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Learning Objectives
 Set up a linear equation to solve a realworld application.
 Use a formula to solve a realworld application.
Josh is hoping to get an \(A\) in his college algebra class. He has scores of \(75\), \(82\), \(95\), \(91\), and \(94\) on his first five tests. Only the final exam remains, and the maximum of points that can be earned is \(100\). Is it possible for Josh to end the course with an \(A\)? A simple linear equation will give Josh his answer.
Many realworld applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talktime; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.
Setting up a Linear Equation to Solve a RealWorld Application
To set up or model a linear equation to fit a realworld application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as \($0.10/mi\), is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write \(0.10x\). This expression represents a variable cost because it changes according to the number of miles driven.
If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges \($0.10/mi\) plus a daily fee of \($50\). We can use these quantities to model an equation that can be used to find the daily car rental cost \(C\).
\(C=0.10x+50 \tag{2.4.1}\)
When dealing with realworld applications, there are certain expressions that we can translate directly into math. Table \(\PageIndex{1}\) lists some common verbal expressions and their equivalent mathematical expressions.
Verbal  Translation to Math Operations 

One number exceeds another by a  \(x,x+a\) 
Twice a number  \(2x\) 
One number is \(a\) more than another number  \(x,x+a\) 
One number is a less than twice another number  \(x,2x−a\) 
The product of a number and \(a\), decreased by \(b\)  \(ax−b\) 
The quotient of a number and the number plus \(a\) is three times the number  \(\dfrac{x}{x+a}=3x\) 
The product of three times a number and the number decreased by \(b\) is \(c\)  \(3x(x−b)=c\) 
How to: Given a realworld problem, model a linear equation to fit it
 Identify known quantities.
 Assign a variable to represent the unknown quantity.
 If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
 Write an equation interpreting the words as mathematical operations.
 Solve the equation. Be sure the solution can be explained in words, including the units of measure.
Example \(\PageIndex{1}\)
Find a linear equation to solve for the following unknown quantities: One number exceeds another number by \( 17\) and their sum is \( 31\). Find the two numbers.
Solution
Let \( x\) equal the first number. Then, as the second number exceeds the first by \(17\), we can write the second number as \( x +17\). The sum of the two numbers is \(31\). We usually interpret the word is as an equal sign.
\[\begin{align*} x+(x+17)&= 31\\ 2x+17&= 31\\ 2x&= 14\\ x&= 7 \end{align*}\]
\[\begin{align*} x+17&= 7 + 17\\ &= 24\\ \end{align*}\]
The two numbers are \(7\) and \(24\).
Exercise \(\PageIndex{1}\)
Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is \(36\), find the numbers.
 Answer

\(11\) and \(25\)
Example \(\PageIndex{2}\): Setting Up a Equation to Solve a RealWorld Application
There are two cell phone companies that offer different packages. Company A charges a monthly service fee of \($34\) plus \($.05/min\) talktime. Company B charges a monthly service fee of \($40\) plus \($.04/min\) talktime.
 Write a linear equation that models the packages offered by both companies.
 If the average number of minutes used each month is \(1,160\), which company offers the better plan?
 If the average number of minutes used each month is \(420\), which company offers the better plan?
 How many minutes of talktime would yield equal monthly statements from both companies?
Solution
a.
The model for Company A can be written as \( A =0.05x+34\). This includes the variable cost of \( 0.05x\) plus the monthly service charge of \($34\). Company B’s package charges a higher monthly fee of \($40\), but a lower variable cost of \( 0.04x\). Company B’s model can be written as \( B =0.04x+$40\).
b.
If the average number of minutes used each month is \(1,160\), we have the following:
\[\begin{align*} \text{Company A}&= 0.05(1.160)+34\\ &= 58+34\\ &= 92 \end{align*}\]
\[\begin{align*} \text{Company B}&= 0.04(1,1600)+40\\ &= 46.4+40\\ &= 86.4 \end{align*}\]
So, Company B offers the lower monthly cost of \($86.40\) as compared with the \($92\) monthly cost offered by Company A when the average number of minutes used each month is \(1,160\).
c.
If the average number of minutes used each month is \(420\), we have the following:
\[\begin{align*} \text{Company A}&= 0.05(420)+34\\ &= 21+34\\ &= 55 \end{align*}\]
\[\begin{align*} \text{Company B}&= 0.04(420)+40\\ &= 16.8+40\\ &= 56.8 \end{align*}\]
If the average number of minutes used each month is \(420\), then Company A offers a lower monthly cost of \($55\) compared to Company B’s monthly cost of \($56.80\).
d.
To answer the question of how many talktime minutes would yield the same bill from both companies, we should think about the problem in terms of \((x,y)\) coordinates: At what point are both the \(x\)value and the \(y\)value equal? We can find this point by setting the equations equal to each other and solving for \(x\).
\[\begin{align*} 0.05x+34&= 0.04x+40\\ 0.01x&= 6\\ x&= 600 \end{align*}\]
Check the \(x\)value in each equation.
\(0.05(600)+34=64\)
\(0.04(600)+40=64\)
Therefore, a monthly average of \(600\) talktime minutes renders the plans equal. See Figure \(\PageIndex{2}\).
Exercise \(\PageIndex{2}\)
Find a linear equation to model this realworld application: It costs ABC electronics company \($2.50\) per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of \($350\) for utilities and \($3,300\) for salaries. What are the company’s monthly expenses?
 Answer

