Transcript Real-world application of systems

BELL-WORK
TB pg 371 #12
CW 2.4 Review
TB pg 353 # 13-14,17-18
TB pg 363 # 16
TB pg 371 # 11
HW 2.4(h)
Due Monday:
Practice Quiz 2.3
HW 2.4(g) Solutions
23.One; (-1,1)
25.One; (3,1)
27.Infinitely many solutions
29.One; (-3,1)
31.One; (-2,-3)
33.One; (-2,-1)
Extra Credit Opportunity
We will not discuss the third method of solving
systems in class.
However, if you read unit 6-3 and correctly
complete PW 6-3 #1-9 using elimination, you will
earn extra credit points!
Due Monday, on a separate sheet of notebook
paper.
Guiding question:
What are some real-world applications of linear systems?
Real-world application of systems
Suppose you plan to start taking aerobics class. One gym charges
a $4 membership fee and $2 per class while another gym
charges a $5 membership fee and $1 per class. After how
many classes will the cost be the same? What is that cost?
Solution:
Write an expression for the cost at gym 1:
y = 4 + 2x
Write an expression for the cost at gym 2:
y=5+x
Solve the system!
(1,6)
Cost will be the same after 1 class with a cost of $6
Complete Got It? 2 on page 361
Real-world application of systems
Tyrel and Dalia bought some pens and pencils. Tyrel bought 4
pens and 5 pencils, which cost him $6.71. Dalia bought 5 pens
and 3 pencils, which cost her $7.12. What is the price of a pen?
What is the price of a pencil?
Suppose you have just enough money, in coins, to pay for a loaf
of bread priced at $1.95. You have 12 coins, all quarters and
dimes. How many quarters do you have? How many dimes
do you have?
Applications of linear systems:
Break-even problems
In the world of business/economics, what is meant by
the break-even point?
Break-even point = the point at which income equals
expenses.
Why is the break even point important?
Because it tells business people whether or not they will
make a profit (if they make above the break-even
point) or suffer a loss (if they make below the breakeven point) in their business.
We can solve a system of linear equations to find the
break-even point.
Applications of linear systems:
Break-even problems
Suppose you have a typing service. You buy a personal
computer for $1750 on which to do your typing. You
charge $5.50 per page for typing. Expenses are $0.50 per
page for ink, paper, electricity, and other expenses. How
many pages must you type to break even?
Let x = number of pages typed, we can form the system:
y = 5.50x
y = 1750 + 0.50x
Solving the system will give the break even point
5.50x = 1750 + 0.50x
5x = 1750
x = 350 pages
Applications of linear systems:
Break-even problems
Suppose an antique car club publishes a newsletter. Expenses are
$0.35 for printing and mailing each copy, plus $770 total for
research and writing. The price of the newsletter is $0.55 per
copy. How many copies must the club sell to break even?
Let x = number of newsletters
y = 0.55x
y = 770 + 0.35x
At the break even point
0.55x = 770 + 0.35x
0.2x = 770
x = 3850
Complete TB pg 386 # 1
Applications of linear systems:
Mixture problems
A chemist has one solution that is 50% acid. She has another
solution that is 25% acid. How many liters of each type of
acid solution should she combine to get 10 liters of a 40%
acid solution?
If x = number of liters of 50% solution
y = number of liters of 25% solution
then we can write:
x + y = 10
0.50x + 0.25y = (0.4)10
and now we simply solve the system:
x + y = 10
0.5x + 0.25y = 4
6 liters of 50% and 4 liters of 25%
Applications of linear systems:
Mixture problems
Suppose you combine ingots of 25% copper alloy and 50% copper
alloy to create 40kg of 45% copper alloy. How many kilograms of
each do you need?
If x = number of kgs of 25% alloy
y = number of kgs of 50% alloy
then we can write:
x + y = 40
0.25x + 0.50y = (0.45)40
and now we simply solve the system:
x + y = 40
0.25x + 0.50y = 18
8 kg of the 25% and 32 kg of the 50%
Complete TB pg 386 # 2
Applications of linear systems:
Resistance problems
In various instances of travel, resistance plays a vital role.
Examples: flying, going up/down stream, etc.
If you fly with a tail wind,
Ground speedtailwind =
air speed + wind speed
G=A+W
If you fly into a head wind,
Ground speedheadwind =
air speed – wind speed
G =A–W
We can use these equations to write and solve systems!
Applications of linear systems:
Resistance problems
A traveler flies from Charlotte, NC, to Los Angeles, CA. At the same
time, another traveler flies from Los Angeles to Charlotte. The air
speed of each plane is the same. The ground speed of the plane to
Charlotte is 550 mph and to LA is 495 mph. What is the air speed?
What is the wind speed?
Which direction is with the tail wind?
So we can write the system:
550 = A + W
495 = A – W
Solve the system! (Round to the nearest tenth if necessary)
A = 522.5 mph
W = 27.5 mph
TB pg 386 # 3
Who wants to answer the Guiding question?
What are some real-world applications of linear systems?