The document provides an example of solving a system of linear equations using the substitution method. It begins with the system 2x + y = 7 and x + y = 5. It solves the second equation for x in terms of y, getting x = 5 - y. This expression for x is then substituted into the first equation, giving 10 - 2y + y = 7, which can be solved to find the value of y, and then substituted back into the original equation to find the value of x. The solution is presented as (2, 3). The document then provides two additional examples demonstrating how to set up and solve systems of equations using the substitution method.
I am Piers L. I am a Calculus Assignment Expert at mathsassignmenthelp.com. I hold a Master's in Mathematics from, the University of Adelaide. I have been helping students with their assignments for the past 6 years. I solve assignments related to Calculus.
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I am Piers L. I am a Calculus Homework Expert at mathhomeworksolver.com. I hold a Master's in Mathematics from, the University of Adelaide. I have been helping students with their homework for the past 6 years. I solve homework related to Calculus.
Visit mathhomeworksolver.com or email support@mathhomeworksolver.com.
You can also call on +1 678 648 4277 for any assistance with Calculus Homework.
I am Piers L. I am a Calculus Assignment Expert at mathsassignmenthelp.com. I hold a Master's in Mathematics from, the University of Adelaide. I have been helping students with their assignments for the past 6 years. I solve assignments related to Calculus.
Visit mathsassignmenthelp.com or email info@mathsassignmenthelp.com.
You can also call on +1 678 648 4277 for any assistance with Calculus Assignment.
I am Piers L. I am a Calculus Homework Expert at mathhomeworksolver.com. I hold a Master's in Mathematics from, the University of Adelaide. I have been helping students with their homework for the past 6 years. I solve homework related to Calculus.
Visit mathhomeworksolver.com or email support@mathhomeworksolver.com.
You can also call on +1 678 648 4277 for any assistance with Calculus Homework.
I am Duncan V. I am a Calculus Homework Expert at mathshomeworksolver.com. I hold a Master's in Mathematics from Manchester, United Kingdom. I have been helping students with their homework for the past 8 years. I solve homework related to Calculus.
Visit mathhomeworksolver.com or email support@mathhomeworksolver.com.
You can also call on +1 678 648 4277 for any assistance with Calculus Homework.
Complementary function, particular integral,homogeneous linear functions with constant variables, Euler Cauchy's equation, Legendre's equation, Method of variation of parameters,Simultaneous first order linear differential equation with constant coefficients,
Successive Differentiation is the process of differentiating a given function successively times and the results of such differentiation are called successive derivatives. The higher order differential coefficients are of utmost importance in scientific and engineering applications.
formulation of first order linear and nonlinear 2nd order differential equationMahaswari Jogia
• Equations which are composed of an unknown function and its derivatives are called differential equations.
• Differential equations play a fundamental role in engineering because many physical phenomena are best formulated mathematically in terms of their rate of change.
• When a function involves one dependent variable, the equation is called an ordinary differential equation (ODE).
• A partial differential equation (PDE) involves two or more independent variables.
Figure 1: CHARACTERIZATION OF DIFFERENTIAL EQUATION
FIRST ORDER DIFFERENTIAL EQUATION:
FIRST ORDER LINEAR AND NON LINEAR EQUATION:
A first order equation includes a first derivative as its highest derivative.
- Linear 1st order ODE:
Where P and Q are functions of x.
TYPES OF LINEAR DIFFERENTIAL EQUATION:
1. Separable Variable
2. Homogeneous Equation
3. Exact Equation
4. Linear Equation
i. SEPARABLE VARIABLE:
The first-order differential equation:
Is called separable provided that f(x,y) can be written as the product of a function of x and a function of y.
Suppose we can write the above equation as
We then say we have “separated” the variables. By taking h(y) to the LHS, the equation becomes:
Integrating, we get the solution as:
Where c is an arbitrary constant.
EXAMPLE 1.
Consider the DE :
Separating the variables, we get
Integrating we get the solution as:
Solutions Manual for An Introduction To Abstract Algebra With Notes To The Fu...Aladdinew
Full download : https://goo.gl/XTCXti Solutions Manual for An Introduction To Abstract Algebra With Notes To The Future Teacher 1st Edition by Nicodemi
I am Duncan V. I am a Calculus Homework Expert at mathshomeworksolver.com. I hold a Master's in Mathematics from Manchester, United Kingdom. I have been helping students with their homework for the past 8 years. I solve homework related to Calculus.
Visit mathhomeworksolver.com or email support@mathhomeworksolver.com.
You can also call on +1 678 648 4277 for any assistance with Calculus Homework.
Complementary function, particular integral,homogeneous linear functions with constant variables, Euler Cauchy's equation, Legendre's equation, Method of variation of parameters,Simultaneous first order linear differential equation with constant coefficients,
Successive Differentiation is the process of differentiating a given function successively times and the results of such differentiation are called successive derivatives. The higher order differential coefficients are of utmost importance in scientific and engineering applications.
formulation of first order linear and nonlinear 2nd order differential equationMahaswari Jogia
• Equations which are composed of an unknown function and its derivatives are called differential equations.
• Differential equations play a fundamental role in engineering because many physical phenomena are best formulated mathematically in terms of their rate of change.
• When a function involves one dependent variable, the equation is called an ordinary differential equation (ODE).