\(C=2.5x+3,650\)
Using a Formula to Solve a RealWorld Application
Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region,
\[A=LW \tag{2.4.2}\]
the perimeter of a rectangle,
\[P=2L+2W \tag{2.4.3}\]
and the volume of a rectangular solid,
\[V=LWH. \tag{2.4.4}\]
When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.
Example \(\PageIndex{3}\): Solving an Application Using a Formula
It takes Andrew \(30\; min\) to drive to work in the morning. He drives home using the same route, but it takes \(10\; min\) longer, and he averages \(10\; mi/h\) less than in the morning. How far does Andrew drive to work?
Solution
This is a distance problem, so we can use the formula \(d =rt\), where distance equals rate multiplied by time. Note that when rate is given in \(mi/h\), time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.
First, we identify the known and unknown quantities. Andrew’s morning drive to work takes \(30\; min\), or \(12\; h\) at rate \(r\). His drive home takes \(40\; min\), or \(23\; h\), and his speed averages \(10\; mi/h\) less than the morning drive. Both trips cover distance \(d\). A table, such as Table \(\PageIndex{2}\), is often helpful for keeping track of information in these types of problems.
\(d\)  \(r\)  \(t\)  

To Work  \(d\)  \(r\)  \(12\) 
To Home  \(d\)  \(r−10\)  \(23\) 
Write two equations, one for each trip.
\[d=r\left(\dfrac{1}{2}\right) \qquad \text{To work} \nonumber\]
\[d=(r10)\left(\dfrac{2}{3}\right) \qquad \text{To home} \nonumber\]
As both equations equal the same distance, we set them equal to each other and solve for \(r\).
\[\begin{align*} r\left (\dfrac{1}{2} \right )&= (r10)\left (\dfrac{2}{3} \right )\\ \dfrac{1}{2r}&= \dfrac{2}{3}r\dfrac{20}{3}\\ \dfrac{1}{2}r\dfrac{2}{3}r&= \dfrac{20}{3}\\ \dfrac{1}{6}r&= \dfrac{20}{3}\\ r&= \dfrac{20}{3}(6)\\ r&= 40 \end{align*}\]
We have solved for the rate of speed to work, \(40\; mph\). Substituting \(40\) into the rate on the return trip yields \(30 mi/h\). Now we can answer the question. Substitute the rate back into either equation and solve for \(d\).
\[\begin{align*}d&= 40\left (\dfrac{1}{2} \right )\\ &= 20 \end{align*}\]
The distance between home and work is \(20\; mi\).
Analysis
Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for \(r\).
\[\begin{align*} r\left (\dfrac{1}{2} \right)&= (r10)\left (\dfrac{2}{3} \right )\\ 6\times r\left (\dfrac{1}{2} \right)&= 6\times (r10)\left (\dfrac{2}{3} \right )\\ 3r&= 4(r10)\\ 3r&= 4r40\\ r&= 40 \end{align*}\]
Exercise \(\PageIndex{3}\)
On Saturday morning, it took Jennifer \(3.6\; h\) to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer \(4\; h\) to return home. Her speed was \(5\; mi/h\) slower on Sunday than on Saturday. What was her speed on Sunday?
 Answer

\(45\; mi/h\)
Example \(\PageIndex{4}\): Solving a Perimeter Problem
The perimeter of a rectangular outdoor patio is \(54\; ft\). The length is \(3\; ft\) greater than the width. What are the dimensions of the patio?
Solution
The perimeter formula is standard: \(P=2L+2W\). We have two unknown quantities, length and width. However, we can write the length in terms of the width as \(L =W+3\). Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure \(\PageIndex{3}\).
Now we can solve for the width and then calculate the length.
\[\begin{align*} P&= 2L + 2W\\ 54&= 2(W+3)+2W\\ 54&= 2W+6+2W\\ 54&= 4W+6\\ 48&= 4W\\ W&= 12 \end{align*}\]
\[\begin{align*} L&= 12+3\\ L&= 15 \end{align*}\]
The dimensions are \(L = 15\; ft\) and \(W = 12\; ft\).
Exercise \(\PageIndex{4}\)
Find the dimensions of a rectangle given that the perimeter is \(110\; cm\) and the length is \(1\; cm\) more than twice the width.
 Answer

