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How to Solve a Linear Function – A Step-by-Step Guide

To solve a linear function, I always begin by identifying its standard form, which is typically expressed as $y = mx + b$ . In this equation, (m) represents the slope of the line , and (b) denotes the y-intercept , where the line crosses the y-axis.

By knowing these components, I can graph the function or work with it algebraically. It’s crucial to remember that in a linear equation , the variables (x) and (y) are to the first power, indicating a constant rate of change and resulting in a straight line when graphed.

Understanding how to manipulate this equation to isolate the variable of interest is the key to finding solutions.

Whether I’m substituting values to find points on the line or reformatting the equation to point-slope or standard form depends on the context of the problem.

Encountering various linear equation word problems sharpens my skills, as they often require a tailored approach for each unique scenario.

Stay tuned, as I’m about to unveil the step-by-step process to confidently tackle any linear function that comes my way.

Steps for Solving Linear Functions

When I approach linear functions , I often think of them as a puzzle where my goal is to find the value of the variable that makes the function true. A linear function takes the form of $f(x) = mx + b$, where $m$ is the slope or rate of change , and $b$ is the y-intercept . To make things easier, I’ve broken down the process into clear steps:

Understanding Function Notation : A linear function is typically written as $f(x)$, which is equivalent to $y$. It represents the output for a given input $x$.

Identifying Key Components :

• Slope ($m$) : Determines how steep the line is on a graph.
• Y-intercept ($b$) : Where the line crosses the y-axis.

Writing the Equation : Start by placing the known values of the slope and y-intercept into their proper places in the equation $y = mx + b$.

Plotting the Graph : On a coordinate plane , plot the y-intercept and use the slope to find a second point. Connect these points to visualize the function as a straight line .

Solving for a Specific Value :

• If you need $y$ for a particular $x$, substitute the $x$ value into the equation and solve for $y$.
• To find an $x$ value for a given $y$, you can rearrange the equation to isolate $x$ and then substitute the $y$ value.

Remember, linear equations involving two lines can be parallel or perpendicular based on their slopes:

• Parallel lines share the same slope ($m$).
• Perpendicular lines have slopes that are negative reciprocals of each other.

The process of solving for a linear function is straightforward if you take it step by step. Just remember that you’re working with constants and a variable to find points that lie on a line, and that’s the heart of linear equations .

Applications in Real-World Contexts

In my day-to-day life, I often encounter situations where linear functions are incredibly useful.

 These functions, which can be written in the form $f(x) = mx + b$, where $m$ is the slope and $b$ is the y-intercept, model relationships with a constant rate of change. Here are a few examples of how I’ve seen linear functions shine in real-world contexts:

Budgeting and Finance: For instance, if I wanted to understand how my savings account grows over time, I could use a linear function . If I start with $200 and save $50 each week, the amount of money, $M$, after $t$ weeks is represented by $M(t) = 50t + 200$.

Cooking and Recipes: When I’m cooking and need to adjust the recipe according to the number of guests, I use linear functions to scale the ingredients. If a recipe calls for 2 cups of flour for 4 cookies, the equation would look like $c = \frac{1}{2}n$, where $c$ is cups of flour and $n$ is the number of cookies.

Travel and Distance: If I go for a jog and keep a steady pace, the distance I cover can be predicted with a linear function . The distance $d$ in miles, at a constant speed $s$, after jogging for $t$ hours, would be $d(t) = st$.

These are simple applications, but they show how understanding linear functions can apply to anything from managing finances to making dinner or exercising, making this concept a valuable tool in my everyday life.

Solving linear functions can be a satisfying experience, as it sharpens my problem-solving skills and enhances my understanding of algebra.

I’ve learned that by following a methodical approach, such as using the slope-intercept form, which is denoted as $f(x) = mx + b$ , handling linear functions becomes much more straightforward.

It’s important to remember the role of $m$ , which represents the slope, and $b$, the y-intercept, in graphing these linear functions .

When I embark on solving linear equations , it’s crucial to apply systematic methods—graphing, substitution, elimination, or matrices—based on the context of the problem. Each method has its own merits, and knowing when to use each can save me time and effort.

Moreover, understanding how to interpret the solutions in real-life scenarios is an integral part of the learning process.

Whether I am predicting outcomes or determining the relationship between variables, linear functions serve as a fundamental tool in various fields, from economics to engineering.

I encourage fellow learners to practice regularly, as this will certainly fortify their abilities to solve and apply linear functions effectively. The more I engage with these concepts, the more intuitive and rewarding they become.

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• What is a linear equation?
• A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
• How do you find the linear equation?
• To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
• What are the 4 methods of solving linear equations?
• There are four common methods to solve a system of linear equations: Graphing, Substitution, Elimination and Matrix.
• How do you identify a linear equation?
• Here are a few ways to identify a linear equation: Look at the degree of the equation, a linear equation is a first-degree equation. Check if the equation has two variables. Graph the equation.
• What is the most basic linear equation?
• The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-Intercept Form

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

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Linear(Simple) Equations: Very Difficult Problems with Solutions

5.11 Linear Programming

Learning objectives.

After completing this section, you should be able to:

• Compose an objective function to be minimized or maximized.
• Compose inequalities representing a system application.
• Apply linear programming to solve application problems.

Imagine you hear about some natural disaster striking a far-away country; it could be an earthquake, a fire, a tsunami, a tornado, a hurricane, or any other type of natural disaster. The survivors of this disaster need help—they especially need food, water, and medical supplies. You work for a company that has these supplies, and your company has decided to help by flying the needed supplies into the disaster area. They want to maximize the number of people they can help. However, there are practical constraints that need to be taken into consideration; the size of the airplanes, how much weight each airplane can carry, and so on. How do you solve this dilemma? This is where linear programming comes into play. Linear programming is a mathematical technique to solve problems involving finding maximums or minimums where a linear function is limited by various constraints.

