Algebra A1 Mr. Brennan

Chapter 7 Systems of Linear Equations and Inequalities

Review

Hamilton-Wenham Regional High School Department of Mathematics

Algebra A1 Mr. Brennan

Chapter 7 ReviewSystems of Linear Equations and Inequalities

Hamilton-Wenham Regional High School Department of Mathematics

Learning Objectives: Chapter 7

Algebra A1 Mr. Brennan

Chapter 7 ReviewSystems of Linear Equations and Inequalities

Hamilton-Wenham Regional High School Department of Mathematics

Learning Objectives: Chapter 7

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

Lesson

7.1Solving Linear Systems

by Graphing

Solution

Solution

Solution

Solution

Lesson

7.2Solving Linear Systems

by Substitution

Solution

Solution

Solution

Solution

Lesson

7.3Solving Linear Systems

by Linear Combinations

Solution

Solution

Solution

Solution

Lesson

7.4Applications of Linear Systems: Exploring Data and Statistics

Solution

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

Solution

Solution

Lesson

7.5Special Types of Linear Systems

Solution

Solution

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.

Solution

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.

Solution

Solution

Lesson

7.6Solving Systems

of Linear Inequalities

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).

Solution

Solution

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.

Solution