• A partial differential equation (PDE) involves two or more independent variables.
Figure 1: CHARACTERIZATION OF DIFFERENTIAL EQUATION
FIRST ORDER DIFFERENTIAL EQUATION:
FIRST ORDER LINEAR AND NON LINEAR EQUATION:
A first order equation includes a first derivative as its highest derivative.
- Linear 1st order ODE:
Where P and Q are functions of x.
TYPES OF LINEAR DIFFERENTIAL EQUATION:
1. Separable Variable
2. Homogeneous Equation
3. Exact Equation
4. Linear Equation
i. SEPARABLE VARIABLE:
The first-order differential equation:
Is called separable provided that f(x,y) can be written as the product of a function of x and a function of y.
Suppose we can write the above equation as
We then say we have “separated” the variables. By taking h(y) to the LHS, the equation becomes:
Integrating, we get the solution as:
Where c is an arbitrary constant.
EXAMPLE 1.
Consider the DE :
Separating the variables, we get
Integrating we get the solution as:
Solutions Manual for An Introduction To Abstract Algebra With Notes To The Fu...Aladdinew
Full download : https://goo.gl/XTCXti Solutions Manual for An Introduction To Abstract Algebra With Notes To The Future Teacher 1st Edition by Nicodemi
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
2. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
3. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
4. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
5. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
6. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
7. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
= 5P + 3D + 10P + 6D
8. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
= 5P + 3D + 10P + 6D
= 15P + 9D
or 15 slices of pizzas and 9 donuts.
9. In math, the phrase "substitute (the expression) back into ...“
means to do the exchange, using the given coupon–expression,
in the targeted equations, or expressions mentioned.
Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
= 5P + 3D + 10P + 6D
= 15P + 9D
or 15 slices of pizzas and 9 donuts.
10. There are two other methods to solve system of equations.
Systems of Linear Equations II
11. There are two other methods to solve system of equations.
Substitution Method
Systems of Linear Equations II
12. There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
Systems of Linear Equations II
13. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
14. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
15. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y
16. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y
17. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y
18. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7 3 = y
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y
19. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7 3 = y
Systems of Linear Equations II
Example B. {
Put 3 = y back to E2 to find x, we get
x + 3 = 5
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y
20. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7 3 = y
Systems of Linear Equations II
Example B. {
Put 3 = y back to E2 to find x, we get
x + 3 = 5 x = 2
Therefore the solution is (2, 3)
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y
21. We use the substitution method when it's easy to solve for
one of the variable in terms of the other.
Systems of Linear Equations II
22. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
Systems of Linear Equations II
23. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
Systems of Linear Equations II
Example C. {
24. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
Systems of Linear Equations II
Example C. {
25. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y.
Systems of Linear Equations II
Example C. {
26. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
Systems of Linear Equations II
Example C. {
27. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
Systems of Linear Equations II
Example C. {
28. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
29. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7
30. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7
y = 2(3) – 7
y = -1
31. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7
y = 2(3) – 7
y = -1 Hence the solution is (3, -1).
33. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line.
Systems of Linear Equations II
34. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines.
Systems of Linear Equations II
35. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically.
Systems of Linear Equations II
36. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
37. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
38. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
39. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0
0
2x + y = 7
40. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
0
2x + y = 7
41. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
E2
42. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
E2
(Check another point.)
43. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
E1
E2
(0, 7)
(7/2, 0)
E1(Check another point.)
44. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x + y = 5
x y
0 7
7/2 0
2x + y = 7
x y
0
0
(0, 7)
(7/2, 0)
E1(Check another point.)
45. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7 x + y = 5
x y
0 5
5 0
(0, 7)
(7/2, 0)
E1(Check another point.)
46. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
(0, 7)
(7/2, 0)
(0, 5)
(5, 0)
E2
x + y = 5
x y
0 5
5 0
E1(Check another point.)
47. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
(0, 7)
(7/2, 0)
(0, 5)
(5, 0)
The intersection
(2, 3) is the solution
E1
E2
x + y = 5
x y
0 5
5 0
E1(Check another point.)
48. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
Systems of Linear Equations II
49. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
50. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
51. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
(0, 7)
(7, 0)
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
52. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
x + y = 5
53. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
(0, 5)
(5, 0)
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
x + y = 5
54. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
(0, 5)
(5, 0)
No intersection.
No solution
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
x + y = 5
55. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
Systems of Linear Equations II
56. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.
57. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.
58. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.
59. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
(5, 0)
(0, 5)
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.
60. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
(5, 0)
(0, 5)
Every point is a solution,
e.g.(0, 5), (2, 3), (5, 0)…
(2, 3)
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.
61. Systems of Linear Equations II
4. {–x + 2y = –12
y = 4 – 2x
Exercise. Solve by the substitution method.
1. {y = 3 – x
2x + y = 4
2. 3. {x = 3 – y
2x – y = 6
{x + y = 3
2x + 6 = y
5. {3x + 4y = 3
x = 6 + 2y
6. { x = 3 – 3y
2x – 9y = –4
10. Graph the inconsistent system
{ x + 3y = 4
2x + 6y = 8
{2x – y = 2
8x – 4y = 6
Problem 7, 8, 9: Solve problem 1, 2, and 3 by graphing.
11. Graph the dependent system