Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab

COMPLETE #1-11 ODD ON PAGE 150 UNDER THE PREREQUISITE

SKILLS TAB

CHAPTER 3:LINEAR SYSTEMS AND

MATRICESBIG IDEAS:1. Solving systems of equations

using a variety of methods

2. Graphing systems of equations and inequalities

3. Using Matrices

LESSON 1: SOLVE LINEAR SYSTEMS BY

GRAPHING

ESSENTIAL QUESTION

How do you solve a system of linear

equations graphically?

VOCABULARY Consistent: A system of equations that has at

least one solution

Inconsistent: A system of equations that has no solutions

Dependent: A consistent system of equations that has infinitely many solutions

Independent: A consistent system that has exactly one solution

EXAMPLE 1 Solve a system graphically

Graph the linear system and estimate the solution. Then check the solution algebraically.

4x + y = 8

2x – 3y = 18Equation 1

Equation 2

SOLUTION

Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, –4). You can check this algebraically as follows.

EXAMPLE 1 Solve a system graphically

Equation 1 Equation 2

4x + y = 8

4(3) + (–4) 8=?

=?12 –4 8

8 = 8

2x – 3y = 18

=?2(3) – 3( – 4) 18

=?6 + 12 18

18 = 18

The solution is (3, –4).

EXAMPLE 2 Solve a system with many solutions

Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.

4x – 3y = 88x – 6y = 16

Equation 1

Equation 2

SOLUTION

The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent.

EXAMPLE 3 Solve a system with no solution

Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent.

2x + y = 42x + y = 1

Equation 1

Equation 2

SOLUTION

The graphs of the equations are two parallel lines. Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent.

GUIDED PRACTICE for Examples 2,3, and 4

Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.

4. 2x + 5y = 64x + 10y = 12

ANSWER Infinitely many solutions; consistent and dependent



5. 3x – 2y = 103x – 2y = 2

ANSWER no solution; inconsistent



6. –2x + y = 5y = –x + 2

ANSWER (–1, 3); consistent and independent

ESSENTIAL QUESTION

How do you solve a system of linear equations

graphically?

Graph the equations. The point at which the graphs

meet is the solution. CHECK your solution!

SPLIT A SHEET OF GRAPH PAPER WITH A PARTNER.

THEN SOLVE THE SYSTEM BY GRAPHING:

X+Y = 22X + Y = 3

LESSON 2: SOLVE LINEAR SYSTEMS

ALGEBRAICALLY

ESSENTIAL QUESTION


equations algebraically?

VOCABULARY Substitution Method: A method of solving a

system of equations by solving one of the equations for one of the variables and then substituting the resulting expression in the other equation(s)

Elimination Method: A method of solving a system of equations by multiplying equations by constants, then adding the revised equations to eliminate a variable

EXAMPLE 1 Use the substitution method

Solve the system using the substitution method.

2x + 5y = –5x + 3y = 3

Equation 1Equation 2

SOLUTION

STEP 1 Solve Equation 2 for x.

x = –3y + 3 Revised Equation 2


STEP 2Substitute the expression for x into Equation 1 and solve for y.

2x +5y = –5

2(–3y + 3) + 5y = –5

y = 11

Write Equation 1.

Substitute –3y + 3 for x.

Solve for y.

STEP 3

Substitute the value of y into revised Equation 2 and solve for x.

x = –3y + 3

x = –3(11) + 3x = –30

Write revised Equation 2.

Substitute 11 for y.

Simplify.


CHECK Check the solution by substituting into the original equations.

2(–30) + 5(11) –5=? Substitute for x and y. =? –30 + 3(11) 3

Solution checks. 3 = 3 –5 = –5

The solution is (– 30, 11).

ANSWER

EXAMPLE 2 Use the elimination method

Solve the system using the elimination method.

3x – 7y = 106x – 8y = 8

Equation 1Equation 2

SOLUTION

Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign.

STEP 1

3x – 7y = 10

6x – 8y = 8

–6x + 14y = 220

6x – 8y = 8


STEP 2Add the revised equations and solve for y. 6y = –12

y = –2STEP 3

Substitute the value of y into one of the original equations. Solve for x.

3x – 7y = 10 3x – 7(–2) = 10

3x + 14 = 10

x = 43 – Solve for x.

Simplify.

Substitute –2 for y.

Write Equation 1.

1. 4x + 3y = –2x + 5y = –9

Solve the system using the substitution or the elimination method.

GUIDED PRACTICE for Examples 1 and 2

The solution is (1,–2).

ANSWER

2. 3x + 3y = –155x – 9y = 3

The solution is ( , –2)3–

ANSWER

Solve the system using the substitution or the elimination method.

GUIDED PRACTICE for Examples 1 and 2

3. 3x – 6y = 9 –4x + 7y = –16

The solution is (11, 4)

ANSWER

ESSENTIAL QUESTION

How do you solve a system of linear equations

algebraically?

By using the 1. Substition Method

2. Elimination Method

ON A HALF SHEET OF GRAPH PAPER (SHARE

WITH A PARTNER) GRAPH THE INEQUALITY:

Y≤X+2

LESSON 3: GRAPH SYSTEMS OF LINEAR

INEQUALITIES

ESSENTIAL QUESTION

How do you find the solution to a system of

linear inequalities?