\(L=37\; cm\), \(W=18\; cm\)
Example \(\PageIndex{5}\): Solving an Area Problem
The perimeter of a tablet of graph paper is \(48\space{in.}^2\). The length is \(6\; in\). more than the width. Find the area of the graph paper.
Solution
The standard formula for area is \(A =LW\); however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.
We know that the length is \(6\; in\). more than the width, so we can write length as \(L =W+6\). Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.
\[\begin{align*} P&= 2L + 2W\\ 48&= 2(W+6)+2W\\ 48&= 2W+12+2W\\ 48&= 4W+12\\ 36&= 4W\\ W&= 9 \end{align*}\]
\[\begin{align*}L&= 9+6\\ L&= 15 \end{align*}\]
Now, we find the area given the dimensions of \(L = 15\; in\). and \(W = 9\; in\).
\[\begin{align*} A&= LW\\ A&=15(9)\\ A&= 135\space{in.}^2 \end{align*}\]
The area is \(135\space{in.}^2\).
Exercise \(\PageIndex{5}\)
A game room has a perimeter of \(70\; ft\). The length is five more than twice the width. How many \(ft^2\) of new carpeting should be ordered?
 Answer

\(250\space{ft}^2\)
Example \(\PageIndex{6}\): Solving a Volume Problem
Find the dimensions of a shipping box given that the length is twice the width, the height is \(8\; \) in, and the volume is \(1,600\space{in.}^3\).
Solution
The formula for the volume of a box is given as \(V =LWH\), the product of length, width, and height. We are given that \(L =2W\), and \(H =8\). The volume is \(1,600\; \text{cubic inches}\).
\[\begin{align*} V&= LWH\\ 1600&= (2W)W(8)\\ 1600&= 16W^2\\ 100&= W^2\\ 10&= W \end{align*}\]
The dimensions are \(L = 20\; in\), \(W= 10\; in\), and \(H = 8\; in\).
Analysis
Note that the square root of \(W^2\) would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.
Media
Access these online resources for additional instruction and practice with models and applications of linear equations.
 Problem solving using linear equations
 Problem solving using equations
 Finding the dimensions of area given the perimeter
 Find the distance between the cities using the distance = rate * time formula
 Linear equation application (Write a cost equation)
Key Concepts
 A linear equation can be used to solve for an unknown in a number problem. See Example.
 Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See Example.
 There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the \(d = rt\) formula. See Example.
 Many geometry problems are solved using the perimeter formula \(P =2L+2W\), the area formula \(A =LW\), or the volume formula \(V =LWH\). See Example, Example, and Example.
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Linear models Linear mixed models Variable selection in linear models Generalized linear models Generalized linear mixed models Semiparametric regression for continuous responses (excluding mixed models) Semiparametric regression for continuous responses (including mixed models). 3, 4, and 10 • A coverage of the entire range of regression models starting with linear models, covering generalized linear and mixed models and also including (generalized) additive models and quantile regression. Structured additive regression models include a variety of special cases, for example, nonparametric and semiparametric regression models, additive models, geoadditive models, and varyingcoefficient models.. 2.26 Munich rent index: scatter plots of the rents in Euro versus living area together with linear quantile regression fits for 11 quantiles (top left panel), quantiles determined from a classical linear model (top right panel), and quantiles determined from a linear model for location and scale (bottom panel). • Predictor: i D ˇ0 C ˇ1 xi1 C : : : C ˇk xi k D lin i : • Remark: Generalized linear models are a broad class of models, with linear models, logit models, and Poisson models as special cases.. 3.2 Modeling Nonlinear Covariate Effects Through Variable Transformation If the continuous covariate z has an approximately nonlinear effect ˇ1 f .z/ with known transformation f , then the model yi D ˇ0 C ˇ1 f .zi / C : : : C "i can be transformed into the linear regression model yi D ˇ0 C ˇ1 xi C : : : C "i ; where xi D f .zi / fN: By subtracting n
CoClustering: Models, Algorithms and Applications ›
Cluster or cocluster analyses are important tools in a variety of scientific areas. The introduction of this book presents a state of the art of already wellestablished, as well as more recent methods of coclustering. The authors mainly deal with the twomode partitioning under different approaches, but pay particular attention to a probabilistic approach. Chapter 1 concerns clustering in general and the modelbased clustering in particular. The authors briefly review the classical clustering methods and focus on the mixture model. They present and discuss the use of different mixtures adapted to different types of data. The algorithms used are described and related works with different classical methods are presented and commented upon. This chapter is useful in tackling the problem of coclustering under the mixture approach. Chapter 2 is devoted to the latent block