As a field, linear programming began in the late 1930s and early 1940s. It was used by many countries during World War II; countries used linear programming to solve problems such as maximizing troop effectiveness, minimizing their own casualties, and maximizing the damage they could inflict upon the enemy. Later, businesses began to realize they could use the concept of linear programming to maximize output, minimize expenses, and so on. In short, linear programming is a method to solve problems that involve finding a maximum or minimum where a linear function is constrained by various factors.

A Mathematician Invents a “Tsunami Cannon”

On December 26, 2004, a massive earthquake occurred in the Indian Ocean. This earthquake, which scientists estimate had a magnitude of 9.0 or 9.1 on the Richter Scale, set off a wave of tsunamis across the Indian Ocean. The waves of the tsunami averaged over 30 feet (10 meters) high, and caused massive damage and loss of life across the coastal regions bordering the Indian Ocean.

Usama Kadri works as an applied mathematician at Cardiff University in Wales. His areas of research include fluid dynamics and non-linear phenomena. Lately, he has been focusing his research on the early detection and easing of the effects of tsunamis. One of his theories involves deploying a series of devices along coastlines which would fire acoustic-gravity waves (AGWs) into an oncoming tsunami, which in theory would lessen the force of the tsunami. Of course, this is all in theory, but Kadri believes it will work. There are issues with creating such a device: they would take a tremendous amount of electricity to generate an AGW, for instance, but if it would save lives, it may well be worth it.

Compose an Objective Function to Be Minimized or Maximized

An objective function is a linear function in two or more variables that describes the quantity that needs to be maximized or minimized.

Example 5.96

Composing an objective function for selling two products.

Miriam starts her own business, where she knits and sells scarves and sweaters out of high-quality wool. She can make a profit of $8 per scarf and $10 per sweater. Write an objective function that describes her profit.

Let x x represent the number of scarves sold, and let y y represent the number of sweaters sold. Let P P represent profit. Since each scarf has a profit of $8 and each sweater has a profit of $10, the objective function is P = 8 x + 10 y P = 8 x + 10 y .

Your Turn 5.96

Example 5.97, composing an objective function for production.

William’s factory produces two products, widgets and wadgets. It takes 24 minutes for his factory to make 1 widget, and 32 minutes for his factory to make 1 wadget. Write an objective function that describes the time it takes to make the products.

Let x x equal the number of widgets made; let y y equal the number of wadgets made; let T T represent total time. The objective function is T = 24 x + 32 y T = 24 x + 32 y .

Your Turn 5.97

Composing inequalities representing a system application.

For our two examples of profit and production, in an ideal world the profit a person makes and/or the number of products a company produces would have no restrictions. After all, who wouldn’t want to have an unrestricted profit? However in reality this is not the case; there are usually several variables that can restrict how much profit a person can make or how many products a company can produce. These restrictions are called constraints .

Many different variables can be constraints. When making or selling a product, the time available, the cost of manufacturing and the amount of raw materials are all constraints. In the opening scenario with the tsunami, the maximum weight on an airplane and the volume of cargo it can carry would be constraints. Constraints are expressed as linear inequalities; the list of constraints defined by the problem forms a system of linear inequalities that, along with the objective function, represent a system application.

Example 5.98

Representing the constraints for selling two products.

Two friends start their own business, where they knit and sell scarves and sweaters out of high-quality wool. They can make a profit of $8 per scarf and $10 per sweater. To make a scarf, 3 bags of knitting wool are needed; to make a sweater, 4 bags of knitting wool are needed. The friends can only make 8 items per day, and can use not more than 27 bags of knitting wool per day. Write the inequalities that represent the constraints. Then summarize what has been described thus far by writing the objective function for profit and the two constraints.

Let x x represent the number of scarves sold, and let y y represent the number of sweaters sold. There are two constraints: the number of items the business can make in a day (a maximum of 8) and the number of bags of knitting wool they can use per day (a maximum of 27). The first constraint (total number of items in a day) is written as:

Since each scarf takes 3 bags of knitting wool and each sweater takes 4 bags of knitting wool, the second constraint, total bags of knitting wool per day, is written as:

In summary, here are the equations that represent the new business:

P = 8 x + 10 y P = 8 x + 10 y ; This is the profit equation: The business makes $8 per scarf and $10 per sweater.

Your Turn 5.98

Example 5.99, representing constraints for production.

A factory produces two products, widgets and wadgets. It takes 24 minutes for the factory to make 1 widget, and 32 minutes for the factory to make 1 wadget. Research indicates that long-term demand for products from the factory will result in average sales of 12 widgets per day and 10 wadgets per day. Because of limitations on storage at the factory, no more than 20 widgets or 17 wadgets can be made each day. Write the inequalities that represent the constraints. Then summarize what has been described thus far by writing the objective function for time and the two constraints.

Let x x equal the number of widgets made; let y y equal the number of wadgets made. Based on the long-term demand, we know the factory must produce a minimum of 12 widgets and 10 wadgets per day. We also know because of storage limitations, the factory cannot produce more than 20 widgets per day or 17 wadgets per day. Writing those as inequalities, we have:

x ≥ 12 x ≥ 12

y ≥ 10 y ≥ 10

x ≤ 20 x ≤ 20

y ≤ 17 y ≤ 17

The number of widgets made per day must be between 12 and 20, and the number of wadgets made per day must be between 10 and 17. Therefore, we have:

12 ≤ x ≤ 20 12 ≤ x ≤ 20

10 ≤ y ≤ 17 10 ≤ y ≤ 17

The system is:

T = 24 x + 32 y T = 24 x + 32 y

T T is the variable for time; it takes 24 minutes to make a widget and 32 minutes to make a wadget.

Your Turn 5.99

Applying linear programming to solve application problems.

There are four steps that need to be completed when solving a problem using linear programming. They are as follows:

Step 1: Compose an objective function to be minimized or maximized.

Step 2: Compose inequalities representing the constraints of the system.

Step 3: Graph the system of inequalities representing the constraints.

