graphing absolute value equations unit 7 section 1...
TRANSCRIPT
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Graphing Absolute Value
Equations
Unit 7 Section 1 How do I make one of those V
graphs?
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Absolute value
transformation
Vocabulary: • Translation: shifts the parent function graph
horizontally and vertically.
• Reflection: it creates a mirror image of the
parent-function graph across the line of
reflection.
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Parent Function y = │x
│
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Vertical Movements?
• Absolute value rules:
– If you have minus outside the absolute value,
you move down on the coordinate plane.
y =│x│- 2
– If you have addition outside the absolute value,
you move up on the coordinate plane.
y =│x│+ 2
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Lets practice! • y = │x│ - 4
• y = │x│+ 3
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Your turn!
• y = │x│ + 7
• y = │x│ - 2
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Horizontal Movements?
• Absolute value special rules:
– If you have minus inside the absolute value,
you move to the right on the coordinate plane.
y =│x - 2 │
– If you have addition inside the absolute value,
you move to the left on the coordinate plane.
y =│x + 2 │
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Lets Practice!
• y =│x - 5│
• y =│x + 6│
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Your Turn!
• y =│x + 4│
• y =│x - 1│
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What about y = │x - 4 │+ 5
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What about y = -│x - 2│
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y = -│x │
Same rule, but now it makes it look down
Reflection of the
Parent-function
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Practice Time!!!
1) y = │x - 1│
2) y = │x +3│
3) y = │x │ + 5
4) y = │x│ - 4
5) y = │x + 3│ + 3
6) y = │x - 1│+ 2
7) y = -│x - 6│
8) y = -│x │- 4
9) │x│+ y = - 3
10) y = -│x + 1│ - 1
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Review
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Review
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Review
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Verbal Comprehension
Analysis • Graph the functions y = │x │ and y = │x -
3 │ together on the same coordinate plane.
• What effect does the 3 have on the graph?
• Graph the functions y = │x │ and y = │x
│- 3 on the same coordinate plane.
• What effect does the 3 have on the graph?
• Graph the functions y = │x │ and y = - │x
│ together on a coordinate plane.
• What effect does the negative sign have on
the graph?
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Practice!!
7-1 Pgs. 327-328 #1-3 all,
#4-34 even
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Solving Systems by Graphing
Section 7-2
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Vocabulary
• System of Linear Equations
• Solutions of a System of Linear Equations
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Definition
• System of Linear Equations - a set of two or more linear
equations containing two or more variables.
– Example:
• Solution of a System of Linear Equations – is an ordered
pair that satisfies each equation in the system. So, if an
ordered pair is a solution, it will make both equations true.
3
2 1
y x
y x
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A system of linear equations is a grouping
of two or more linear equations where each
equation contains one or more variables.
A solution of a system of equations consists of values for the variables that satisfy each equation of the system. When we are solving a system of two linear equations containing two unknowns, we represent the solution as an ordered pair (x, y), a point.
y = – 4x – 6
y = 2x
5x + 3y = – 8
x – 4y = 7
A brace is used to remind us that we are dealing
with a system of equations.
Solutions
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Tell whether the ordered pair is a solution of the given system.
(5, 2);
The ordered pair (5, 2) makes both equations true.
(5, 2) is the solution of the system.
3x – y = 13
2 – 2 0
0 0
0 3(5) – 2 13
15 – 2 13
13 13
3x – y = 13
Substitute 5 for
x and 2 for
y.
Example: Identifying
Solutions to a System
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If an ordered pair does not satisfy the first equation in
the system, there is no reason to check the other
equations.
Helpful Hint
Solutions
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Tell whether the ordered pair is a solution of the given system.
(–2, 2);
–x + y = 2
x + 3y = 4
Substitute –2
for x and 2
for y.
x + 3y = 4
–2 + 6 4
4 4
4 –2 + (3)2
–x + y = 2
4 2
2 –(–2) + 2
The ordered pair (–2, 2) makes one equation true, but
not the other. (–2, 2) is not a solution of the system.
Example: Identifying
Solutions to a System
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Tell whether the ordered pair is a solution of the given system.
(1, 3); 2x + y = 5
–2x + y = 1
2x + y = 5
2(1) + 3 5
2 + 3 5
5 5
The ordered pair (1, 3) makes both equations true.
Substitute 1 for x
and 3 for y.
–2x + y = 1
–2(1) + 3 1
–2 + 3 1
1 1
(1, 3) is the solution of the system.
Your Turn:
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Tell whether the ordered pair is a solution of the given system.
(2, –1); x – 2y = 4
3x + y = 6
The ordered pair (2, –1) makes one equation true, but not the other.
(2, –1) is not a solution of the system.
3x + y = 6
3(2) + (–1) 6
6 – 1 6
5 6
x – 2y = 4
2 – 2(–1) 4
2 + 2 4
4 4
Substitute 2 for x
and –1 for y.
Your Turn:
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Wren and Jenni are reading the same book. Wren is on
page 14 and reads 2 pages every night. Jenni is on page 6
and reads 3 pages every night. After how many nights
will they have read the same number of pages? How
many pages will that be?
