3.1 solve linear systems by graphing note: definitely need graph paper for your notes today
TRANSCRIPT
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3.1 Solve Linear Systems by Graphing
Note: Definitely need graph paper for your notes today
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System of Two Linear Equations
• Also called a linear system
• Simply two linear equations
• Solution of the system: an ordered pair that satisfies both equations
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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 = 18
Equation 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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SOLUTION
GUIDED PRACTICE for Example 1
Graph the linear system and estimate the solution. Then check the solution algebraically.1. 3x + 2y = – 4
x + 3y = 13x + 2y = – 4x + 3y = 1
Equation 1Equation 2
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–2, 1). You can check this algebraically as follows.
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GUIDED PRACTICE for Example 1
Equation 1 Equation 23x + 2y = –4
=?–6 + 2 –4
x + 3y = 1
–2 + 3 1=?
1 = 1
The solution is (–2, 1).
=?3(–2) + 2(1) –4
–4 = –4
(–2 ) + 3( 1) 1=?
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SOLUTION
GUIDED PRACTICE for Example 1
Graph the linear system and estimate the solution. Then check the solution algebraically.
Equation 1Equation 2
2. 4x – 5y = – 102x – 7y = 4
2x – 7y = 44x – 5y = – 10
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–5, –2). You can check this algebraically as follows.
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GUIDED PRACTICE for Example 1
Equation 1 Equation 2 2x –7y = 4
–10 + 14 4=?
4 = 4
The solution is (–5, –2).
–10 = –10
4x – 5y = –10=?4(–5) – 5(–2 ) –10 2(–5) – 7(–2 ) 4=?
=?–20 + 10 –10
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SOLUTION
GUIDED PRACTICE for Example 1
Graph the linear system and estimate the solution. Then check the solution algebraically.
Equation 1Equation 2
3. 8x – y = 83x + 2y = – 16
8x – y = 83x + 2y = – 16
Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (0, –8). You can check this algebraically as follows.
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GUIDED PRACTICE for Example 1
Equation 1 Equation 2
=? 0 + 8 8 0 – 16 –16=?
–16 = –16
The solution is (0, –8).
8(0) – (–8 ) 8=?
8 = 8
8x – y = 8 3x + 2y = –163(0) + 2(–8) –16=?
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How to do it with a graphing calculator
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Classifying Systems page 154
• Consistent: at least one solution
• Inconsistent: no solution (lines never intersect)
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Types of Consistent Systems
• Independent- exactly one solution (graphs have one intersection point)
• Dependent- infinitely many solutions (two graphs are the same line)
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Keep in Mind…
• If it’s inconsistent- parallel lines, neither independent or dependent
• If it’s consistent- either one intersection point or two equations that represent the same line- classify as either independent or dependent
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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 = 8
8x – 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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If it is an inconsistent system (no solution)
• The lines are parallel– The lines have the same _______.
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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 = 4
2x + 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 4 – 6
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EXAMPLE 4 Standardized Test Practice
SOLUTION
Equation 1 (Option A)
y = 1 x + 30
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Equation 2 (Option B)
EXAMPLE 4 Standardized Test Practice
y = x2.5
To solve the system, graph the equations y = x + 30 and y = 2.5x, as shown at the right.
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EXAMPLE 4 Standardized Test Practice
Notice that you need to graph the equations only in the first quadrant because only nonnegative values of x and y make sense in this situation.
The lines appear to intersect at about the point (20, 50). You can check this algebraically as follows.
Equation 1 checks.
Equation 2 checks.
50 = 20 + 30
50 = 2.5(20)
ANSWERThe total costs are equal after 20 rides.
The correct answer is B.
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Extra Word Problem Example
• A soccer league offers two options for membership plans. Option A incluces an initial fee of $40 and costs $5 for each game played. Option B costs $10 for each game played. After how many games will the total cost of the two options be the same?
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Review Questions
• How do you solve a linear system by graphing?
• How can you tell just by looking at the system that the same graph is shown twice?
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Homework
• 1, 2, 6-11, 15, 16, 20 – 25, 28-31, 35 – 39