Step 4: Find the value of the objective function at each corner point of the graphed region.

The first two steps you have already learned. Let’s continue to use the same examples to illustrate Steps 3 and 4.

Example 5.100

Solving a linear programming problem for two products.

Three friends start their own business, where they knit and sell scarves and sweaters out of high-quality wool. They can make a profit of $8 per scarf and $10 per sweater. To make a scarf, 3 bags of knitting wool are needed; to make a sweater, 4 bags of knitting wool are needed. The friends can only make 8 items per day, and can use not more than 27 bags of knitting wool per day. Determine the number of scarves and sweaters they should make each day to maximize their profit.

Step 1: Compose an objective function to be minimized or maximized. From Example 5.98 , the objective function is P = 8 x + 10 y P = 8 x + 10 y .

Step 2: Compose inequalities representing the constraints of the system. From Example 5.98 , the constraints are x + y ≤ 8 x + y ≤ 8 and 3 x + 4 y ≤ 27 3 x + 4 y ≤ 27 .

Step 3: Graph the system of inequalities representing the constraints. Using methods discussed in Graphing Linear Equations and Inequalities , the graphs of the constraints are shown below. Because the number of scarves ( x x ) and the number of sweaters ( y y ) both must be non-negative numbers (i.e., x ≥ 0 x ≥ 0 and y ≥ 0 y ≥ 0 ), we need to graph the system of inequalities in Quadrant I only. Figure 5.99 shows each constraint graphed on its own axes, while Figure 5.100 shows the graph of the system of inequalities (the two constraints graphed together). In Figure 5.100 , the large shaded region represents the area where the two constraints intersect. If you are unsure how to graph these regions, refer back to Graphing Linear Equations and Inequalities .

Step 4: Find the value of the objective function at each corner point of the graphed region. The “graphed region” is the area where both of the regions intersect; in Figure 5.101 , it is the large shaded area. The “corner points” refer to each vertex of the shaded area. Why the corner points? Because the maximum and minimum of every objective function will occur at one (or more) of the corner points. Figure 5.101 shows the location and coordinates of each corner point.

Three of the four points are readily found, as we used them to graph the regions; the fourth point, the intersection point of the two constraint lines, will have to be found using methods discussed in Systems of Linear Equations in Two Variables , either using substitution or elimination. As a reminder, set up the two equations of the constraint lines:

For this example, substitution will be used.

Substituting 8 - x 8 - x into the first equation for y y , we have

Now, substituting the 5 in for x x in either equation to solve for y y . Choosing the second equation, we have:

Therefore, x = 5 x = 5 , and y = 3 y = 3 .

To find the value of the objective function, P = 8 x + 10 y P = 8 x + 10 y , put the coordinates for each corner point into the equation and solve. The largest solution found when doing this will be the maximum value, and thus will be the answer to the question originally posed: determining the number of scarves and sweaters the new business should make each day to maximize their profit.

The maximum value for the profit P P occurs when x = 5 x = 5 and y = 3 y = 3 . This means that to maximize their profit, the new business should make 5 scarves and 3 sweaters every day.

Your Turn 5.100

People in mathematics, leonid kantorovich.

Leonid Vitalyevich Kantorovich was born January 19, 1912, in St. Petersburg, Russia. Two major events affected young Leonid’s life: when he was five, the Russian Revolution began, making life in St. Petersburg very difficult; so much so that Leonid’s family fled to Belarus for a year. When Leonid was 10, his father died, leaving his mother to raise five children on her own.

Despite the hardships, Leonid showed incredible mathematical ability at a young age. When he was only 14, he enrolled in Leningrad State University to study mathematics. Four years later, at age 18, he graduated with what would be equivalent to a Ph.D. in mathematics.

Although his primary interests were in pure mathematics, in 1938 he began working on problems in economics. Supposedly, he was approached by a local plywood manufacturer with the following question: how to come up with a work schedule for eight lathes to maximize output, given the five different kinds of plywood they had at the factory. By July 1939, Leonid had come up with a solution, not only to the lathe scheduling problem but to other areas as well, such as an optimal crop rotation schedule for farmers, minimizing waste material in manufacturing, and finding optimal routes for transporting goods. The technique he discovered to solve these problems eventually became known as linear programming. He continued to use this technique for solving many other problems involving optimization, which resulted in the book The Best Use of Economic Resources , which was published in 1959. His continued work in linear programming would ultimately result in him winning the Nobel Prize of Economics in 1975.

Check Your Understanding

• P = 20 t − 10 c
• P = 20 c + 10 t
• P = 20 t + 10 c
• P = 20 c − 10 t
• P = 150 w + 180 b
• P = 150 b + 180 w
• P = 180 w + 150 b
• P = 150 w − 180 b
• P = 2.50 f + 6.75 c
• P = 2.50 f + 6.75 t
• P = 2.50 t + 6.75 f
• None of these
• 20 t + 10 c ≤ 70
• 15 t + 4 c ≥ 70
• 15 t + 4 c ≤ 70
• 20 t + 10 c ≤ 12
• 20 t + 10 c ≥ 12
• w + b ≤ 945
• 10 w + 15 b ≥ 945
• 10 w + 15 b ≤ 945
• 150 w + 180 w ≤ 945
• 150 w + 180 b ≤ 1,635
• 25 w + 30 b ≤ 1,635
• w + b ≤ 1,635
• 30 w + 25 b ≤ 1,635
• ( 0 , 0 ) , ( 0 , 12 ) , ( 10 , 2 ) , ( 12 , 0 )
• ( 0 , 0 ) , ( 0 , 12 ) , ( 2 , 10 ) , ( 4 2 3 , 0 )
• ( 0 , 0 ) , ( 17.5 , 0 ) , ( 2 , 10 ) , ( 12 , 0 )

Section 5.11 Exercises

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Business Calculus with Excel

Mike May, S.J., Anneke Bart

Section 1.1 Linear Functions and models

Subsection 1.1.1 equations of a line, kinds of linear equations..