Example: Writing a
System of Equations
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Understand the Problem
The answer will be the number of nights it takes for the
number of pages read to be the same for both girls.
List the important information:
Wren on page 14 Reads 2 pages a night
Jenni on page 6 Reads 3 pages a night
Example: Continued
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2 Write a System of Equations
Write a system of equations, one equation to represent the
number of pages read by each girl. Let x be the number of
nights and y be the total pages read.
Total
pages is
number
read every
night plus already
read.
Wren y = 2 x + 14
Jenni y = 3 x + 6
Example: Continued
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Solve 3
(8, 30)
Nights
Graph y = 2x + 14 and y = 3x + 6. The lines appear to intersect at
(8, 30). So, the number of pages read will be the same at 8 nights
with a total of 30 pages.
Example: Continued
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Verify the Solution 4
Check (8, 30) using both equations.
After 8 nights, Wren will have read 30 pages:
After 8 nights, Jenni will have read 30 pages:
3(8) + 6 = 24 + 6 = 30
2(8) + 14 = 16 + 14 = 30
Example: Continued
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Video club A charges $10 for membership and $3
per movie rental. Video club B charges $15 for
membership and $2 per movie rental. For how
many movie rentals will the cost be the same at
both video clubs? What is that cost?
Your Turn:
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1 Understand the Problem
The answer will be the number of movies rented for
which the cost will be the same at both clubs.
List the important information:
• Rental price: Club A $3 Club B $2
• Membership: Club A $10 Club B $15
Continued
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2 Write a System of Equations
Write a system of equations, one equation to represent the
cost of Club A and one for Club B. Let x be the number of
movies rented and y the total cost.
Total
cost is price rentals plus
membership
fee. times
Club A y = 3 + 10
Club B y = 2 + 15
x
x
Continued
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Solve 3
Graph y = 3x + 10 and y = 2x + 15. The lines appear to
intersect at (5, 25). So, the cost will be the same for 5 rentals
and the total cost will be $25.
Continued
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Verify the Solution 4
Check (5, 25) using both equations.
Number of movie rentals for Club A to reach $25:
Number of movie rentals for Club B to reach $25:
2(5) + 15 = 10 + 15 = 25
3(5) + 10 = 15 + 10 = 25
Continued
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All solutions of a linear equation are on its graph. To
find a solution of a system of linear equations, you
need a point that each line has in common. In other
words, you need their point of intersection.
y = 2x – 1
y = –x + 5
The point (2, 3) is where the two
lines intersect and is a solution of
both equations, so (2, 3) is the
solution of the systems.
System Solution on a Graph
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Steps for Obtaining the Solution of a System of
Linear Equations by Graphing
Step 1: Graph the first equation in the system.
Step 2: Graph the second equation in the system.
Step 3: Determine the point of intersection, if any.
Step 4: Verify that the point of intersection
determined in Step 3 is a solution of the
system. Remember to check the point in both
equations.
Solving a System by
Graphing
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Solve the system by graphing. Check your answer. y = x
y = –2x – 3 Graph the system.
The solution appears to be at
(–1, –1).
The solution is (–1, –1).
Check
Substitute (–1, –1) into the
system. y = x
y = –2x – 3
•
y = x
(–1) (–1)
–1 –1
y = –2x – 3
(–1) –2(–1) –3
–1 2 – 3
–1 – 1
Example: System solution by
Graphing
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Solve the system by graphing. Check your answer.
Additional Example 2B: Solving a System Equations by Graphing
y = x – 6
y = 1/3x + 1
y = x – 6
Graph the system.
y + 1
3 x =– 1
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Additional Example 2B Continued
y = x – 6
y + x = –1
+ – 1
–1
–1
–1 – 1
y = x – 6
– 6
The solution is
Check Substitute into the system
Solve the system by graphing. Check your answer.
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Solve the system by graphing. Check your answer.
Check It Out! Example 2a
y = –2x – 1
y = x + 5 Graph the system.
The solution appears to be (–2, 3).
Check Substitute (–2, 3) into the system.
y = x + 5
3 –2 + 5
3 3
y = –2x – 1
3 –2(–2) – 1
3 4 – 1
3 3 The solution is (–2, 3).
y = x + 5
y = –2x – 1
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Solve the system by graphing. Check your answer.
Check It Out! Example 2b
2x + y = 4
Graph the system.
Rewrite the second equation in
slope-intercept form.
2x + y = 4
–2x – 2x
y = –2x + 4
The solution appears to be (3, –2).
y = –2x + 4
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Solve the system by graphing. Check your answer.
Check It Out! Example 2b Continued
2x + y = 4 Check Substitute (3, –2) into the system.
2x + y = 4
2(3) + (–2) 4
6 – 2 4
4 4
–2 (3) – 3
–2 1 – 3
–2 –2
The solution is (3, –2).
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• There are three possible outcomes or
solutions when graphing two linear
equations in a plane.
• One point of intersection, so one solution.
• Parallel lines, so no solution.
• Same lines, so infinite # of solutions.
System Possible Solutions
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IDENTIFYING THE NUMBER OF SOLUTIONS
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
x
Lines intersect
one solution
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NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
x
Lines are parallel
no solution
IDENTIFYING THE NUMBER OF SOLUTIONS
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NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
x
Lines coincide
infinitely many
solutions
IDENTIFYING THE NUMBER OF SOLUTIONS
(the coordinates
of every point on
the line)
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NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
IDENTIFYING THE NUMBER OF SOLUTIONS
CONCEPT
SUMMARY
y
x
y
x
Lines intersect one solution
Lines are parallel no solution
y
x
Lines coincide infinitely many solutions
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Example: A Linear System with No Solution
Show that this linear system
has no solution.
Rewrite each equation
in slope-intercept form.
5 4 3 2 1 0 1 2 3 4 5 1
1
2
3
4
5
6
y –2 x 5 Revised Equation 1
Graph the linear system.
y –2 x 1 Revised Equation 2
2 x y 5 Equation 1
2 x y 1 Equation 2
The lines are parallel; they
have the same slope but
different y-intercepts. Parallel
lines never intersect, so the
system has no solution.
y 2x 1
y 2x 5
METHOD 1: GRAPHING
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Example: A Linear System with Infinite Solutions
Show that this linear system
has infinitely many solutions.
– 2 x y 3 Equation 1
– 4 x 2y 6 Equation 2
y 2 x 3 Revised Equation 1
y 2 x 3 Revised Equation 2
Rewrite each equation
in slope-intercept form.
Graph the linear system.
From these graphs you can see that the equations represent the same line. Any point on the line is a solution.
5 4 3 2 1 0 1 2 3 4 5 1
1
2
3
4
5
6
METHOD 1: GRAPHING
– 4x 2y 6
–2x y 3
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Assignment
• 7-2 Pg. 343-344 #1-23 even, 29-33 all
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Solving Systems Using
Substitution
Section 7-3
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Sometimes it is difficult to identify the exact solution
to a system by graphing. In that case, you can use an
algebraic method called substitution to solve a
system.
Substitution is used to reduce the system to one
equation that has only one variable. Then you can
solve this equation by the methods taught in Chapter 2.
Substitution Method
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Substitution Method
• Sometimes it is difficult to identify the exact
solution to a system by graphing.
• In this case, you can use a method called
substitution.
• Substitution is used to reduce the system to one
equation that has only one variable.
• Then you can solve this equation for the one
variable and substitute again to find the other
variable.
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Solving Systems of Equations by Substitution
Step 2
Step 3
Step 4
Step 5
Step 1 Solve for one variable in at least one equation, if
necessary.
Substitute the resulting expression into the other
equation.
Solve that equation to get the value of the first
variable.
Substitute that value into one of the original equations
and solve for the other variable.
Write the values from steps 3 and 4 as an ordered pair,
(x, y), and check.
Substitution Method
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Solve the system by substitution.
y = 3x
y = x – 2
Step 1 y = 3x
y = x – 2
Both equations are solved for y.
Step 2 y = x – 2
3x = x – 2
Substitute 3x for y in the second
equation.
Now solve this equation for x. Subtract x from both sides and then divide by 2.
Step 3 –x –x
2x = –2 2x = –2
2 2
x = –1
Example: Substitution
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Solve the system by substitution.
Step 4 y = 3x Write one of the original
equations.
Substitute –1 for x. y = 3(–1)
y = –3
Step 5 (–1, –3)
Check Substitute (–1, –3) into both equations in the system.
Write the solution as an ordered pair.
y = 3x
–3 3(–1)
–3 –3
y = x – 2
–3 –1 – 2
–3 –3
Example: Continued
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You can substitute the value of one variable into
either of the original equations to find the value of
the other variable.
Helpful Hint
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Solve the system by substitution.
y = x + 1
4x + y = 6
Step 1 y = x + 1 The first equation is solved for y.
Step 2 4x + y = 6
4x + (x + 1) = 6 Substitute x + 1 for y in the second
equation.
Subtract 1 from both sides.
5x = 5
5 5
x = 1
Step 3 –1 –1
5x = 5
5x + 1 = 6 Simplify. Solve for x.
Divide both sides by 5.
Write the second equation.
Example: Substitution
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Solve the system by substitution.
Step 4 y = x + 1 Write one of the original
equations.
Substitute 1 for x. y = 1 + 1
y = 2
Step 5 (1, 2) Write the solution as an ordered pair.
Example: Continued
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Solve the system by substitution.
x + 2y = –1
x – y = 5
Step 1 x + 2y = –1 Solve the first equation for x by
subtracting 2y from both sides.
Step 2 x – y = 5
(–2y – 1) – y = 5
Substitute –2y – 1 for x in the second
equation.
–3y – 1 = 5 Simplify.
−2y −2y
x = –2y – 1
Example: Substitution
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Step 3 –3y – 1 = 5
Add 1 to both sides. +1 +1
–3y = 6
–3y = 6
–3 –3
y = –2
Solve for y.
Divide both sides by –3.
Step 4 x – y = 5
x – (–2) = 5
x + 2 = 5 –2 –2
x = 3
Step 5 (3, –2)
Write one of the original equations.
Substitute –2 for y.
Subtract 2 from both sides.
Write the solution as an ordered pair.
Example: Continued
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Solve the system by substitution. Check your answer.
y = x + 3
y = 2x + 5
Both equations are solved for y. Step 1 y = x + 3
y = 2x + 5
Substitute 2x + 5 for y in the first
equation.
Solve for x. Subtract x and 5 from both
sides. –x – 5 –x – 5
x = –2
Step 3 2x + 5 = x + 3
Step 2
2x + 5 = x + 3
y = x + 3
Your Turn:
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Solve the system by substitution. Check your answer.
Step 4 y = x + 3 Write one of the original
equations.
Substitute –2 for x. y = –2 + 3
y = 1
Step 5 (–2, 1) Write the solution as an ordered pair.
Check Substitute (–2, 1) into both equations in the system.
y = x + 3
1 (–2) + 3
1 1
y = 2x + 5
1 2(–2) + 5 1 –4 + 5
1 1
Continued
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Solve the system by substitution.
x = 2y – 4
x + 8y = 16
The first equation is solved for x. Step 1 x = 2y – 4
Substitute 2y – 4 for x in the second
equation.
Simplify. Then solve for y.
(2y – 4) + 8y = 16
x + 8y = 16 Step 2
Step 3 10y – 4 = 16 Add 4 to both sides. +4 +4
10y = 20
y = 2
10y 20
10 10 = Divide both sides by 10.
Your Turn:
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Step 4 x + 8y = 16 Write one of the original equations.
Substitute 2 for y. x + 8(2) = 16
x + 16 = 16
x = 0
– 16 –16
Simplify.
Subtract 16 from both sides.
Step 5 (0, 2) Write the solution as an
ordered pair.
Continued
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Solve the system by substitution.
2x + y = –4
x + y = –7
Solve the second equation for x by
subtracting y from each side.
Substitute –y – 7 for x in the first
equation.
Distribute 2.
2(–y – 7) + y = –4
x = –y – 7 Step 2
Step 1 x + y = –7 – y – y
x = –y – 7
2(–y – 7) + y = –4
–2y – 14 + y = –4
Your Turn:
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Combine like terms. Step 3
+14 +14
–y = 10
–2y – 14 + y = –4
Add 14 to each side.
–y – 14 = –4
y = –10
Step 4 x + y = –7 Write one of the original equations.
Substitute –10 for y. x + (–10) = –7
x – 10 = – 7
Continued
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x – 10 = –7 Step 5
+10 +10
x = 3
Add 10 to both sides.
Step 6 (3, –10) Write the solution as an
ordered pair.
Continued
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When you solve one equation for a variable, you
must substitute the value or expression into the other
original equation, not the one that had just been
solved. You have to use both equations.
Caution
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y + 6x = 11
3x + 2y = –5 Solve by substitution.
Solve the first equation for y by
subtracting 6x from each side. Step 1 y + 6x = 11 – 6x – 6x
y = –6x + 11
Substitute –6x + 11 for y in the
second equation.
Distribute 2 to the expression in
parentheses.
3x + 2(–6x + 11) = –5
3x + 2y = –5 Step 2
3x + 2(–6x + 11) = –5
Example: Substitution Involving
Distribution
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Step 3 3x + 2(–6x) + 2(11) = –5
–9x + 22 = –5
Simplify. Solve for x.
Subtract 22 from both
sides. –9x = –27
– 22 –22
Divide both sides by –
9. –9x = –27
–9 –9
x = 3
3x – 12x + 22 = –5
3x + 2(–6x + 11) = –5
Continued
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Step 4 y + 6x = 11
Substitute 3 for x. y + 6(3) = 11
Subtract 18 from each side.
y + 18 = 11
–18 –18
y = –7
Step 5 (3, –7) Write the solution as an ordered
pair.
Simplify.
Write one of the original equations.
Continued
x = 3
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–2x + y = 8
3x + 2y = 9 Solve by substitution.
Solve the first equation for y by
adding 2x to each side.
Step 1 –2x + y = 8 + 2x +2x
y = 2x + 8
Substitute 2x + 8 for y in the second
equation. 3x + 2(2x + 8) = 9
3x + 2y = 9 Step 2
Distribute 2 to the expression in
parentheses. 3x + 2(2x + 8) = 9
Your Turn:
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Step 3 3x + 2(2x) + 2(8) = 9
7x + 16 = 9
Simplify. Solve for x.
Subtract 16 from both
sides. 7x = –7
–16 –16
Divide both sides by 7. 7x = –7
7 7
x = –1
3x + 4x + 16 = 9
Continued
3x + 2(2x + 8) = 9
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Step 4 –2x + y = 8
Substitute –1 for x. –2(–1) + y = 8
y + 2 = 8
–2 –2
y = 6
Step 5 (–1, 6) Write the solution as an
ordered pair.
Subtract 2 from each side.
Simplify.
Write one of the original
equations.
Continued x = –1
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Example: A Linear System with No Solution
Show that this linear system has no solution.
2 x y 5 Equation 1
2 x y 1 Equation 2
Because Equation 2 can be revised
to y –2 x 1, you can substitute –2 x 1 for y in Equation 1.
2 x y 5
2 x (–2 x 1) 5 Substitute –2 x 1 for y.
1 5 Simplify. False statement!
Once the variables are eliminated, the statement is not true regardless
of the values of x and y. This tells you the system has no solution.
Write Equation 1.
METHOD: SUBSTITUTION
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Substitute 2 x + 3 for y in Equation 2.
Show that this linear system
has infinitely many solutions.
– 2 x y 3 Equation 1
– 4 x 2y 6 Equation 2 METHOD: SUBSTITUTION
You can solve Equation 1 for y.
– 4 x 2( 2 x + 3) 6
– 4 x + 4x 6 6 Simplify.
6 6 True statement!
The variables are eliminated and you are left with a statement that is true
regardless of the values of x and y. This result tells you that the linear system
has infinitely many solutions.
Example: A Linear System with Many Solutions
y 2 x + 3
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Your Turn:
Solve the system.
1)
2)
3 10
9 3 30
x y
y x
4 9
4 6
y x
y x
30 = 30 (true statement)
Infinite Solutions
6 = -9 (false statement)
No Solution
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Assignment
• 7-3 Pg. 349-351 #1 & 2, 6-32 even, #45
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Solving Systems Using
Elimination
Section 7-4
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Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together.
Remember that an equation stays balanced if you add equal amounts to both sides. So, if 5x + 2y = 1, you can add 5x + 2y to one side of an equation and 1 to the other side and the balance is maintained.
Elimination Method
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Since –2y and 2y have opposite coefficients, the y-term is eliminated. The result is one equation that has only one variable: 6x = –18.
When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients.
Elimination Method
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Solving Systems of
Equations • So far, we have solved systems using
graphing and substitution. These notes show
how to solve the system algebraically using
ELIMINATION with addition and
subtraction.
• Elimination is easiest when the equations are
in standard form.
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Step 1: Put the equations in Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Add or subtract the
equations.
Step 4: Plug back in to find the
other variable.
Step 5: Check your solution.
Standard Form: Ax + By = C
Look for variables that have the
same coefficient.
Solve for the variable.
Substitute the value of the variable
into the equation.
Substitute your ordered pair into
BOTH equations.
Elimination Procedure
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x + y = 5
3x – y = 7
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
They already are!
The y’s have the same
coefficient.
Step 3: Add or subtract the
equations.
Add to eliminate y.
x + y = 5
(+) 3x – y = 7
4x = 12
x = 3
Example: Elimination by
Addition
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Step 4: Plug back in to find the
other variable.
x + y = 5
(3) + y = 5
y = 2
Step 5: Check your solution.
(3, 2)
(3) + (2) = 5
3(3) - (2) = 7
The solution is (3, 2). What do you think the answer would
be if you solved using substitution?
x + y = 5
3x – y = 7
Example: Continued
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y + 3x = –2
2y – 3x = 14 Solve by elimination.
Write the system so that like
terms are aligned. 2y – 3x = 14
y + 3x = –2
Add the equations to
eliminate the x-terms. 3y + 0 = 12
3y = 12 Simplify and solve for y.
Divide both sides by 3.
y = 4
Your Turn:
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y + 3x = –2 Write one of the original
equations.
4 + 3x = –2 Substitute 4 for y. Subtract 4 from both sides. –4 –4
3x = –6
Divide both sides by 3. 3x = –6 3 3
x = –2 Write the solution as an
ordered pair. (–2, 4)
Continued
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3x – 4y = 10
x + 4y = –2 Solve by elimination.
3x – 4y = 10 Write the system so that like
terms are aligned. Add the equations to
eliminate the y-terms.
4x = 8 Simplify and solve for x.
x + 4y = –2
4x + 0 = 8
Divide both sides by 4. 4x = 8
4 4
x = 2
Your Turn:
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x + 4y = –2 Write one of the original
equations.
2 + 4y = –2 Substitute 2 for x.
–2 –2
4y = –4
4y –4
4 4 y = –1
(2, –1)
Subtract 2 from both sides.
Divide both sides by 4.
Write the solution as an ordered pair.
Continued
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More Elimination
• When two equations each contain the same
term, you can subtract one equation from
the other to solve the system.
• To subtract an equation add the opposite of
each term.
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4x + y = 7
4x – 2y = -2
Step 1: Put the equations in
Standard Form. They already are!
Step 2: Determine which
variable to eliminate.
The x’s have the same
coefficient.
Step 3: Add or subtract the
equations.
Subtract to eliminate x.
4x + y = 7
(-) 4x – 2y = -2
3y = 9
y = 3
Remember to
“keep-change-
change”
Example: Elimination by
Subtracting
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Step 4: Plug back in to find the
other variable.
4x + y = 7
4x + (3) = 7
4x = 4
x = 1
Step 5: Check your solution.
(1, 3)
4(1) + (3) = 7
4(1) - 2(3) = -2
4x + y = 7
4x – 2y = -2
Example: Continued
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3x + 3y = 15
–2x + 3y = –5 Solve by elimination.
3x + 3y = 15
–(–2x + 3y = –5)
3x + 3y = 15 + 2x – 3y = +5
Add the opposite of each
term in the second
equation.
Eliminate the y term.
Simplify and solve for x.
5x + 0 = 20
5x = 20
x = 4
Your Turn:
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Write one of the original
equations. Substitute 4 for x.
Subtract 12 from both sides.
3x + 3y = 15
3(4) + 3y = 15
12 + 3y = 15
–12 –12 3y = 3
y = 1
Simplify and solve for y.
Write the solution as an
ordered pair.
(4, 1)
Continued
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2x + y = –5
2x – 5y = 13 Solve by elimination.
Add the opposite of each
term in the second
equation.
–(2x – 5y = 13)
2x + y = –5
2x + y = –5
–2x + 5y = –13
Eliminate the x term.
Simplify and solve for y.
0 + 6y = –18
6y = –18
y = –3
Your Turn:
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Write one of the original
equations. 2x + y = –5
2x + (–3) = –5 Substitute –3 for y.
Add 3 to both sides.
2x – 3 = –5
+3 +3
2x = –2 Simplify and solve for x.
x = –1
Write the solution as an
ordered pair.
(–1, –3)
Continued
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Assignment
• 7-4 Part 1 Pg. 356-358: #2-8 even
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Solving Systems Using
Elimination
Section 7-4 Part 2
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Further Elimination
• In Part 1 of this lesson, you was that to
eliminate a variable, its coefficients must
have a sum or difference of zero.
• In some cases, you will first need to
multiply one or both of the equations by a
number so that one variable has opposite
coefficients, so you can add or subtract to
eliminate the variable.
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Step 1: Put the equations in Standard Form.
Step 2: Determine which
variable to eliminate.
Step 3: Multiply the equations
and solve.
Step 4: Plug back in to find the
other variable.
Step 5: Check your solution.
Standard Form: Ax + By = C
Look for variables that have the
same coefficient.
Solve for the variable.
Substitute the value of the variable
into the equation.
Substitute your ordered pair into
BOTH equations.
Further Elimination Procedure
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2x + 2y = 6
3x – y = 5
Step 1: Put the equations in
Standard Form.
Step 2: Determine which
variable to eliminate.
They already are!
None of the coefficients are the
same!
Find the least common multiple
of each variable.
LCM = 6x, LCM = 2y
Which is easier to obtain?
2y
(you only have to multiply
the bottom equation by 2)
Example: Multiplying One
Equation
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Step 4: Plug back in to find the
other variable.
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y = 1
2x + 2y = 6
3x – y = 5
Step 3: Multiply the equations
and solve.
Multiply the bottom equation by 2
2x + 2y = 6
(2)(3x – y = 5)
8x = 16
x = 2
2x + 2y = 6
(+) 6x – 2y = 10
Example: Continued
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Step 5: Check your solution.
(2, 1)
2(2) + 2(1) = 6
3(2) - (1) = 5
2x + 2y = 6
3x – y = 5
Solving with multiplication adds one more step
to the elimination process.
Example: Continued
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x + 4y = 7
4x – 3y = 9
Step 1: Put the equations in
Standard Form. They already are!
Step 2: Determine which
variable to eliminate.
Find the least common multiple
of each variable.
LCM = 4x, LCM = 12y
Which is easier to obtain?
4x
(you only have to multiply
the top equation by -4 to make
them inverses)
Example: Multiplying One
Equation
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x + 4y = 7
4x – 3y = 9
Step 4: Plug back in to find the
other variable.
x + 4(1) = 7
x + 4 = 7
x = 3
Step 3: Multiply the equations
and solve.
Multiply the top equation by -4
(-4)(x + 4y = 7)
4x – 3y = 9)
y = 1
-4x – 16y = -28
(+) 4x – 3y = 9
-19y = -19
Example: Continued
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Step 5: Check your solution.
(3, 1)
(3) + 4(1) = 7
4(3) - 3(1) = 9
x + 4y = 7
4x – 3y = 9
Example: Continued
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x + 2y = 11
–3x + y = –5
Solve the system by elimination.
Multiply each term in the
second equation by –2 to
get opposite y-coefficients.
x + 2y = 11
–2(–3x + y = –5)
x + 2y = 11
+(6x –2y = +10) Add the new equation to
the first equation. 7x + 0 = 21
7x = 21
x = 3
Simplify and solve for x.
Your Turn:
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Write one of the original
equations. x + 2y = 11
Substitute 3 for x. 3 + 2y = 11
Subtract 3 from each side. –3 –3
2y = 8
y = 4
Simplify and solve for y.
Write the solution as an
ordered pair.
(3, 4)
Continued
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Solve the system by elimination.
3x + 2y = 6
–x + y = –2
3x + 2y = 6
3(–x + y = –2)
3x + 2y = 6
+(–3x + 3y = –6)
0 + 5y = 0
Multiply each term in the second
equation by 3 to get opposite
x-coefficients.
Add the new equation to
the first equation.
Simplify and solve for y. 5y = 0
y = 0
Your Turn:
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Write one of the original
equations. –x + y = –2
Substitute 0 for y. –x + 3(0) = –2
–x + 0 = –2
–x = –2
Simplify and solve for x.
Write the solution as an
ordered pair. (2, 0)
x = 2
Continued
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3x + 4y = -1
4x – 3y = 7
Step 1: Put the equations in
Standard Form. They already are!
Step 2: Determine which
variable to eliminate.
Find the least common multiple
of each variable.
LCM = 12x, LCM = 12y
Which is easier to obtain?
Either! I’ll pick y because the signs
are already opposite.
Example: Multiplying Both
Equations
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3x + 4y = -1
4x – 3y = 7
Step 4: Plug back in to find the
other variable.
3(1) + 4y = -1
3 + 4y = -1
4y = -4
y = -1
Step 3: Multiply the equations
and solve.
Multiply both equations
(3)(3x + 4y = -1)
(4)(4x – 3y = 7)
x = 1
9x + 12y = -3
(+) 16x – 12y = 28
25x = 25
Example: Continued
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Step 5: Check your solution.
(1, -1)
3(1) + 4(-1) = -1
4(1) - 3(-1) = 7
3x + 4y = -1
4x – 3y = 7
Example: Continued
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–5x + 2y = 32
2x + 3y = 10
Solve the system by elimination.
2(–5x + 2y = 32)
5(2x + 3y = 10)
Multiply the first equation
by 2 and the second
equation by 5 to get
opposite x-coefficients –10x + 4y = 64
+(10x + 15y = 50) Add the new equations.
Simplify and solve for y. 19y = 114
y = 6
Your Turn:
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Write one of the original
equations. 2x + 3y = 10
Substitute 6 for y. 2x + 3(6) = 10
Subtract 18 from both sides. –18 –18
2x = –8
2x + 18 = 10
x = –4 Simplify and solve for x.
Write the solution as an
ordered pair.
(–4, 6)
Continued
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Solve the system by elimination.
2x + 5y = 26
–3x – 4y = –25
3(2x + 5y = 26)
+(2)(–3x – 4y = –25)
Multiply the first equation
by 3 and the second
equation by 2 to get
opposite x-coefficients 6x + 15y = 78
+(–6x – 8y = –50) Add the new equations.
Simplify and solve for y. y = 4
0 + 7y = 28
Your Turn:
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Write one of the original
equations. 2x + 5y = 26
Substitute 4 for y. 2x + 5(4) = 26
Simplify and solve for x.
Write the solution as an
ordered pair. (3, 4)
x = 3
2x + 20 = 26 –20 –20
2x = 6
Subtract 20 from both
sides.
Your Turn:
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NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
IDENTIFYING THE NUMBER OF SOLUTIONS
CONCEPT
SUMMARY
y
x
y
x
Lines intersect one solution
Lines are parallel no solution
y
x
Lines coincide infinitely many solutions
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Identifying The Number of
Solutions
• If both variable terms are eliminated as you
solve a system of equations, the answer is
either no solution or infinite solutions.
– No solution: get a false statement when solving
the system.
– Infinite solutions: get a true statement when
solving the system.
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Show that this linear system
has infinitely many solutions.
– 2 x y 3 Equation 1
– 4 x 2y 6 Equation 2 METHOD: Elimination
You can multiply Equation 1 by –2.
4x – 2y – 6 Multiply Equation 1 by –2.
– 4 x 2y 6 Write Equation 2.
0 0 Add Equations. True statement!
The variables are eliminated and you are left with a statement that is true
regardless of the values of x and y. This result tells you that the linear system
has infinitely many solutions.
A Linear System with Infinite Solutions
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Show that this linear system
has no solution.
2 x y 5 Equation 1
2 x y 1 Equation 2
You can multiply Equation 1 by –1.
-2 x - y -5 Multiply Equation 1 by –1.
2 x y 1 Write Equation 2.
0 - 4 Add Equations. False statement!
The variables are eliminated and you are left with a statement that is false
regardless of the values of x and y. This result tells you that the linear system
has no solution.
A Linear System with No Solution
METHOD: Elimination
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Your Turn:
Solve the systems using elimination.
1)
2)
2 5 7
2 12
5 5
x y
y x
12 8 20
3 2 5
x y
x y
False Statement
No Solution
True Statement
Infinite Solutions
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Summary
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Summary of Methods for Solving Systems
7.3 The Elimination Method
Substitution
The value of one variable is
known and can easily be
substituted into the other
equation.
6x + y = 10
y = 5 Example
Suggested
Method
Why
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Summary of Methods for Solving Systems
7.3 The Elimination Method
Elimination
eliminate ‘y’ 5
Add the two equations
2x – 5y = –20
4x + 5y = 14 Example
Suggested
Method
Why
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Summary of Methods for Solving Systems
7.3 The Elimination Method
Elimination
9a – 2b = –11
8a + 4b = 25 Example
Suggested
Method
Why eliminate ‘b’ 4
Multiply first equation by 2
Add the equations
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Assignment
• 7-4 Part 2 Pg. 356-358 : #10-14 even, 18-28
even and #30
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Lesson 7-5 Graphing Linear Inequalities
•Graph a linear inequality
in two variables
•Model a real life situation
with a linear inequality.
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Graphing an Inequality in Two Variables
Graph x < 2
Step 1: Start by graphing the line x =
2
Now what points
would give you less
than 2?
Since it has to be x < 2
we shade everything to
the left of the line.
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Graphing a Linear Inequality
Sketch a graph of y 3
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Graphing a Linear Inequality
Sketch a graph of y < -3
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Some Helpful Hints •If the sign is > or < the line is
dashed
•If the sign is or the line will be
solid
When dealing with just x and y.
•If the sign > or the shading
either goes up or to the right
•If the sign is < or the shading
either goes down or to the left
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Using What We Know Sketch a graph of x + y < 3
Step 1: Put into
slope intercept form
y <-x + 3
Step 2: Graph the
line y = -x + 3
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Using What We Know Sketch a graph of 2x + 3y < 6
Step 1: Put into
slope intercept form
y < -2/3 x + 2
Step 2: Graph the
line y = -2/3 x + 2
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Using What We Know Sketch a graph of y > 2x - 3
Step 1: Put into
slope intercept form
It already is!
Step 2: Graph the
line y = 2x - 3
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When dealing with slanted lines
•If it is > or then you shade above
•If it is < or then you shade below
the line
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Look at the two graphs. Determine the following:
A. The equation of each line.
B. How the graphs are similar.
C. How the graphs are different.
A. The equation of each line is y = x + 3.
B. The lines in each graph are the same and represent all of
the solutions to the equation y = x + 3.
C. The graph on the right is shaded above the line and this
means that all of these points are solutions as well.
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Pick a point from the shaded region
and test that point in the equation y =
x + 3.
Point: (-4, 5)
y x
3
5 4 3
5 1
This is incorrect. Five is
greater than or equal to
negative 1.
5 1
5 1 5 1 or
If a solid line is used, then the equation would be 5 -1.
If a dashed line is used, then the equation would be 5 > -1.
The area above the line is shaded.
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Pick a point from the shaded region
and test that point in the equation y =
-x + 4.
Point: (1, -3)
y x
4
3 1 4
3 3
This is incorrect. Negative
three is less than or equal to 3.
3 3
3 3 3 3 or
If a solid line is used, then the equation would be -3 3.
If a dashed line is used, then the equation would be -3 < 3.
The area below the line is shaded.
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1. Write the inequality in slope-intercept form.
2. Use the slope and y-intercept to plot two points.
3. Draw in the line. Use a solid line for less than or equal to ()
or greater than or equal to (). Use a dashed line for less than
(<) or greater than (>).
4. Pick a point above the line or below the line. Test that point in
the inequality. If it makes the inequality true, then shade the
region that contains that point. If the point does not make the
inequality true, shade the region on the other side of the line.
5. Systems of inequalities – Follow steps 1-4 for each inequality.
Find the region where the solutions to the two inequalities
would overlap and this is the region that should be shaded.
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Graph the following linear system of inequalities.
y x
y x
2 4
3 2
x
y
Use the slope and y-intercept to
plot two points for the first
inequality.
Draw in the line. For use a
solid line.
Pick a point and test it in the
inequality. Shade the
appropriate region.
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Graph the following linear system of inequalities.
y x
y x
2 4
3 2y x
2 4 Point (0,0)
0 2(0) - 4
0 -4The region above the line should be
shaded.
x
y
Now do the same for the second
inequality.
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Graph the following linear system of inequalities.
y x
y x
2 4
3 2
x
y
Use the slope and y-intercept to
plot two points for the second
inequality.
Draw in the line. For < use a
dashed line.
Pick a point and test it in the
inequality. Shade the
appropriate region.
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Graph the following linear system of inequalities.
y x
y x
2 4
3 2
x
y y x
3 2
3
Point (-2,-2)
-2 (-2) + 2
-2 < 8
The region below the line should be
shaded.
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Graph the following linear system of inequalities.
y x
y x
2 4
3 2
x
y
The solution to this system of
inequalities is the region where
the solutions to each inequality
overlap. This is the region above
or to the left of the green line and
below or to the left of the blue
line.
Shade in that region.
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Graph the following linear systems of inequalities.
1. y x
y x
4
2
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y x
y x
4
2
x
y Use the slope and y-intercept to
plot two points for the first
inequality.
Draw in the line.
Shade in the appropriate
region.
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x
y
y x
y x
4
2
Use the slope and y-intercept to
plot two points for the second
inequality.
Draw in the line.
Shade in the appropriate
region.
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x
y
y x
y x
4
2
The final solution is the region
where the two shaded areas
overlap (purple region).