Transcript Document

Unit 3 – Chapter 7
Unit 3
• Section 7.1 – Solve Linear Equations by graphing
• Section 7.2 – Solve Linear Equations by
substitution.
• Section 7.3 – Solve Linear Equations by Adding or
Subtracting
• Section 7.4 – Solve Linear Systems by multiplying
first
• Section 7.5 – Solve Special Types of Linear
Systems
• Section 7.6 – Solve systems of linear inequalities
Warm-Up – 7.1
Lesson 7.1, For use with pages 426-434
1. Graph the equation –2x + y = 1.
ANSWER
2. It takes 3 hours to mow a lawn and 2 hours to trim
hedges. You spend 16 hours doing yard work.
What are 2 possible numbers of lawns you mowed
and hedges you trimmed?
ANSWER
2 lawns and 5 hedges, or
4 lawns and 2 hedges
Vocabulary – 7.1
• System of Linear Equations
• 2 or more linear equations with the same variables
• Solution of a Linear Equation
• The solution set that MAKES ALL THE
EQUATIONS TRUE AT THE SAME TIME!!
• Usually a single point!
• Consistent Independent System
• A linear system that has EXACTLY one solution.
Notes – 7.1 – Solving Linear Systems by Graphing
•Remember the steps to convert English to Mathlish:
1. Read and highlight key words
2. DEFINE THE VARIABLES (MOST
CRITICAL STEP!!)
3. Write Mathlish sentence left to right (careful
with subtraction and division!)
•3 Step Process To Solve Linear Eqns by graphing
1.Write equations in slope-intercept form
2.Graph them on calculator
3.Find intersection of lines
•CHECK ANSWERS!
Examples 7.1
EXAMPLE 2
Use the graph-and-check method
Solve the linear system:
–x+y=–7
Equation 1
x + 4y = – 8
Equation 2
SOLUTION
STEP 1
Graph both equations.
EXAMPLE 2
Use the graph-and-check method
STEP 2
Estimate the point of intersection. The two lines
appear to intersect at (4, – 3).
STEP 3
Check whether (4, – 3) is a solution by substituting 4 for
x and – 3 for y in each of the original equations.
Equation 1
–x+y=–7
?
–(4) + (– 3) = – 7
–7=–7
Equation 2
x + 4y = – 8
?
4 + 4(– 3) = – 8
–8= –8
EXAMPLE 2
Use the graph-and-check method
ANSWER
Because (4, – 3) is a solution of each equation, it is a
solution of the linear system.
Use the graph-and-check
method
EXAMPLE
2
for Examples 1 and
2
GUIDED PRACTICE
Solve the linear system by graphing. Check your solution.
1. – 5x + y = 0
5x + y = 10
Put eqns in slopeIntercept form
y = 5x
5x + y = -5x + 10
The Intersection point is at (1,5).
EXAMPLE 3
Standardized Test Practice
The parks and recreation
department in your town
offers a season pass for $90.
•
As a season pass
holder, you pay $4 per
session to use the
town’s tennis courts.
•
Without the season
pass, you pay $13 per
session to use the
tennis courts.
•
Write a system of
equations to model
these situations and
find out where they are
equal.
EXAMPLE 3
Standardized Test Practice
SOLUTION
Write a system of equations where y is the total cost
(in dollars) for x sessions.
EQUATION 1
y
=
13
x
EXAMPLE 3
Standardized Test Practice
EQUATION 2
y
=
90
+
4
ANSWER
The correct answer is y = 13x and y = 4x + 90.
x
GUIDED PRACTICE
for Example 3
4. Solve the linear system in Example 3 to find the
number of sessions after which the total cost with a
season pass, including the cost of the pass, is the
same as the total cost without a season pass.
SOLUTION
Let the number of sessions be x
So, 13 x = 90 + 4
x
13x = 90 + 4x
9x = 90
x = 10
ANSWER
10 sessions
GUIDED PRACTICE
for Example 3
5. WHAT IF? In Example 3, suppose a season pass
costs $135. After how many sessions is the total cost
with a season pass, including the cost of the pass, the
same as the total cost without a season pass?
SOLUTION
Let the number of sessions be x
So, 13 x = 135 + 4
x
13x = 135 + 4x
9x = 135
x = 15
ANSWER
15 sessions
EXAMPLE 4
Solve a multi-step problem
RENTAL BUSINESS
A business rents in-line skates
and bicycles. During one day,
the business has a total of 25
rentals and collects $450 for the
rentals. Find the number of pairs
of skates rented and the number
of bicycles rented.
EXAMPLE 4
Solve a multi-step problem
SOLUTION
STEP 1
Write a linear system. Let x be the number
of pairs of skates rented, and let y be the
number of bicycles rented.
x + y =25
Equation for number of rentals
15x + 30y = 450 Equation for money collected from rentals
STEP 2
Graph both equations.
EXAMPLE 4
Solve a multi-step problem
STEP 3
Estimate the point of
intersection. The two lines
appear to intersect at (20, 5).
STEP 4
Check whether (20, 5) is a
solution.
20 + 5=? 25
25 = 25
15(20) + 30(5) =? 450
450 = 450
ANSWER
The business rented 20 pairs of skates and 5 bicycles.
Solve a for
multi-step
problem
EXAMPLE
4
Example
4
GUIDED PRACTICE
6. In Example 4, suppose the business has a total of
20 rentals and collects $ 420. Find the number of
bicycles rented.
SOLUTION
STEP 1
Write a linear system. Let x be the number
of pairs of skates rented, and let y be the
number of bicycles rented.
x + y =20
Equation for number of rentals.
15x + 30y = 420 Equation for money collected from
rentals.
Solve a for
multi-step
problem
EXAMPLE
4
Example
4
GUIDED PRACTICE
STEP 2
Graph both equations.
STEP 3
Estimate the point of intersection. The two lines
appear to intersect at (12, 8).
STEP 4
Check whether (12, 8) is a solution.
x + y = 20
12 + 8 = 20
20 = 20
15x + 30y = 420
15(12) + 30(8) = 420
420 = 420
Solve a for
multi-step
problem
EXAMPLE
4
Example
4
GUIDED PRACTICE
ANSWER
The business rented number of bicycle is 8.
Warm-Up – 7.2
Lesson 7.2, For use with pages 435-441
Solve the equation.
1. 6a – 3 + 2a = 13
ANSWER
a=2
2. 4(n + 2) – n = 11
ANSWER
n=1
Lesson 7.2, For use with pages 435-441
Solve the equation.
3. You burned 8 calories per minute on a treadmill
and 10 calories per minute on an elliptical trainer
for a total of 560 calories in 60 minutes. How many
minutes did you spend on each machine?
HINT
ANSWER
Define the variables
Create two equations (one for total minutes
And one for total calories)
EQN 1 – 8x + 10y = 560
EQN 2 – x + y = 60
treadmill: 20 min,
elliptical trainer: 40 min
Lesson 7.2, For use with pages 435-441
Solve the system of equations BUT YOU CAN’T USE THE
CALCULATOR OR GRAPH IT!!! WORK WITH YOUR
GROUP!!.
1. 2x + 3y = 40
2. y = x + 5
HINT!
Do the Dance!!!!
ANSWER
X=5
Y = 10
Vocabulary – 7.2
• None!!
Notes – 7.2 – Solving Systems w/Substitution
• If it’s EASY to get one of the variables in an
equation by itself, substitution may be the easiest way
to solve the system of equations.
•To Solve Linear Systems with Substitution
1. Get ONE of the variables in ONE of the
equations by itself.
2. Substitute that variable into the OTHER
equation.
3. Solve the equation from #2.
4. Plug the answer back into the original equation.
Examples 7.2
EXAMPLE 2
Use the substitution method
Solve the linear system:
x – 2y = – 6
4x + 6y = 4
Equation 1
Equation 2
SOLUTION
STEP 1
Solve Equation 1 for x.
x – 2y = – 6
x =2y – 6
Write original Equation 1.
Revised Equation 1
EXAMPLE 2
Use the substitution method
STEP 2
Substitute 2y – 6 for x in Equation 2 and solve for y.
4x + 6y = 4
4(2y – 6) + 6y = 4
8y – 24 + 6y = 4
14y – 24 = 4
14y = 28
y=2
Write Equation 2.
Substitute 2y – 6 for x.
Distributive property
Simplify.
Add 24 to each side.
Divide each side by 14.
EXAMPLE 2
Use the substitution method
STEP 3
Substitute 2 for y in the revised Equation 1 to find the
value of x.
Revised Equation 1
x = 2y – 6
x = 2(2) – 6
Substitute 2 for y.
x=–2
Simplify.
ANSWER The solution is (– 2, 2).
EXAMPLE
2
Use the substitution method
GUIDED PRACTICE
CHECK
Substitute –2 for x and 2 for y in each of the original
equations.
Equation 1
Equation 2
x – 2y = – 6
4x + 6y = 4
?
–2 –2 (2)= – 6
?
4( –2 )+ 6 (2 ) = 4
–6 = –6
4 = 4
EXAMPLE
1
for Examples
1 and 2
Use the substitution
method
GUIDED PRACTICE
Solve the linear system using the substitution method.
1.
y = 2x + 5
Equation 1
3x + y = 10
Equation 2
SOLUTION
STEP 1
Solve for y. Equation 1 is already solved for y.
EXAMPLE
2
for Examples
1 and 2
Use the substitution
method
GUIDED PRACTICE
STEP 2
Substitute 2x + 5 for y in Equation 2 and solve for x.
3x + y = 10
3x + (2x + 5) = 10
5x + 5 = 10
5x = 5
x=2
Write Equation 2.
Substitute 2x+5 for x.
Simplify.
EXAMPLE
2
GUIDED PRACTICE
for Examples 1 and 2
STEP 3
Substitute 1 for x in the revised Equation 1 to find the
value of y.
y = 2x + 5 = 2(1) + 5 = 7
ANSWER The solution is ( 1, 7).
Use the substitution
method
for Examples
Example
11 and 2
EXAMPLE
1
GUIDED PRACTICE
CHECK
Substitute 1 for x and 7 for y in each of the original
equations.
Equation 1
Equation 2
y = 2x + 5
3x + y = 10
7 =? 2(1) + 5
7 = 7
? 11
3(1)+7 =
10 = 10
EXAMPLE
2
for Examples
1 and 2
Use the substitution
method
GUIDED PRACTICE
2.
x– y= 3
Equation 1
x + 2y = – 6
Equation 2
SOLUTION
STEP 1
Solve Equation 1 for x.
x– y =3
x=y +3
Write original Equation 1.
Revised Equation 1
EXAMPLE
2
for Examples
1 and 2
Use the substitution
method
GUIDED PRACTICE
STEP 2
Substitute y + 3 for x in Equation 2 and solve for y.
x + 2y = – 6
( y + 3) + 2y = – 6
3y + 3 = – 6
3y = – 9
y=–3
Write Equation 2.
Substitute y + 3 for x.
Simplify.
Add 3 to each side.
Divide each side by 3.
EXAMPLE
2
for Examples
1and 2
Use the substitution
method
GUIDED PRACTICE
STEP 3
Substitute – 3 for y in the revised Equation 1 to find the
value of x.
Revised Equation 1
x=y+3
x= –3+3
Substitute – 3 for y.
x=0
Simplify.
ANSWER The solution is ( 0, – 3 ).
Use the substitution method
EXAMPLE
2
GUIDED PRACTICE
for Examples 1 and 2
GUIDED PRACTICE
CHECK
Substitute 0 for x and – 3 for y in each of the original
equations.
Equation 1
x– y=3
?
0 – (–3) = 3
3= 3
Equation 2
x + 2y = – 6
?
0 + 2 (–3 ) = – 6
–6 = –6
EXAMPLE 3
Solve a multi-step problem
WEBSITES
Many businesses pay website hosting companies to
store and maintain the computer files that make up
their websites. Internet service providers also offer
website hosting. The costs for website hosting
offered by a website hosting company and an Internet
service provider are shown in the table. Find the
number of months after which the total cost for
website hosting will be the same for both companies.
EXAMPLE 3
Solve a multi-step problem
SOLUTION
STEP 1
Write a system of equations. Let y be the total
cost after x months.
Equation 1: Internet service provider
y
=
10
+ 21.95
x
EXAMPLE 3
Solve a multi-step problem
Equation 2: Website hosting company
y
=
22.45
x
The system of equations is:
y = 10 + 21.95x
Equation 1
y = 22.45x
Equation 2
EXAMPLE 3
Solve a multi-step problem
STEP 2
Substitute 22.45x for y in Equation 1 and solve
for x.
Write Equation 1.
y = 10 + 21.95x
22.45x = 10 + 21.95x
0.5x = 10
x = 20
Substitute 22.45x for y.
Subtract 21.95x from each side.
Divide each side by 0.5.
ANSWER
The total cost will be the same for both companies
after 20 months.
GUIDED PRACTICE
4.
for Example 3
In Example 3, what is the total cost for website hosting
for each company after 20 months?
SOLUTION
Let y be the total cost.
y
=
=
22.45
449
20
GUIDED PRACTICE
for Example 3
ANSWER
The total cost for website hosting for each company
after 20 months is $ 449.
EXAMPLE 4
Solve a mixture problem
ANTIFREEZE
For extremely cold temperatures, an automobile
manufacturer recommends that a 70% antifreeze and
30% water mix be used in the cooling system of a car.
How many quarts of pure (100%) antifreeze and a 50%
antifreeze and 50% water mix should be combined to
make 11 quarts of a 70% antifreeze and 30% water mix?
SOLUTION
STEP 1
Write an equation for the total number of quarts and
an equation for the number of quarts of antifreeze. Let
x be the number of quarts of 100% antifreeze, and let y
be the number of quarts of a 50% antifreeze and 50%
water mix.
EXAMPLE 4
Solve a mixture problem
Equation 1: Total number of quarts
x + y = 11
Equation 2: Number of quarts of antifreeze
y quarts of
50% –50% mix
x quarts of
100% antifreeze
1
x
+
0.5
11 quarts of
70% – 30% mix
y
=
0.7(11)
x + 0.5y = 7.7
EXAMPLE 4
Solve a mixture problem
The system of equations is: x + y =11
Equation 1
x + 0.5y = 7.7 Equation 2
STEP 2
Solve Equation 1 for x.
x + y = 11
x = 11 – y
Write Equation 1.
Revised Equation 1
STEP 3
Substitute 11 – y for x in Equation 2 and solve
for y.
x + 0.5y = 7.7
Write Equation 2.
EXAMPLE 4
Solve a mixture problem
(11 – y) = 0.5y = 7.7
y = 6.6
Substitute 11 – y for x.
Solve for y.
STEP 4
Substitute 6.6 for y in the revised Equation 1 to
find the value of x.
x = 11 – y = 11 – 6.6 = 4.4
ANSWER
Mix 4.4 quarts of 100% antifreeze and 6.6 quarts of a 50%
antifreeze and 50% water mix to get 11 quarts of a 70%
antifreeze and 30% water mix.
GUIDED PRACTICE
6.
for Example 4
WHAT IF ? How many quarts of 100% antifreeze
and a 50% antifreeze and 50% water mix should be
combined to make 16 quarts of a 70% antifreeze
and 30% water mix?
SOLUTION
STEP 1
Write an equation for the total number of quarts and
an equation for the number of quarts of antifreeze. Let
x be the number of quarts of 100% antifreeze, and let y
be the number of quarts of a 50% antifreeze and 50%
water mix.
GUIDED PRACTICE
for Example 4
Equation 1: Total number of quarts
x + y = 16
Equation 2: Number of quarts of antifreeze
y quarts of
50% –50% mix
x quarts of
100% antifreeze
1
x
+
11 quarts of
70% – 30% mix
0.5 y
=
0.7(16)
x + 0.5y = 11.2
GUIDED PRACTICE
for Example 4
The system of equations is: x + y =16
x + 0.5y = 11.2
STEP 2
Solve Equation 1 for x.
x + y = 16
x = 16 – y
Equation 1
Equation 2.
Write Equation 1.
Revised Equation 1
GUIDED PRACTICE
for Example 4
STEP 3
Substitute 16 – y for x in Equation 1 and solve
for x.
x + 0.5y = 7.7
Write Equation 2.
(16 – y) + 0.5y = 7.7
y = 9.6
Substitute 16 – y for x.
Solve for y.
GUIDED PRACTICE
for Example 4
STEP 4
Substitute 9.6 for y in the revised Equation 1 to
find the value of x.
x = 16 – y = 16 – 9.6 = 6.4
ANSWER
Mix 6.4 quarts of 100% antifreeze and 9.6 quarts of a 50%
Antifreeze.
Warm-Up – 7.3
Lesson 7.2, For use with pages 435-441
Solve the linear systems by GRAPHING!!!
1. x + y = -2
2. -x + y = 6
ANSWER
x = -4
y=2
2. x – y = 0
3. 5x + 2y = -7
ANSWER
x = -1
y = -1
Lesson 7.2, For use with pages 435-441
Solve the linear systems by SUBSTITUTION!!!!!!
1. y = x – 4
2. -2x + y = 18
ANSWER
x = -22
y = -26
2. 5x – 4y = 27
3. -2x + y = 3
ANSWER
x= -13
y = -23
Lesson 7.3, For use with pages 443-450
1. Solve the linear system using substitution.
2x + y = 12
3x – 2y = 11
ANSWER
(5, 2)
2. One auto repair shop charges $30 for a diagnosis
and $25 per hour for labor. Another auto repair shop
charges $35 per hour for labor. For how many hours
are the total charges for both of the shops the same?
HINT: FIND EQUATIONS FOR TOTAL AND SUBSTITUTE!
ANSWER
3h
Lesson 7.3, For use with pages 443-450
1. Add the two equations together (combine like
terms) and solve for x and y.
2x + 3y = 11
-2x + 5y = 13
ANSWER
(1,3)
Vocabulary – 7.3 - REVIEW
• System of Linear Equations
• 2 or more linear equations with the same variables
• Solution of a Linear Equation
• The solution set that MAKES ALL THE
EQUATIONS TRUE AT THE SAME TIME!!
• Consistent Independent System
• A linear system that has EXACTLY one solution.
Notes – 7.3 – Solving systems with elimination
• If we have two equations and two variables, how
many solutions should we USUALLY have??
•What’s the goal of solving every algebra eqn. you
will ever see?
•You can eliminate variables from some systems by
adding or subtracting equations to eliminate variables.
•RULES/HINTS TO MAKE PROCESS EASIER!
1. Remember the goal: You are trying to eliminate
one variable!
2. Line up like terms under each other.
3. NEVER subtract. Add the negative instead!
Examples 7.3
EXAMPLE 1
Use addition to eliminate a variable
Solve the linear system:
2x + 3y = 11
– 2x + 5y = 13
Equation 1
Equation 2
SOLUTION
STEP 1
STEP 2
STEP 3
Add the equations to
eliminate one variable.
Solve for y.
2x + 3y = 11
– 2x +5y = 13
8y = 24
y=3
Substitute 3 for y in either equation and
Solve for x.
EXAMPLE 1
Use addition to eliminate a variable
2x + 3y = 11
2x + 3(3) = 11
x=1
ANSWER
The solution is (1, 3).
Write Equation 1
Substitute 3 for y.
Solve for x.
EXAMPLE 2
Use subtraction to eliminate a variable
Solve the linear system:
4x + 3y = 2
5x + 3y = – 2
Equation 1
Equation 2
SOLUTION
STEP 1
STEP 2
Subtract the equations to
eliminate one variable.
Solve for x.
4x + 3y = 2
5x + 3y = – 2
–x
= 4
x = 4
STEP 3 Substitute 4 for x in either equation and solve
for y.
EXAMPLE 2
Use subtraction to eliminate a variable
4x + 3y = 2
Write Equation 1.
Substitute – 4 for x.
4(– 4) + 3y = 2
y=2
Solve for y.
ANSWER
The solution is (– 4, 6).
EXAMPLE 3
Arrange like terms
Solve the linear system:
8x – 4y = –4
4y = 3x + 14
Equation 1
Equation 2
SOLUTION
STEP 1
STEP 2
STEP 3
STEP 4
Rewrite Equation 2 so that the like terms are
arranged in columns.
8x – 4y = –4
8x – 4y = –4
4y = 3x + 14
3x + 4y = 14
5x
= 10
Add the equations.
Solve for x.
x=2
Substitute 2 for x in either equation and
solve for y.
EXAMPLE 3
Arrange like terms
4y = 3x + 14
4y = 3(2) + 14
y=5
Write Equation 2.
Substitute 2 for x.
Solve for y.
ANSWER
The solution is (2, 5).
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system:
1.
4x – 3y = 5
– 2x + 3y = – 7
Equation 1
Equation 2
SOLUTION
STEP 1
STEP 2
STEP 3
4x – 3y = 5
– 2x +3y = – 7
2x = – 2
x=–1
Substitute – 1 for y in either equation and
Solve for x.
Add the equations to
eliminate one variable.
Solve for x.
GUIDED PRACTICE
for Example 1,2 and 3
4x – 3y = 5
2(– 1) – 3y = 5
y=–3
Write Equation 1.
ANSWER
The solution is (– 1, – 3).
Substitute – 1 for x.
Solve for x.
GUIDED PRACTICE
CHECK
for Example 1,2 and 3
Substitute 1 for x and 3 for y in each of the.
original equation
4x – 3y = 5
– 2x + 3y = – 7
4(– 1) – 3(– 3) =? 5
? –7
– 2(– 1) + 5(– 3) =
5=5
–7=–7
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system:
4.
7x – 2y = 5
7x – 3y = 4
Equation 1
Equation 2
SOLUTION
STEP 1
STEP 2
Subtract the equations to
eliminate one variable.
Solve for y.
7x – 2y = 5
7x – 3y = 4
y= 1
STEP 3
Substitute 1 for y in either and solve for x.
GUIDED PRACTICE
for Example 1,2 and 3
Write Equation 1.
7x – 2y = 5
Substitute 1 for y.
7x – 2(1) = 5
Solve for x.
x=1
ANSWER
The solution is (1, 1).
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system:
5.
3x + 4y = – 6
2y = 3x + 6
Equation 1
Equation 2
SOLUTION
STEP 1
Rewrite Equation 2 so that the like terms are
arranged in columns.
3x + 4y = – 6
3x + 4y = – 6
2y = 3x + 6
3x – 2y = – 6
STEP 2
STEP 3
STEP 4
6y = 0
Subtract the equations.
Solve for y.
y=0
Substitute 0 for y in either equation and
solve for x.
GUIDED PRACTICE
2y = 3x + 6
2(0) = 3x + 6
x=–2
for Example 1,2 and 3
Write Equation 2.
Substitute 0 for y.
Solve for x .
ANSWER
The solution is (– 2, 0).
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system:
6.
2x + 5y = 12
5y = 4x + 6
Equation 1
Equation 2
SOLUTION
STEP 1
Rewrite Equation 2 so that the like terms are
arranged in columns.
2x + 5y = 12
2x + 5y = 12
5y = 4x + 6
– 4x + 5y = 6
STEP 2
STEP 3
STEP 4
Subtract the equations.
6x
= 6
Solve for x.
x=1
Substitute 1 for x in either equation and
solve for y.
GUIDED PRACTICE
2x + 5y = 12
2(1) + 5y = 6
x= 2
for Example 1,2 and 3
Write Equation 2.
Substitute 1 for x.
Solve for y .
ANSWER
The solution is (1, 2).
EXAMPLE 4
Write and solve a linear system
KAYAKING
During a kayaking trip, a kayaker travels 12 miles
upstream (against the current) and 12 miles
downstream (with the current), as shown. The speed
of the current remained constant during the trip. Find
the average speed of the kayak in still water and the
speed of the current.
EXAMPLE 4
Write and solve a linear system
STEP 1
Write a system of equations. First find the speed of
the kayak going upstream and the speed of the kayak
going downstream.
Upstream: d = rt
12 = r 3
4=r
Downstream: d = rt
12 = r 2
6=r
EXAMPLE 4
Write and solve a linear system
Use the speeds to write a linear system. Let x be the
average speed of the kayak in still water, and let y be
the speed of the current.
Equation 1: Going upstream
x
–
y
=
4
EXAMPLE 4
Write and solve a linear system
Equation 2: Going downstream
x
+
y
=
6
EXAMPLE 4
Write and solve a linear system
STEP 2
Solve the system of equations.
x – y= 4
Write Equation 1.
x+y=6
Write Equation 2.
2x
= 10
x=5
Add equations.
Solve for x.
Substitute 5 for x in Equation 2 and solve for y.
EXAMPLE 4
Write and solve a linear system
5+y=6
y=1
Substitute 5 for x in Equation 2.
Subtract 5 from each side.
GUIDED PRACTICE
for Example 4
WHAT IF? In Example 4, suppose it takes the
kayaker 5 hours to travel 10 miles upstream and 2
hours to travel 10 miles downstream. The speed
of the current remains constant during the trip.
Find the average speed of the kayak in still water
and the speed of the current.
7.
SOLUTION
STEP 1
Write a system of equations. First find the speed of
the kayak going downstream.
GUIDED PRACTICE
Upstream: d = rt
10 = r 5
2=r
for Example 4
Downstream: d = rt
10 = r 2
5=r
Use the speeds to write a linear system. Let x be the
average speed of the kayakar in still water, and let y
be the speed of the current.
GUIDED PRACTICE
for Example 4
Equation 1: Going upstream
x
–
y
=
2
=
5
Equation 2: Going downstream
x
+
y
GUIDED PRACTICE
for Example 4
STEP 2
Solve the system of equations.
x – y= 2
Equation 1.
x+y=5
Equation 2.
2x
=7
x = 3.5
Add equations.
Solve for x.
Substitute 3.5 for x in Equation 2 and solve for y.
GUIDED PRACTICE
3.5 + y = 6
y = 1.5
for Example 4
Substitute 3.5 for x in Equation 2.
Subtract 3.5 from each side.
ANSWER
The average speed of the kayakar in still water is 3.5
miles per hour, and the speed of the current is 1.5 mile
per hour.
Warm-Up – 7.4
Lesson 7.4, For use with pages 451-457
Solve the linear system.
1. 4x – 3y = 15
2x – 3y = 9
ANSWER
(3, – 1)
2. –2x + y = – 8
2x – 2y = 8
ANSWER
(4, 0)
Lesson 7.4, For use with pages 451-457
Solve the linear system.
3. You can row a canoe 10 miles upstream in 2.5 hours
and 10 miles downstream in 2 hours. What is the
average speed of the canoe in still water?
ANSWER
4.5 mi/h
Multiply the second equation by -2 and rewrite it.
Then use it to solve the system.
1. 8x – 6y = 30
2. 2x – 3y = 9
ANSWER
(3, – 1)
Vocabulary – 7.4
• Least Common Multiple
• Smallest POSITIVE number that is a multiple of two
or more factors
Notes – 7.4 – Solving systems by multiplying first.
• In order to add equations and eliminate a variable,
two of the coefficients must be opposite signs.
• Learned 3 ways to solve systems of linear eqns:
1. Graphing
• Easiest when I can get y by itself and have a calculator!
2. Substitution
• Easiest when I can get one variable by itself.
3. Elimination
•
Easiest when I can get opposite coefficients.
• There is a 4th way - multiply and then eliminate.
• To get coefficients with opposite signs, you can
multiply one or more equations by constants.
• May need to identify LCM of two coefficients.
Examples 7.4
EXAMPLE 1
Multiply one equation, then add
Solve the linear system:
6x +5y = 19
Equation 1
2x +3y = 5
Equation 2
SOLUTION
STEP 1 Multiply: Equation 2 by –3 so that the
coefficients of x are opposites.
6x + 5y = 19
2x + 3y = 5
STEP 2 Add: the equations.
6x + 5y = 19
–6x – 9y = –15
–4y = 4
EXAMPLE 1
Multiply one equation, then add
STEP 3 Solve: for y.
y = –1
STEP 4 Substitute: –1 for y in either of the original
equations and solve for x.
ANSWER
The solution is (4, –1).
EXAMPLE 2
Multiply both equations, then subtract
Solve the linear system:
4x + 5y = 35
Equation 1
2y = 3x – 9
Equation 2
SOLUTION
STEP 1
Arrange: the equations so that like terms are in
columns.
4x + 5y = 35
–3x + 2y = –9
Write Equation 1.
Rewrite Equation 2.
EXAMPLE 2
Multiply both equations, then subtract
STEP 2
Multiply: Equation 1 by 2 and Equation 2 by 5 so that
the coefficient of y in each equation is the least
common multiple of 5 and 2, or 10.
4x + 5y = 35
8x + 10y = 70
–3x + 2y = –9
–15x +10y = –45
STEP 3
Subtract: the equations.
STEP 4
Solve: for x.
ANSWER
The solution is (5, 3).
23x
= 115
x=5
GUIDED PRACTICE
for Examples 1 and 2
Solve the linear system using elimination:
1.
6x – 2y = 1
Equation 1
–2x + 3y = –5
Equation 2
SOLUTION
ANSWER
The solution is (–0.5, –2).
y = –2
GUIDED PRACTICE
2.
for Examples 1 and 2
2x + 5y = 3
Equation 1
3x + 10y = –3
Equation 2
SOLUTION
ANSWER
The solution is (9, –3).
GUIDED PRACTICE
for Examples 1 and 2
3. 3x – 7y = 5
9y = 5x +5
SOLUTION
ANSWER
The solution is (–10, –5).
Equation 1
Equation 2
EXAMPLE 3
Standardized Test Practice
Darlene is making a quilt that has alternating stripes of
regular quilting fabric and sateen fabric. She spends
$76 on a total of 16 yards of the two fabrics at a fabric
store. Write a system of equations can be used to find
the amount x (in yards) of regular quilting fabric and
the amount y (in yards) of sateen fabric she purchased?
EXAMPLE 3
Standardized Test Practice
SOLUTION
Write a system of equations where x is the number of
yards of regular quilting fabric purchased and y is the
number of yards of sateen fabric purchased.
Equation 1: Amount of fabric
x
+
y
=
16
EXAMPLE 3
Standardized Test Practice
Equation 2: Cost of fabric
4
x
+
The system of equations is:
x + y = 16
4x +6y = 76
ANSWER
6
y
=
Equation 1
Equation 2
The correct answer is x = 10 and y = 6.
76
GUIDED PRACTICE
4.
for Example 3
SOCCER A sports equipment store is having a sale
on soccer balls. A soccer coach purchases 10
soccer balls and 2 soccer ball bags for $155. Another
soccer coach purchases 12 soccer balls and 3
soccer ball bags for $189. Find the cost of a soccer
ball and the cost of a soccer ball bag.
SOLUTION
Write a system of equations where x is the cost of
soccer ball and y is the cost of soccer ball bag.
for Example 3
GUIDED PRACTICE
Equation 1:
Cost of
soccer ball
10x
+
+
Cost of
soccer bag
2y
=
=
Total Cost
155
for Example 3
GUIDED PRACTICE
Equation 2:
Cost of
soccer ball
12x
+
Cost of
soccer bag
+
3y
=
=
Total Cost
189
The system of equations is:
10x +2y = 155
Equation 1
12x +3y = 189
Equation 2
GUIDED PRACTICE
for Example 3
STEP 1
Multiply equation 1 by – 3 and equation 2 by 2 so that
the coefficient of y in each equation is the least
common multiple of – 3 and 2 .
STEP 2
STEP 3
STEP 4
10x +2y = 155
– 30x – 6y = –465
12x +3y = 189
24x + 6y = 378
Add the equation
Solve for x
6x
x
= 87
= 14.50
GUIDED PRACTICE
for Example 3
STEP 5
Substitute 14.50 for x in either of the original
equations and solve for y.
10x + 2y = 155
10(14.50) + 2y = 155
y=5
Write Equation 1.
Substitute 14.50 for x.
solve for y
ANSWER
Cost of soccer ball is $ 14.50 and soccer ball bag is $5.
Warm-Up – 7.5
Lesson 7.5, For use with pages 459-465
1. Solve the linear system.
2x + 3y = – 9
x – 2y = 6
ANSWER
(0, – 3)
2. You buy 8 pencils for $8 at the bookstore. Standard
pencils cost $.85 and specialty pencils cost $1.25.
How many specialty pencils did you buy?
ANSWER
3 specialty pencils
Lesson 7.5, For use with pages 459-465
1. Solve the linear system by GRAPHING. Describe the
lines on your whiteboard.
x + y = -2
y = -x+5
1. Solve the linear system by SUBSTITUTION. What do
you get?
x + y = -2
y = -x+5
1. Solve the linear system by ELIMINATION. What do
you get?
x + y = -2
y = -x+5
ANSWER
No solution.
Vocabulary – 7.5
• Consistent Independent System
• System of equations with ONE solution
• Inconsistent System
• System of equations with NO solution.
• Consistent Dependent System
• System of equations with INFINITE solutions.
Notes – 7.5–Special Types of Systems
• Systems can have one soln, no soln, or infinite.
• Easiest ways to check for solutions:
1. Graph them (put them in slope-intercept form)
1. Intersect = 1 solution
2. Parallel = No solution
3. Same line = Infinite solutions
2. Check if equations are multiples of each other.
1. Yes = infinite solutions
2. No = Check some more!
3. Eliminate the variables (using Add. or Mult.)
1. Always False statement = No solutions
2. Always True statement = Infinite solutions
Notes – 7.5–Special Types of Systems
Examples 7.4
EXAMPLE 1
A linear system with no solution
Solve the linear system by graphing and by elimination!
3x + 2y = 10
Equation 1
3x + 2y = 2
Equation 2
SOLUTION
METHOD 1
Graphing
Graph the linear system.
Answer: NO SOLUTION.
EXAMPLE 1
METHOD 2
A linear system with no solution
Elimination
Subtract the equation.
3x + 2y = 10
3x + 2y = 2
0 = 8
This is a false statement.
ANSWER
The variables are eliminated and you are left with a
false statement regardless of the values of x and y.
This tells you that the system has no solution.
EXAMPLE 2
A linear system with infinitely many solutions
Solve the system by graphing and substitution.
x – 2y = – 4
Equation 1
y = 1x + 2
2
Equation 2
SOLUTION
METHOD 1 Graphing
Graph the linear system.
EXAMPLE 2
METHOD 2
A linear system with infinitely many solutions
Substitution
Substitute 1 x + 2 for y in Equation 1 and solve for x.
2
x – 2y = – 4
x – 2 1x + 2 = – 4
2
–4= –4
Write Equation 1
Substitute 1 x + 2 for y.
2
Simplify.
ANSWER
The variables are eliminated and you are left with a
true statement regardless of the values of x and y.
This tells you that the system has infinite solutions.
GUIDED PRACTICE
for Examples 1 and 2
Tell whether the linear system has no solution or
infinitely many solutions. Explain.
1.
5x + 3y = 6
– 5x – 3y = 3
METHOD 2
Equation 1
Equation 2
Elimination
Subtract the equations.
5x + 3y = 6
– 5x – 3y = 3
0=9
This is a false statement.
GUIDED PRACTICE
for Examples 1 and 2
ANSWER
The variables are eliminated and you are left with a
false statement regardless of the values of x and y.
This tells you that the system has no solution.
GUIDED PRACTICE
2.
for Examples 1 and 2
y = 2x – 4
Equation 1
– 6x + 3y = – 12
METHOD 2
Equation 2
Elimination
Substitute 2x – 4 for y in Equation 2 and solve for x.
– 6x + 3y = – 12
– 6x + 3(2x – 4) = – 12
– 12 = – 12
Write Equation 2
Substitute (2x – 4) for y.
Simplify.
GUIDED PRACTICE
for Examples 1 and 2
ANSWER
The variables are eliminated and you are left with a
true statement regardless of the values of x and y. This
tells you that the system has infinitely many solution.
Warm-Up – 7.6
Lesson 7.6, For use with pages 466-472
1. Graph y < 2 x – 1.
3
ANSWER
1. Solve -3x >= 12
ANSWER: x <= -4
Lesson 7.6, For use with pages 466-472
2. You are running one ad that costs $6 per day and
another that costs $8 per day. You can spend no
more than $120. Graph this inequality. HINT:
WRITE THE EQUATION FIRST!
2. Graph the following
on a number line:
x <= 5 and x >= 0
ANSWER:
-2 -1 0 1 2 3 4 5 6 7
Vocabulary – 7.6
• System of Linear Inequalities
• Two or more linear inequalities in the same variables.
• Solution of a system of linear inequalities
• An ordered pair that makes ALL the inequalities true at
the same time.
• Graph of a system of linear inequalitites
• Graph of all the solutions of the system.
Notes–7.6–Solve Systems of linear inequalities.
•REVIEW
•Graphing inequalities - similar to graphing lin.eqns
1. Play the pretend game and let equation be =.
2. Dotted line is < or >
3. Solid line is <= or >=
4. Pick a point NOT ON THE LINE, check the
answer, and shade the correct side of the line.
•TO GRAPH SYSTEMS OF INEQUALITIES
1.Graph each inequality
2.Find the AREA where solutions intersect.
3.Pick a point and check your solution.
Examples 7.6
EXAMPLE 1
Graph a system of two linear inequalities
Graph the system of inequalities. y > – x – 2
y < 3x + 6
SOLUTION
Graph both inequalities in the same
coordinate plane. The graph of the
system is the intersection of the two
half-planes, which is shown as the
darker shade of blue.
Inequality 1
Inequality 2
EXAMPLE 1
CHECK
Graph a system of two linear inequalities
Choose a point in the dark blue region, such
as (0, 1). To check this solution, substitute 0
for x and 1 for y into each inequality.
?
1>0–2
1>–2
?
1>0+6
1>6
EXAMPLE 2
Graph a system of three linear inequalities
Graph the system of inequalities. y > – 1
x>2
x + 2y < 4
SOLUTION
Inequality 1
Inequality 2
Inequality 3
Graph all three inequalities in the same coordinate
plane. The graph of the system is the triangular region
shown.
GUIDED PRACTICE
for Examples 1 and 2
Graph the system of linear inequalities.
1.
y<x–4
y>–x+3
ANSWER
GUIDED PRACTICE
for Examples 1 and 2
Graph the system of linear inequalities.
2.
y>–x+2
y<4
x<3
ANSWER
GUIDED PRACTICE
for Examples 1 and 2
Graph the system of linear inequalities.
3.
y>–x
y> x–4
y<5
ANSWER
EXAMPLE 3
Write a system of linear inequalities
Write a system of inequalities for the shaded region.
SOLUTION
INEQUALITY 1: One boundary line
for the shaded region is y = 3.
Because the shaded region is
above the solid line, the inequality
is y > 3.
INEQUALITY 2: Another boundary line for the shaded
region has a slope of 2 and a y-intercept of 1. So, its
equation is y = 2x + 1. Because the shaded region is
above the dashed line, the inequality is y > 2x + 1.
EXAMPLE 3
Write a system of linear inequalities
ANSWER
The system of inequalities for the shaded region is:
y>3
Inequality 1
y > 2x + 1 Inequality 2
EXAMPLE 4
Write and solve a system of linear inequalities
BASEBALL
The National Collegiate Athletic Association (NCAA)
regulates the lengths of aluminum baseball bats used
by college baseball teams. The NCAA states that the
length (in inches) of the bat minus the weight (in
ounces) of the bat cannot exceed 3. Bats can be
purchased at lengths from 26 to 34 inches.
a. Write and graph a system of linear inequalities that
describes the information given above.
b. A sporting goods store sells an aluminum bat that
is 31 inches long and weighs 25 ounces. Use the
graph to determine if this bat can be used by a player
on an NCAA team.
EXAMPLE 4
Write and solve a system of linear inequalities
SOLUTION
a. Let x be the length (in inches) of the bat, and let y be
the weight (in ounces) of the bat. From the given
information, you can write the following inequalities:
x–y<3
The difference of the bat’s length and weight can
be at most 3.
x ≥ 26
The length of the bat must be at least 26 inches.
x ≤ 34
The length of the bat can be at most 34 inches.
y≥0
The weight of the bat cannot be a negative number.
Graph each inequality in the system. Then identify the
region that is common to all of the graphs of the
inequalities. This region is shaded in the graph shown.
EXAMPLE 4
Write and solve a system of linear inequalities
b. Graph the point that
represents a bat that is 31 inches
long and weighs 25 ounces.
ANSWER
Because the point falls outside the solution
region, the bat cannot be used by a player on an
NCAA team.
GUIDED PRACTICE
for Examples 3 and 4
Write a system of inequalities that defines the shaded
region.
4.
ANSWER
x ≤ 3, y > 2 x 1
3
GUIDED PRACTICE
for Examples 3 and 4
Write a system of inequalities that defines the shaded
region.
5.
ANSWER
y ≤ 4, x < 2
GUIDED PRACTICE
for Examples 3 and 4
6. WHAT IF? In Example 4, suppose a Senior League
(ages 10–14) player wants to buy the bat described in
part (b). In Senior League, the length (in inches) of the
bat minus the weight (in ounces) of the bat cannot
exceed 8. Write and graph a system of inequalities to
determine whether the described bat can be used by
the Senior League player.
ANSWER
x
y ≤ 8, x ≥ 26, x ≤ 34, y ≥ 0
Review – Ch. 7 – PUT
HW QUIZZES HERE
Daily Homework Quiz
1.
For use after Lesson 7.1
Use the graphing to solve the linear system
3x – y = 5
– x + 3y = 5
ANSWER
(2.5,2.5)
Daily Homework Quiz
2.
For use after Lesson 7.1
Solve the linear system by graphing.
2x + y = – 3
– 6x + 3y = 3
ANSWER
(–1, –1)
Daily Homework Quiz
3.
For use after Lesson 7.1
A pet store sells angel fish for $6 each and clown
loaches for $4 each . If the pet store sold 8 fish for
$36, how many of each type of fish did it sell?
ANSWER
2 angel fish and 6 clown loaches
Daily Homework Quiz
For use after Lesson 7.2
Solve the linear system using substitution
1.
–5x – y = 12
3x – 5y = 4
(–2, –2)
ANSWER
2. 2x + 9y = –4
x – 2y = 11
ANSWER
(7, –2 )
Daily Homework Quiz
3.
For use after Lesson 7.2
You are making 6 quarts of fruit punch for a party.
You want the punch to contain 80% fruit juice. You
have bottles of 100% fruit juice and 20% fruit juice.
How many quarts of 20% fruit juice should you mix
to make 6 quarts of 80% fruit juice?
ANSWER
4.5 quarts of 100% fruit juice and 1.5 quarts of
20% fruit juice
Daily Homework Quiz
For use after Lesson 7.3
Solve the linear system using elimination.
1.
–5x +y = 18
3x – y = –10
ANSWER
2.
4x + 2y = 14
4x – 3y = –11
ANSWER
3.
(–4, –2)
(1, 5)
2x – y = –14
y = 3x + 6
ANSWER
(8, 30)
Daily Homework Quiz
4.
For use after Lesson 7.3
x + 4y = 15
2y = x – 9
ANSWER
(11, 1)
5. A business center charges a flat fee to send faxes
plus a fee per page.You send one fax with 4 pages
for $5.36 and another fax with 7 pages for $7.88.
Find the flat fee and the cost per page to send a fax.
HINT: Find two points and then utilize the slopeintercept form of a line.
ANSWER
flat fee: $2, price per page: $.84
Daily Homework Quiz
For use after Lesson 7.4
Solve the linear system using elimination.
1.
8x + 3y = 12
– 2x + y = 4
ANSWER
2.
(0,4)
– 3x + 2y = 7
5x – 4y = – 15
ANSWER
3.
(1,5)
– 7x – 3y = 11
4x – 2y = 16
ANSWER
(1, – 6)
Daily Homework Quiz
For use after Lesson 7.4
4. A recreation center charges nonmembers $3 to use
the pool and $5 to use the basketball courts. A
person pays $42 to use the recreation facilities 12
times. How many times did the person use the
pool.
ANSWER
9 times
Daily Homework Quiz
For use after Lesson 7.5
Without solving the linear system, tell whether
the linear system has one solution, no solution,
or infinitely many solutions.
1. 4x + 2y = 12
y = –2x + 8
ANSWER
2.
no solution
– 2x + 5y = 5
y= 2x+1
5
ANSWER
infinitely many solutions
Daily Homework Quiz
For use after Lesson 7.3
3. A group of 12 students and 3 teachers pays $57 for
admission to a primate research center. Another
group of 14 students and 4 teachers pays $69. Find
the cost of one student ticket.
ANSWER
$3.50
Daily Homework Quiz
1.
For use after Lesson 7.6
Write a system of inequalities for the shaded
region.
ANSWER
x < 2, y > x =1
Daily Homework Quiz
2.
For use after Lesson 7.6
A bibliography can refer to at most 8 articles, at
most 4 books, and at most 8 references in all. Write
and graph a system of inequalities that models the
situation.
ANSWER
x = articles, y = books; x< 8, y < 4,
x + y < 8, x > 0, and y >0
Warm-Up – X.X
Vocabulary – X.X
• Holder
• Holder 2
• Holder 3
• Holder 4
Notes – X.X – LESSON TITLE.
• Holder
•Holder
•Holder
•Holder
•Holder
Examples X.X