\(p=mq+b\) the slope-intercept form of a line \(p-p_0=m(q-q_0)\) the point-slope form of a line

Example 1.1.1 . Supply and Demand Curves.

\begin{equation*} m=\frac{\text{rise}}{\text{run}}=\frac{(p_1-p_0)}{(q_1-q_0 )} \end{equation*}

\(a x+b y+c=0\) The general form of a line \(y-y_0=\frac{y_1-y_0}{x_1-x_0 }(x-x_0)\) The 2-point form of a line

\(a q+b p+c=0\) The general form of a line \(p-p_0=\frac{p_1-p_0}{q_1-q_0 }(q-q_0)\) The 2-point form of a line

Example 1.1.3 . Finding four versions of a line.

Example 1.1.4 . finding a line from two points..

• Find the equation of the line.
• Find the cost of making 200 widgets.
• We have one point of the form (quantity, cost) at (0, 400) from the fixed cost. This point happens to be an intercept. The slope of the line is \(m=11/1=11\text{.}\) We will let \(C\) stand for Cost and \(q\) stand for quantity. The general formula for a line with these variables will have the form \begin{equation*} C=m q+b. \end{equation*} In this example \(m = 11\) and \(b = 400\text{,}\) hence the equation of the line is \begin{equation*} C=11 q+400 \end{equation*}
• Using the equation form part a) we see that the cost of producing 200 widgets is \begin{equation*} C=11 (200)+400=\$2,600 \end{equation*}

Example 1.1.5 . A nonlinear function.

• How much is she paid if she works 56 hours in a week?
• What is she paid for working 30 hours in a week?
• For this example we will use “designer variables”, longer variables that make the equation easier to read. The output will be Pay, and the input variable — the number of hours worked — will be hrs. We are told that Pay = 500 when hrs = 40. The slope of the line for work beyond 40 hours is \(m=18\text{.}\) Another way to think of this is to say that there is a fixed Pay of $500 and a variable Pay for any hours in excess of 40: i.e. (hrs - 40). Thus, the equation of the line, according to the point-slope form is \begin{equation*} \text{Pay} = \text{variable pay} + \text{Fixed pay}\text{.} \end{equation*} This gives us \begin{equation*} \text{Pay}=m(\text{hrs}-40)+500=18*(\text{hrs}-40)+500\text{.} \end{equation*}
• The pay for working 56 hours is 18(56-40)+500=$788.
• The pay for working 30 hours is $500. This is a trick question part of the problem. From the text of the problem, the linear model only works for overtime, with a flat rate applying to less than 40 hours per week. Comment: The function should be written as a piecewise defined function.

Reading Questions 1.1.2 Reading Check

1 . reading check, linear functions and models..

• In this class we will sometimes use \(q\) and \(p\) instead of \(x\) and \(y\text{.}\)
• \(y=mx+b\) is often referred to as the general form of the line.
• \(y=mx+b\) is often referred to as the slope-intercept form of the line.
• The demand function is a line with a positive slope.
• If they are not labeled, there is no way to tell which is the supply curve and which is the demand curve.
• \(y-y_1=m(x-x_1)\) is often referred to as the point-slope form of the line.
• The supply curve has a positive slope.
• The slope can be defined as rise over run.
• None of the above

Exercises 1.1.3 Exercises 1.1 Linear Functions and models

Exercise group..