model proposed in the mixture approach context. The authors discuss this model in detail and present its interest regarding coclustering. Various algorithms are presented in a general context. Chapter 3 focuses on binary and categorical data. It presents, in detail, the appropriated latent block mixture models. Variants of these models and algorithms are presented and illustrated using examples. Chapter 4 focuses on contingency data. Mutual information, phisquared and modelbased coclustering are studied. Models, algorithms and connections among different approaches are described and illustrated. Chapter 5 presents the case of continuous data. In the same way, the different approaches used in the previous chapters are extended to this situation. Contents 1. Cluster Analysis. 2. ModelBased CoClustering. 3. CoClustering of Binary and Categorical Data. 4. CoClustering of Contingency Tables. 5. CoClustering of Continuous Data. About the Authors Gérard Govaert is Professor at the University of Technology of Compiègne, France. He is also a member of the CNRS Laboratory Heudiasyc (Heuristic and diagnostic of complex systems). His research interests include latent structure modeling, model selection, modelbased cluster analysis, block clustering and statistical pattern recognition. He is one of the authors of the MIXMOD (MIXtureMODelling) software. Mohamed Nadif is Professor at the University of ParisDescartes, France, where he is a member of LIPADE (Paris Descartes computer science laboratory) in the Mathematics and Computer Science department. His research interests include machine learning, data mining, modelbased cluster analysis, coclustering, factorization and data analysis. Cluster Analysis is an important tool in a variety of scientific areas. Chapter 1 briefly presents a state of the art of already wellestablished as well more recent methods. The hierarchical, partitioning and fuzzy approaches will be discussed amongst others. The authors review the difficulty of these classical methods in tackling the high dimensionality, sparsity and scalability. Chapter 2 discusses the interests of coclustering, presenting different approaches and defining a cocluster. The authors focus on coclustering as a simultaneous clustering and discuss the cases of binary, continuous and cooccurrence data. The criteria and algorithms are described and illustrated on simulated and real data. Chapter 3 considers coclustering as a modelbased coclustering. A latent block model is defined for different kinds of data. The estimation of parameters and coclustering is tackled under two approaches: maximum likelihood and classification maximum likelihood. Hard and soft algorithms are described and applied on simulated and real data. Chapter 4 considers coclustering as a matrix approximation. The trifactorization approach is considered and algorithms based on update rules are described. Links with numerical and probabilistic approaches are established. A combination of algorithms are proposed and evaluated on simulated and real data. Chapter 5 considers a coclustering or biclustering as the search for coherent coclusters in biological terms or the extraction of coclusters under conditions. Classical algorithms will be described and evaluated on simulated and real data. Different indices to evaluate the quality of coclusters are noted and used in numerical experiments.
Other approaches 7. Modelbased clustering and the mixture model 11. Application to mixture models 18. Clustering and the mixture model 20. The two approaches 20. CoClustering of Binary and Categorical Data 79. Bernoulli latent block model and algorithms 84. Block model for contingency tables 133. Poisson latent block model 137. Parsimonious Gaussian latent block models 161. Gaussian block mixture model 168
2.3 Modeling with Linear Functions  Precalculus  OpenStax ›
When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we typically follow the same pro...
When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function.. Using a Linear Model to Investigate a Town’s Population A town’s population has been growing linearly.. Find the linear function that models the town’s population PP as a function of the year, t,t, where tt is the number of years since the model began.. Find the linear function that models the baby’s weight WW as a function of the age of the baby, in months, t.t.. Find the linear function that models the number of people inflicted with the common cold CC as a function of the year, t.t.. ⓔ Find an equation for the population, P , P , of the school t years after 2000. ⓕ Using your equation, predict the population of the school in 2011.. ⓔ Find an equation for the population, P P of the town t t years after 2000. ⓕ Using your equation, predict the population of the town in 2014.. ⓐ Find a linear equation for the monthly cost of the cell plan as a function of x , the number of monthly minutes used.. ⓐ Find a formula for the owl population, P. P. Let the input be years since 2003. ⓑ What does your model predict the owl population to be in 2012?