Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab
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Transcript of Complete #1-11 Odd on Page 150 under the Prerequisite Skills Tab
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COMPLETE #1-11 ODD ON PAGE 150 UNDER THE PREREQUISITE
SKILLS TAB
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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
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LESSON 1: SOLVE LINEAR SYSTEMS BY
GRAPHING
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ESSENTIAL QUESTION
How do you solve a system of linear
equations graphically?
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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
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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.
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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).
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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.
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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.
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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
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GUIDED PRACTICE for Examples 2,3, and 4
Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.
5. 3x – 2y = 103x – 2y = 2
ANSWER no solution; inconsistent
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GUIDED PRACTICE for Examples 2,3, and 4
Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.
6. –2x + y = 5y = –x + 2
ANSWER (–1, 3); consistent and independent
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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!
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SPLIT A SHEET OF GRAPH PAPER WITH A PARTNER.
THEN SOLVE THE SYSTEM BY GRAPHING:
X+Y = 22X + Y = 3
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LESSON 2: SOLVE LINEAR SYSTEMS
ALGEBRAICALLY
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ESSENTIAL QUESTION
How do you solve a system of linear
equations algebraically?
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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
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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
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EXAMPLE 1 Use the substitution method
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.
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EXAMPLE 1 Use the substitution method
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
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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
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EXAMPLE 2 Use the elimination method
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.
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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
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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
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ESSENTIAL QUESTION
How do you solve a system of linear equations
algebraically?
By using the 1. Substition Method
2. Elimination Method
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ON A HALF SHEET OF GRAPH PAPER (SHARE
WITH A PARTNER) GRAPH THE INEQUALITY:
Y≤X+2
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LESSON 3: GRAPH SYSTEMS OF LINEAR
INEQUALITIES
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ESSENTIAL QUESTION
How do you find the solution to a system of
linear inequalities?
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VOCABULARY No new vocab!!
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EXAMPLE 1 Graph a system of two inequalities
Graph the system of inequalities.
y > –2x – 5 Inequality 1
y < x + 3 Inequality 2
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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.
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EXAMPLE 2 Graph a system with no solution
Graph the system of inequalities.
2x + 3y < 6 Inequality 1
y < – x + 423 Inequality 2
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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.
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GUIDED PRACTICE for Examples 1, 2 and 3
Graph the system of inequalities.
1. y < 3x – 2y > – x + 4
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GUIDED PRACTICE for Examples 1, 2 and 3
2. 2x – y > 412
4x – y < 5
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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.
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SOLVE THE SYSTEM USING SUBSTITUTION OR
ELIMINATION:
3X + 4Y = -253X – 2Y = -1
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LESSON 4: SOLVE SYSTEMS OF LINEAR
EQUATIONS IN THREE VARIABLES
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ESSENTIAL QUESTION
How do you solve a system of linear
equations in three variables?
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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
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EXAMPLE 1 Use the elimination method
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
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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
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6x – y + 4z = –1
EXAMPLE 1 Use the elimination method
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.
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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
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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.
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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
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EXAMPLE 3 Solve a three-variable system with many solutions
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
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EXAMPLE 3 Solve a three-variable system with many solutions
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.
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GUIDED PRACTICE for Examples 1, 2 and 3
Solve the system.1. 3x + y – 2z = 10
6x – 2y + z = –2 x + 4y + 3z = 7
ANSWER (1, 3, –2)
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GUIDED PRACTICE for Examples 1, 2 and 3
2. x + y – z = 22x + 2y – 2z = 65x + y – 3z = 8
ANSWER no solution
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GUIDED PRACTICE for Examples 1, 2 and 3
3. x + y + z = 3x + y – z = 3
2x + 2y + z = 6
ANSWERInfinitely many solutions
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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