• Find the slope of the line determined by the points.
• Give the equation of the line determined by the points.
• Give the value of \(p\) predicted for \(q_0\) by the line.
• First find the slope: \(m= \frac{\text{change in }p}{\text{change in }q} = \frac{17-5}{6-2}=\frac{12}{4}=3\)
• Method 1: use the point-slope equation: \(p-p_0=m (q-q_0)\text{.}\) We can choose either one of the points, so in this case we will find the line using the point \((q_0,p_0 )=(2,5)\text{.}\) This gives the equation \(p-5=3 (q-2)\text{.}\) Rewrite this as \(p=3q-1\)
• Method 2: use the slope- intercept equation \(p=m q+b\text{.}\) Use \((q,p)=(2,5)\) and \(m = 3\) and solve for \(b\text{:}\) \(5=3 (2)+b\text{.}\) And solving for \(b\) we have that \(b= -1\text{,}\) and hence \(p=3q-1\)
• Evaluate at the given point. \(p(4)=3*4-1=11\)
• First find the slope: \(m= \frac{\text{change in }p}{\text{change in }q} = \frac{5-10}{40-20}=-\frac{5}{20}=-\frac{1}{4}\)
• Using \(p=m (q-q_0)+p_0\) with \((q_0,p_0 )=(20, 10)\) and \(m = -\frac{1}{4}\text{,}\) we get \(p=-\frac{1}{4}(q-20)+10\text{.}\) Solving for \(p\) we get \(p =-\frac{1}{4}q+15\text{.}\)
• Evaluate at the given point. \(p(12)=-\frac{1}{4}(12)+15=12\text{.}\)
• First find the slope: \(m= \frac{\text{change in }p}{\text{change in }q} = \frac{578-6}{273-412}=-\frac{572}{139}\)
• Using \(p=m (q-q_0)+p_0\) with \((q_0,p_0 )=(412, 6)\) and \(m = -\frac{572}{139}\text{,}\) we get \(p=-\frac{572}{139}(q-412)+6\text{.}\) (We can combine the constant terms — the \(6\) and the \(-\frac{572}{139}*(-412)\text{,}\) but leaving the equation in this form is acceptable.)
• Evaluate at the given point. \(p(309)=-\frac{572}{139}(309-412)+6 =-\frac{572}{139}(-103)+6=429\frac{119}{139}\)
• Give the equation of the line determined by that information.
• Using the line, give the predicted value of \(p\) for the given \(q_0\text{.}\)
• Give the value of \(q\) for which the predicted value of \(p\) is \(0\text{.}\)
• Using \(p=m q+b\text{,}\) we get \(3=3 (6)+b\text{,}\) so \(b= -15\) and the equation of the line is \(p=3 q-15\text{.}\)
• If \(q_0=4\text{,}\) then \(p=3 q-15=3(4)-15= -3\)
• The predicted value of \(p\) is \(0\text{,}\) when \(0=3 q-15\text{.}\) Solving for \(q\) we get \(q=5\text{.}\) [Comment: in part a) we could have used the point-slope for as well.]
• Using \(p=m (q-q_0 )+p_0\) we get \(p=12.7 (q-22)+183\text{.}\)
• When \(q_0=46\text{,}\) we get that \(p=12.7 (46-22)+183=12.7(24)+183\text{.}\) And hence \(p=487.8\text{.}\)
• The predicted value of \(p\) is \(0\text{,}\) when \(0=183+12.7 (q-22)\text{.}\) Solving for \(q\) we get: \(12.7 (q-22)= -183\) This implies \((q-22)= \frac{-183}{12.7}\text{.}\) Therefore, \begin{equation*} q=22 -\frac{183}{12.7}=7.59\text{.} \end{equation*}
• Using \(p=m (q-q_0 )+p_0\) we get \(p=0 (q-12.3)+9.8\) or \(p=9.8\text{.}\)
• When \(q_0=74\text{,}\) we get that \(p=9.8\text{.}\) (This is a constant function, and the output is going to be \(9.8\) no matter what value of \(q\) we plug in.)
• The predicted value of \(p\) is never \(0\text{,}\) because the predicted value of p is always 9.8.
• Give the slope of the line or say that the slope is undefined.
• Give the intercepts of the line with the axes.
• Give two points that are on the line but not on the axes.
• Solve for \(p\) so we can read off the slope more easily: \(p= -2/3 q+2\text{.}\) Hence the slope is \(m= -2/3\)
• To get the \(p\) -intercept we set \(q=0\text{,}\) so that \(3 p=6\text{,}\) and we get \(p=2\text{.}\) To get the \(q\) -intercept we set \(p=0\text{,}\) so that \(2q=6\text{,}\) and we get \(q=3\text{.}\)
• We can pick any two values for \(q\) (except \(0\) and \(3\) because those are intercepts), so arbitrarily pick some values: say \(q=1\) and \(q=-1\text{.}\) If \(q=1\text{,}\) then \(3 p+2=6\text{,}\) and hence \(p=4/3\) so the point would be \((q,p)=(1,4/3)\text{.}\) If \(q=-1\text{,}\) then \(3 p-2=6\text{,}\) and hence \(p=8/3\) so the point would be \((q,p)=(-1,8/3 )\text{.}\)
• The slope is \(0\)
• The y-intercept is \(y = 5\text{.}\) There is no \(x\) -intercept
• No matter what \(x\) we choose the \(y\) value will always be \(5\text{.}\) So we can pick any two values of \(x\) (not \(0\) to avoid the intercept). For instance \((x,y)=(-20,5)\text{,}\) \((x,y)=(-2,5)\text{,}\) \((x,y)=(7,5)\text{,}\) etc are all points on the line.
• The slope is \(4\)
• To get the \(y\) -intercept we set \(x=0\text{,}\) so that \(y=4(-3)+9=-3\text{,}\) To get the \(x\) -intercept we set \(y=0\text{,}\) so that \(0=4(x-3)+9\text{.}\) Solve for \(x\) and we get \(x-3=-\frac{9}{4}\) and \(x=3-\frac{9}{4}=\frac{3}{4}\)
• Avoiding the intercepts we can choose \(x\) to be any number other than \(x = 0\) and \(x=3/4\text{.}\) We may look for nice values of \(x\) that make the arithmetic come out nice: for example choose \(x=3\) and \(x=4\text{.}\) When \(x=3\text{,}\) \(y=9\text{,}\) and we get the point \((x,y)=(3,9)\text{.}\) When \(x=4\text{,}\) \(y=13\text{,}\) and we get the point \((x,y)=(4,13)\text{.}\)

Grade 8 Mathematics Module: “Solving Problems Involving Linear Functions”

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

In this module, you will be acquainted with solving problems involving linear functions. The scope of this module enables you to use it in many different learning situations. The lesson is arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

This module contains:

• Lesson 1: Solving Problems Involving Linear Functions

After going through this module, you are expected to:

1. identify steps in modeling and solving word problems involving linear functions;

2. create linear functions that represent relation between quantities; and

3. apply the concepts of linear function in solving real-life problems.

Grade 8 Mathematics Quarter 2 Self-Learning Module: “Solving Problems Involving Linear Functions”

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Solve Linear Equations in Disguise Worksheets

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Printable “Algebra” worksheets: Write Equations Using Symbols Write Linear & Nonlinear Expressions Write & Solve Linear Equations Solve Linear Equations (Distributive) Solve Linear Equations in Disguise Solve Linear Equations Word Problems

Examples, solutions, videos, and worksheets to help Grade 7 and Grade 8 learn how to rewrite and solve equations that are not obviously linear equations using properties of equality.

How to solve “linear equations in disguise”?

“Linear equations in disguise” refer to equations that don’t initially appear to be linear but can be manipulated into a linear form. These equations might involve fractions, products, or other operations, but after simplification, they reveal their linear nature.

Steps to Solve Linear Equations in Disguise

• Simplify the Equation:
• Remove any fractions by multiplying both sides by the least common denominator (LCD).
• Distribute any factors across parentheses.
• Combine like terms.
• Isolate the Variable:
• Move all terms involving the variable to one side of the equation.
• Move all constant terms to the other side.
• Solve for the Variable:
• Divide or multiply as needed to solve for the variable.
• Check the Solution:
• Substitute your solution back into the original equation to ensure it satisfies the equation.

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Grade 9/10 Math Circles Linear Diophantine Equations Part 1 - Problem Set

Calculate the following gcds using the Euclidean algorithm.

\(\gcd(12, 57)\)

\(\gcd(212, 37)\)

\(\gcd(4389, 2919)\)

Using the Euclidean algorithm, we find \[\begin{align*} 57&=12\cdot 4+9\\ 12&=9\cdot 1+3\\ 9&=3\cdot3+0\end{align*}\] Thus, \(\gcd(12, 57) = 3\) .

Using the Euclidean algorithm, we find \[\begin{align*} 212&=37\cdot 5+27\\ 37&=27\cdot 1+10\\ 27&=10\cdot 2+7\\ 10&=7\cdot 1+3\\ 7&=3\cdot 2+1\\ 3&=1\cdot 3+0\end{align*}\] Thus, \(\gcd(212, 37) = 1\) .

Using the Euclidean algorithm, we find \[\begin{align*} 4389&=2919\cdot 1+1470\\ 2919&=1470\cdot 1+1449\\ 1470&=1449\cdot 1+21\\ 1449&=21\cdot 69+0\end{align*}\] Thus, \(\gcd(4389, 2919) = 21\) .

For each of the following linear Diophantine equations, find a solution where \(x\) and \(y\) are integers, or explain why one does not exist.

\(4389x+2919y=21\)

\(4389x+2919y=231\)

\(212x - 37 y =1\)

\(12x + 57y = 124\)

\(12x + 57y = 423\)

Since \(\gcd(4389, 2919)= 21\) , this linear Diophantine equation has a solution. To find a solution, we will work backwards through the steps in the Euclidean algorithm in our solution to Problem \(1\) . \[\begin{align*} 21&=\underline{1470}-1\cdot\underline{1449} & \text{from (3)}\\ &=\underline{1470}-1\cdot(\underline{2919}-1\cdot\underline{1470})& \text{from (2)}\\ &=2\cdot\underline{1470}-1\cdot\underline{2919}\\ &=2\cdot(\underline{4389}-1\cdot\underline{2919})-1\cdot\underline{2919} & \text{from (1)}\\ &=2\cdot\underline{4389}-3\cdot\underline{2919}\end{align*}\] That is, \(4389(2) + 2919(-3) =21\) . Thus, one solution to the linear Diophantine equation \(4389x + 2919y = 21\) is \(x=2\) , \(y=-3\) .

Since \(\gcd(4389, 2919) = 21\) and \(231 = 21\times 11\) , this linear Diophantine equation has a solution.

From part (a), we know that \[4389(2)+2919(-3)=21\] Multiplying both sides of the equation by \(11\) : \[11(4389(2)+2919(-3))=11(21)\] or \[4389(22)+2919(-33)=231\] Thus, a solution to the linear Diophantine equation \(4389x+2919y=231\) is \(x=22\) , \(y=-33\) .

This is equivalent to solving the linear Diophantine equation \(212x + 37(-y) = 1\) , or equivalently, \(212x+37z = 1\) , where \(y=-z\) .

From Problem 1, we know that \(\gcd(212, 37) = 1\) , and so the linear Diophantine equation \(212x+37z = 1\) has a solution. To find a solution, we will work backwards through the steps in the Euclidean algorithm in our solution to Problem \(1\) . \[\begin{align*} 1&=\underline{7}-2\cdot\underline{3}& \text{from (5)}\\ &=\underline{7}-2\cdot(\underline{10}-1\cdot\underline{7})& \text{from (4)}\\ &=3\cdot\underline{7}-2\cdot\underline{10}\\ &=3\cdot(\underline{27}-2\cdot\underline{10})-2\cdot\underline{10}& \text{from (3)}\\ &=3\cdot\underline{27}-8\cdot\underline{10}\\ &=3\cdot\underline{27}-8\cdot(\underline{37} - 1\cdot\underline{27})& \text{from (2)}\\ &=11\cdot\underline{27}-8\cdot\underline{37}\\ &=11\cdot(\underline{212} - 5\cdot\underline{37})-8\cdot\underline{37}& \text{from (1)}\\ &=11\cdot\underline{212}-63\cdot\underline{37}\\ &= 212(11) + 37(-63)\end{align*}\] Thus, one solution to the linear Diophantine equation \(212x+37z = 1\) is \(x=11\) , \(z=-63\) .

Therefore, a solution to \(212x - 37 y =1\) is \(x=11\) , \(y=63\) .

Since \(\gcd(12, 57)=3\) , and \(3\) does not divide \(124\) , there is no solution to this linear Diophantine equation where \(x\) and \(y\) are both integers.

Since \(\gcd(12, 57)=3\) and \(423 = 3 \times 141\) , this linear Diophantine equation has a solution. To find a solution, we will first work backwards through the steps in the Euclidean algorithm in our solution to Problem \(1\) . \[\begin{align*} 3&=\underline{12}-1\cdot\underline{9}& \text{from (2)}\\ &=\underline{12}-1\cdot(\underline{57}-4\cdot\underline{12})& \text{from (1)}\\ &=5\cdot\underline{12}-1\cdot\underline{57}\end{align*}\] Thus, \[12(5) + 57(-1) = 3\] Multiplying both sides of the equation by \(141\) gives, \[\begin{align*} 141(12(5) + 57(-1)) &= 141(3) \\ 12(141\cdot5) + 57(141\cdot(-1)) &= 423 \\ 12(705) + 57(-141) &= 423 \end{align*}\] Thus, one solution to the linear Diophantine equation \(12x + 57y = 423\) is \(x=705\) , \(y=-141\) .

Can \(1000\) be expressed as the sum of two integers, one of which is divisible by \(11\) and the other by \(17\) ? If so, find examples of such integers. If not, explain why.

This problem is equivalent to asking if there is there a solution to the linear Diophantine equation \(11x + 17y = 1000\) .

We know that there is a solution only if \(\gcd(11, 17)\) divides \(1000\) .