VOCABULARY No new vocab!!

EXAMPLE 1 Graph a system of two inequalities

Graph the system of inequalities.

y > –2x – 5 Inequality 1

y < x + 3 Inequality 2

EXAMPLE 1 Graph a system of two inequalities

STEP 2 Identify the region that is common to both graphs. It is the region that is shaded purple.

SOLUTION

STEP 1 Graph each inequality in the system. Use red for y > –2x – 5 and blue for y ≤ x + 3.

EXAMPLE 2 Graph a system with no solution


2x + 3y < 6 Inequality 1

y < – x + 423 Inequality 2

EXAMPLE 2 Graph a system with no solution

STEP 2 Identify the region that is common to both graphs. There is no region shaded both red and blue. So, the system has no solution.

SOLUTION

STEP 123

Graph each inequality in the system. Use red for 2x + 3y < 6 and blue for y > – x + 4.

GUIDED PRACTICE for Examples 1, 2 and 3


1. y < 3x – 2y > – x + 4


2. 2x – y > 412

4x – y < 5

ESSENTIAL QUESTIONHow do you find the

solution to a system of linear inequalities?

The solution to a system of linear inequalities is found

by graphing each inequality. The solution is

the overlapping shaded region.

SOLVE THE SYSTEM USING SUBSTITUTION OR

ELIMINATION:

3X + 4Y = -253X – 2Y = -1

LESSON 4: SOLVE SYSTEMS OF LINEAR

EQUATIONS IN THREE VARIABLES

ESSENTIAL QUESTION


equations in three variables?

VOCABULARY Ordered Pair: A set of two numbers (x,y) that

represent a point in space

Ordered Triple: A set of three numbers of the form (x,y,z) that represent a point in space


Solve the system. 4x + 2y + 3z = 1 Equation 1

2x – 3y + 5z = –14 Equation 26x – y + 4z = –1 Equation 3

SOLUTION

STEP 1Rewrite the system as a linear system in two variables.

4x + 2y + 3z = 1

12x – 2y + 8z = –2

Add 2 times Equation 3

to Equation 1.

16x + 11z = –1 New Equation 1

EXAMPLE 1

2x – 3y + 5z = –14

–18x + 3y –12z = 3

Add – 3 times Equation 3to Equation 2.

–16x – 7z = –11 New Equation 2

STEP 2 Solve the new linear system for both of its variables.

16x + 11z = –1 Add new Equation 1

and new Equation 2. –16x – 7z = –11

4z = –12z = –3 Solve for z.x = 2 Substitute into new

Equation 1 or 2 to find x.

Use the elimination method

6x – y + 4z = –1


STEP 3Substitute x = 2 and z = – 3 into an original equation and solve for y.

Write original Equation 3.

6(2) – y + 4(–3) = –1 Substitute 2 for x and –3 for z.

y = 1 Solve for y.

EXAMPLE 2 Solve a three-variable system with no solution

Solve the system. x + y + z = 3 Equation 14x + 4y + 4z = 7 Equation 2

3x – y + 2z = 5 Equation 3

SOLUTION

When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation. Add – 4 times Equation 1

to Equation 2.

–4x – 4y – 4z = –12

4x + 4y + 4z = 7

0 = –5 New Equation 1

EXAMPLE 2 Solve a three-variable system with no solution

Because you obtain a false equation, you can conclude that the original system has no solution.

EXAMPLE 3 Solve a three-variable system with many solutions

Solve the system. x + y + z = 4 Equation 1x + y – z = 4 Equation 2

3x + 3y + z = 12 Equation 3

SOLUTIONSTEP 1 Rewrite the system as a linear

system in two variables.Add Equation 1

to Equation 2.

x + y + z = 4

x + y – z = 4

2x + 2y = 8 New Equation 1


x + y – z = 4 Add Equation 2

3x + 3y + z = 12 to Equation 3.

4x + 4y = 16 New Equation 2

Solve the new linear system for both of its variables.

STEP 2

Add –2 times new Equation 1

to new Equation 2.

Because you obtain the identity 0 = 0, the system has infinitely many solutions.

–4x – 4y = –16

4x + 4y = 16


STEP 3 Describe the solutions of the system. One way to do this is to divide new Equation 1 by 2 to get x + y = 4, or y = –x + 4. Substituting this into original Equation 1 produces z = 0. So, any ordered triple of the form (x, –x + 4, 0) is a solution of the system.


Solve the system.1. 3x + y – 2z = 10

6x – 2y + z = –2 x + 4y + 3z = 7

ANSWER (1, 3, –2)


2. x + y – z = 22x + 2y – 2z = 65x + y – 3z = 8

ANSWER no solution


3. x + y + z = 3x + y – z = 3

2x + 2y + z = 6

ANSWERInfinitely many solutions

ESSENTIAL QUESTION

How do you solve a system of linear equations in three

variables?

•Rewrite as a system of two variables by eliminating

one variable•Solve for each variable

•Substitute to find the third variable

Documents

Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab