y = 16

The second piece of information we have is that the total number of legs in the farmhouse is 60. Since we know that cows have four legs each and chickens have two legs each, we have enough information to make another equation. This one will look like this: 2*x + 4*y = 60

Now we have a system of linear equations with two equations and two variables. The only thing left to do now is to solve the system. We will solve it here for you, but if you need to remind yourself how to do that step by step, read the article called Systems of linear equations . x = 16 – y 32 – 2y + 4y = 60 2y = 28 y = 14 x = 2

We can now see that there are two chickens and 14 cows in the farmhouse. The next example will be a bit harder,

Example 2: Rodney’s Kitchen Supplies makes and sells spoons and forks. It costs the store $2 to buy the supplies needed to make a fork, and $1 for the supplies needed to make a spoon. The store sells the forks for $4 and the spoons for 5$. Last month Rodney’s Kitchen Supplies spent $39 on supplies and sold the all of the forks and spoons that were made last month using those supplies for $93. How many forks and spoons did they make?

As we did in the first example, we will first designate symbols to available variables. So, the number of forks made will be represented with x and the number of spoons with y. Again, we have enough information to make two equations. The total cost of making a particular number of forks (x), which cost $2 to make each, and a particular number of spoons (y), which cost $1 to make each, is $39. So that will be our first equation and it will look like this: 2*x + y = 39 The other piece of information tells us that if we sell that number of forks (x) for $4 each and that number of spoons (y) for $5 each, we will make $93. And that will be our second equation: 4*x + 5*y = 93 This was the hard part. Now all we have to do is to solve this linear system to find how many spoons and how many forks did we make last month. y = 39 – 2x 4x + 5*(39 – 2x) = 93 4x + 195 – 10x = 93 -6x = 93 – 195 -6x = -102 |: (-6) x = 17 y = 5

We can see that last month the store made and sold 17 forks and five spoons.

linear systems problems

Although they can seem complicated, mastery and understanding of linear systems and associated word problems will come with a bit of practice. With experience you will be able to recognize their elements and solve even complicated systems with ease. Feel free to use the math worksheets below to practice solving this type of linear systems.

Linear systems – word problems exams for teachers

Linear systems – word problems worksheets for studets.

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A screenshot from an AI-generated video of woolly mammoths.

Sora: OpenAI launches tool that instantly creates video from text

Model from ChatGPT maker ‘simulates physical world in motion’ up to a minute long based on users’ subject and style instructions

OpenAI revealed a tool on Thursday that can generate videos from text prompts.

The new model, nicknamed Sora after the Japanese word for “sky”, can produce realistic footage up to a minute long that adheres to a user’s instructions on both subject matter and style. According to a company blogpost, the model is also able to create a video based on a still image or extend existing footage with new material.

“We’re teaching AI to understand and simulate the physical world in motion, with the goal of training models that help people solve problems that require real-world interaction,” the blogpost reads.

One video included among several initial examples from the company was based on the prompt: “A movie trailer featuring the adventures of the 30-year-old space man wearing a red wool knitted motorcycle helmet, blue sky, salt desert, cinematic style, shot on 35mm film, vivid colors.”

The company announced it had opened access to Sora to a few researchers and video creators. The experts would “red team” the product – test it for susceptibility to skirt OpenAI’s terms of service, which prohibit “extreme violence, sexual content, hateful imagery, celebrity likeness, or the IP of others”, per the company’s blogpost. The company is only allowing limited access to researchers, visual artists and film-makers, though CEO Sam Altman responded to users’ prompts on Twitter after the announcement with video clips he said were made by Sora. The videos bear a watermark to show they were made by AI.

Introducing Sora, our text-to-video model. Sora can create videos of up to 60 seconds featuring highly detailed scenes, complex camera motion, and multiple characters with vibrant emotions. https://t.co/7j2JN27M3W Prompt: “Beautiful, snowy… pic.twitter.com/ruTEWn87vf — OpenAI (@OpenAI) February 15, 2024

The company debuted the still image generator Dall-E in 2021 and generative AI chatbot ChatGPT in November 2022, which quickly accrued 100 million users. Other AI companies have debuted video generation tools, though those models have only been able to produce a few seconds of footage that often bears little relation to their prompts. Google and Meta have said they are in the process of developing generative video tools, though they have not released them to the public. On Wednesday, it announced an experiment with adding deeper memory to ChatGPT so that it could remember more of its users’ chats.

https://t.co/uCuhUPv51N pic.twitter.com/nej4TIwgaP — Sam Altman (@sama) February 15, 2024

OpenAI did not disclose how much footage was used to train Sora or where the training videos may have originated, other than telling the New York Times that the corpus contained videos that were both publicly available and licensed from copyright owners. The company has been sued multiple times for alleged copyright infringement in the training of its generative AI tools, which digest gargantuan amounts of material scraped from the internet and imitate the images or text contained in those datasets.

  • Artificial intelligence (AI)

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