Using the Euclidean algorithm, \[\begin{align*} 17&=11\cdot 1+6\\ 11&=6\cdot 1+5\\ 6&=5\cdot 1+1\\ 5&=1\cdot 5+0\end{align*}\] Thus, \(\gcd(11, 17) = 1\) . Since \(1000\) is divisible by \(1\) , there is a solution to this linear Diophantine equation.

We begin by first finding a solution to the linear Diophantine equation \(11x + 17y = 1\) . Working backwards through our steps in the Euclidean algorithm, we have \[\begin{align*} 1&=\underline{6}-1\cdot\underline{5}& \text{from (3)}\\ &=\underline{6}-1\cdot(\underline{11}-1\cdot\underline{6})& \text{from (2)}\\ &=2\cdot\underline{6}-1\cdot\underline{11}\\ &=2\cdot(\underline{17} - 1\cdot\underline{11})-1\cdot\underline{11}& \text{from (1)}\\ &=2\cdot\underline{17}-3\cdot\underline{11}\end{align*}\] Thus, \(11(-3) + 17(2) = 1\) .

Multiplying both sides of this equation by \(1000\) gives \[\begin{align*} 1000(11(-3) + 17(2)) &= 1000(1) \\ 11(-3\cdot1000) + 17(2\cdot1000) &= 1000 \\ 11(-3000) + 17(2000) &= 1000 \end{align*}\] Thus, one solution to the linear Diophantine equation \(11x + 17y = 1000\) is \(x=-3000\) , \(y=2000\) .

Therefore, \(1000\) can be expressed as the sum of \(-33000\) (which is divisible by \(11\) ) and \(34000\) (which is divisible by \(17\) ).

Can \(1000\) be expressed as the sum of two integers, one of which is divisible by \(9\) and the other by \(12\) ? If so, find examples of such integers. If not, explain why.

This problem is equivalent to asking if there is a solution to the linear Diophantine equation \(9x + 12y = 1000\) .

We know that there is a solution only if \(\gcd(9, 12)\) divides \(1000\) . By inspection, \(\gcd(9, 12) = 3\) . Since \(1000\) is not divisible by \(3\) , there is no integer solution to the linear Diophantine equation \(9x + 12y = 1000\) .

Therefore, \(1000\) cannot be expressed as the sum of two integers, one of which is divisible by \(9\) and the other by \(12\) .

Use the Euclidean algorithm to find a solution to \(25x+10y=215\) , the linear Diophantine equation from Example \(1\) in the Lesson. Does your answer make sense in the context of the problem? If not, how can we find a solution that does make sense?

When solving a linear Diophantine equation, our first step is always to find the gcd of the coefficients using the Euclidean algorithm: \[\begin{align*} 25&=10(2)+5\\ 10&=5(2)+0\end{align*}\] We end the Euclidean algorithm to find that \(\gcd(25,10)=5\) .

Does \(5\) divide \(215\) ? Yes! This means that the linear Diophantine equation \(25x+10y=215\) has a solution. To find it, we work backwards through the Euclidean algorithm.

Since our algorithm ended so quickly, it is easy to see that \(5=25-10(2)\) .

By multiplying both sides of this equation by \(\frac{215}{5}=43\) , we get \[25(43)+10(-86)=215\]

The answer does not make sense in the context of the problem. This answer suggests that we should use \(43\) quarters and \(-86\) dimes to total \(\$2.15\) .

What we are really looking for is a solution to the linear Diophantine equation \[25x+10y=215\] where both \(x\) and \(y\) are non-negative integers.

One way to rectify this problem is the following: for every \(5\) dimes we are told to subtract, we just add \(2\) fewer quarters (since \(5\) dimes and \(2\) quarters both total \(50\) cents). So by taking away \(36=2(18)\) quarters, we can add \(90=5(18)\) dimes. This means that \(x=43-36=7\) and \(y=-86+90=4\) is another solution to the linear Diophantine equation \[25x+10y=215\] That is, Sara can buy the coffee with \(7\) quarters and \(4\) dimes. We will further discuss constraints like this in the second lesson on linear Diophantine equations.

Find an integer \(n\) , which, when divided by \(78\) leaves a remainder of \(37\) , and when divided by \(29\) leaves a remainder of \(17\) .

Since \(n\) is the same in both equations, it must be the case that \(78x+37=29y+17.\) By rearranging the terms, we arrive at the linear Diophantine equation \[78x-29y=-20.\]

If we can find a solution to this linear Diophantine equation, we can use equation \((1)\) or \((2)\) to obtain \(n\) .

Our first step is to find \(\gcd(78,29)\) . Using the Euclidean algorithm, \[\begin{align*} 78&=29(2)+20\\ 29&=20(1)+9\\ 20&=9(2)+2\\ 9&=2(4)+1\\ 2&=1(2)+0\end{align*}\]

Thus we know that \(\gcd(78,29)=\gcd(29,20)=\gcd(20,9)=\gcd(9,2)=\gcd(2,1)=\gcd(1,0)=1\) .

We end the Euclidean algorithm to find that \(\gcd(78,29)=1\) . Since \(1\) divides \(-20\) , a solution to the linear Diophantine equation \(78x - 29y = -20\) exists. To find a solution, we have to work backwards through the Euclidean algorithm.

From \((6)\) : \[1= 9-\underline{2}(4)\] Substituting for \(2\) using \((5)\) : \[1= 9-[20-9(2)](4)\] Simplifying: \[1= \underline{9}(9)-20(4)\] Substituting for \(9\) using \((4)\) : \[1= [29-20(1)](9)-20(4)\] Simplifying: \[1=29(9)-\underline{20}(13)\] Substituting for \(20\) using \((4)\) : \[1=29(9)-[78-29(2)](13)\] Simplifying: \[1=78(-13)+29(35)\] This means that \(78(-13)+29(35)=1\) .

Multiplying both sides of the equation \(78(-13)+29(35)=1\) by \(-20\) , the equation becomes \[78(260)+29(-700)=-20\] Or equivalently, \[78(260)-29(700)=-20\]

Hence, a solution to the linear Diophantine equation \(78x - 29y = -20\) is \(x=260, y=700\) .

