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Algebra A1 Mr. Brennan
Chapter 7 Systems of Linear Equations and Inequalities
Review
Hamilton-Wenham Regional High School Department of Mathematics
![Page 2: Algebra A1Mr. Brennan Chapter 7 Systems of Linear Equations and Inequalities Review Hamilton-Wenham Regional High SchoolDepartment of Mathematics](https://reader031.vdocuments.mx/reader031/viewer/2022032722/56649cf45503460f949c3027/html5/thumbnails/2.jpg)
Algebra A1 Mr. Brennan
Chapter 7 ReviewSystems of Linear Equations and Inequalities
Hamilton-Wenham Regional High School Department of Mathematics
Learning Objectives: Chapter 7
![Page 3: Algebra A1Mr. Brennan Chapter 7 Systems of Linear Equations and Inequalities Review Hamilton-Wenham Regional High SchoolDepartment of Mathematics](https://reader031.vdocuments.mx/reader031/viewer/2022032722/56649cf45503460f949c3027/html5/thumbnails/3.jpg)
Algebra A1 Mr. Brennan
Chapter 7 ReviewSystems of Linear Equations and Inequalities
Hamilton-Wenham Regional High School Department of Mathematics
Learning Objectives: Chapter 7
![Page 4: Algebra A1Mr. Brennan Chapter 7 Systems of Linear Equations and Inequalities Review Hamilton-Wenham Regional High SchoolDepartment of Mathematics](https://reader031.vdocuments.mx/reader031/viewer/2022032722/56649cf45503460f949c3027/html5/thumbnails/4.jpg)
The review for chapter 7 has three types of slides
1. Review material with goals and definitions2. Examples with solutions (13)3. Practice problems There are 30 practice problems for Chapter 7.
Work out the problems as you encounter them.
The answer to each question is displayed after each question.
Systems of Linear Equations and InequalitiesChapter
7
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Lesson
7.1Solving Linear Systems
by Graphing
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Solution
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Solution
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Solution
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Solution
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Lesson
7.2Solving Linear Systems
by Substitution
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Solution
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Solution
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Solution
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Solution
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Lesson
7.3Solving Linear Systems
by Linear Combinations
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Solution
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Solution
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Solution
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Solution
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Lesson
7.4Applications of Linear Systems: Exploring Data and Statistics
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Solution
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SolutionSample answers are given here, there are multiple ways to solve each of these problems.
a) Substitution method because the coefficient of x is one in the second equation, so x = 5 – 3y can be substituted into the first equation. Or 2x -3 = y can be used to substitute for y in the second equation.
b) Use linear combinations – none of the variables have a coefficient of 1, or -1, though you could divide both sides of the first equation by 4 and get x + y = 4, then use substitution.
c) Substitution method because the coefficient of x is one in the first equation
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Solution
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Solution
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Lesson
7.5Special Types of Linear Systems
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Solution
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Solution
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Solution
Multiple methods can be used for any of these problems.
1. Using substitution you would end up showing -4 = -1 which is a false statement, so the system has no solution.
Using a graphing method you would see that the lines are parallel, so the system has no solution.
2. Using substitution you would end up showing 5 = 7/3 which is a false statement, so the system has no solution.
Using a graphing method you would see that the lines are parallel, so the system has no solution.
3. Using substitution you would end up showing 1 = -1 which is a false statement, so the system has no solution.
Using a graphing method you would see that the lines are parallel, so the system has no solution.
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Solution
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SolutionMultiple methods can be used for any of these problems.
4. Using substitution you would end up showing 0 = 0 which is a true statement, so the system has infinitely many solutions.
Using a graphing method you would see that the lines coincide, so the system has infinitely many solutions.
5. Using substitution you would end up showing 0 = 0 which is a true statement, so the system has infinitely many solutions.
Using a graphing method you would see that the lines coincide, so the system has infinitely many solutions.
6. Using substitution you would end up showing 0 = 0 which is a true statement, so the system has infinitely many solutions.
Using a graphing method you would see that the lines coincide, so the system has infinitely many solutions.
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Solution
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Solution
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Lesson
7.6Solving Systems
of Linear Inequalities
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Solution
When graphing a system of linear inequalities, find each corner point (or vertex). The graph of the system for Example 1 has four corner points: (0, 0), (2, 2), (3, 2), and (3, -3).
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Solution
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Solution
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Solution
The graph of the system of inequalities is shown. Any point in the shaded region is a solution of the system. Because you cannot buy a fractionof a disc or video, only ordered pairs of integersin the shaded region will answer the problem.
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Solution