We may now use either equation \((1)\) or equation \((2)\) to get \(n\) . By substituting \(x=260\) into equation \((1)\) , we get \(n=20\,317\) .

Suppose \(a\) , \(b\) , and \(c\) are given integers. If \(ax^2 + by^2 = c\) has a solution where \(x\) and \(y\) are integers, is it true that \(\gcd(a,b)\) divides \(c\) ? Why or why not?

Suppose \(a\) , \(b\) , and \(c\) are given integers. If \(\gcd(a,b)\) divides \(c\) , is it true that \(ax^2 + by^2 = c\) always has a solution where \(x\) and \(y\) are integers? Why or why not?

Yes, this must be true. Suppose it is not true. That is, suppose \(c\) is not divisible by \(\gcd(a,b)\) . Let’s divide both sides of the equation \(ax^2 + by^2 = c\) by \(\gcd(a,b)\) . Since both \(a\) and \(b\) are divisible by \(\gcd(a,b)\) , after dividing by \(\gcd(a,b)\) , the left side can be written as \(nx^2+my^2\) for some integers \(n\) and \(m\) . Thus, after dividing by \(\gcd(a,b)\) , we have \[nx^2 + my^2 = \frac{c}{\gcd(a,b)}\] Here, \(n\) , \(x\) , \(m\) , and \(y\) are integers, and so \(nx^2 + my^2\) is an integer. But \(\frac{c}{\gcd(a,b)}\) is not an integer, since \(c\) is not divisible by \(\gcd(a,b)\) . So this last equality is impossible! Therefore, if \(c\) is not divisible by \(\gcd(a,b)\) , there cannot be a solution to \(ax^2 + by^2 = c\) where \(x\) and \(y\) are integers. Thus, if there is a solution to \(ax^2 + by^2 = c\) where \(x\) and \(y\) are integers, it must be the case that \(\gcd(a,b)\) divides \(c\) .

No, this is not necessarily true. Consider when \(a=1\) , \(b=1\) , and \(c=3\) . We know that \(\gcd(a,b) = \gcd(1,1) = 1\) . Also, \(c=3\) is divisible by \(1\) . However, there are no integers that satisfy the equation \(x^2 + y^2 = 3\) .

Use the division algorithm and the important fact to show that \(\gcd(k+1,k)=1\) for any integer \(k\geq 1\) .

Use the division algorithm and the important fact to show that \[\gcd(7k+6,6k+5)=1\] for any integer \(k\geq 1\) .

Recall from the important fact, that if \(a=bq+r\) , then \(\gcd(a,b)=\gcd(b,r)\) . How can we use this to solve our problem? Well, we can write \[(k+1)=k(1)+1\] Here \(k+1\) is playing the role of \(a\) , \(k\) is playing the role of \(b\) , and \(1\) is playing the roles of \(q\) and \(r\) . Thus, \[\gcd(k+1,k)=\gcd(k,1)=1\]

We’re going to try an approach similar to that in part (a). First, let’s use the division algorithm once to write \[(7k+6)=(6k+5)(1)+(k+1)\] Here, \(7k+6\) is playing the role of \(a\) , \(6k+5\) is playing the role of \(b\) , \(1\) is playing the role of \(q\) , and \(k\) is playing the role of \(r\) . We deduce from the important fact that \[\gcd(7k+6,6k+5)=\gcd(6k+5,k+1)\]

Again, we’ll apply the division algorithm to our new pair to get \[(6k+5)=(k+1)(5)+k\] Now the roles of \(a,b,q\) and \(r\) are played by \(6k+5\) , \(k+1\) , \(5\) , and \(k\) , respectively. The important fact tells us that \[\gcd(6k+5,k+1)=\gcd(k+1,k)\]

From part (a), we know that \(\gcd(k+1,k) = 1\) .

Putting this all together, we get \[\gcd(7k+6,6k+5)=\gcd(6k+5,k+1)=\gcd(k+1,k)=1\]

Estimation of Oil Reservoir Transmissivity and Storativity Fields Using a Radial Basis Function Network Based on Inverse Problem Solving

• Published: 28 August 2024
• Volume 45 , pages 2067–2075, ( 2024 )

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• V. P. Kosyakov 1 &
• D. Yu. Legostaev 1  

In the oil industry, there is a noticeable trend towards using proxy models to simulate various levels of complexity in order to make operational predictions. In particular, machine learning techniques are being actively developed in the context of the digitalization and automation of production processes. This paper proposes a method for combining a physically relevant fluid flow model with machine learning techniques to address the challenges of history-matching and prediction. The approach is demonstrated using synthetic oil reservoir models. The synthetic model has significant zonal inhomogeneities in the permeability and storativity fields. The simplified single-phase flow through a porous medium model was used for the proposed approach. This model was matched to the historical values of the development parameters by restoring the fields of the reservoir parameters. Properties fields were reconstructed using a radial basis functions network and a fully connected linear layer. Based on the reconstructed field, interwell connectivity coefficients were calculated, which corresponded qualitatively and quantitatively to the true interwell connectivity coefficients. The predictive characteristics of the proposed approach were evaluated by split the historical dataset into training and test time intervals.

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The research was supported by the Russian Scientific Foundation (project no. 24-21-00468), https://rscf.ru/project/24-21-00468/ .

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Kosyakov, V.P., Legostaev, D.Y. Estimation of Oil Reservoir Transmissivity and Storativity Fields Using a Radial Basis Function Network Based on Inverse Problem Solving. Lobachevskii J Math 45 , 2067–2075 (2024). https://doi.org/10.1134/S199508022460225X

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Received : 01 March 2024

Revised : 19 March 2024

Accepted : 02 April 2024

Published : 28 August 2024

Issue Date : May 2024

DOI : https://doi.org/10.1134/S199508022460225X

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• flow through porous medium
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