Download Julie Pavlock

Julie Pavlock
Thursday 4/26/07
Lesson Title: Solve this!
Grade Level: 9th
Materials: paper, pencils, chalkboard (whiteboard), chalk (markers), worksheets,
calculators
Lesson Overview: Students will learn how to find answers to real – world problems.
Students will need to use systems of equations and inequalities to arrive at a correct
answer for the different problems. Some problems will come form previous Math A
exams.
Lesson Objectives:
 Students will analyze given problems and be able to select the right method
needed to solve the problem.
 Students will develop the ability to adapt math equations to real – world word
problems using the same form of equations that were previously given.
 Students will formulate different conjectures about how to solve these real –
world problems.
New York State Standards:
4E. Model real – world problems with systems of equations and inequalities
3A. Use addition, subtraction, multiplication, division, and exponentiation with
real numbers and algebraic expressions
Anticipatory Set: Students will be given different equations based on the systems of
equations and inequalities. The student will have about five minutes to complete the
worksheet called what’s the answer. There will be two questions for the equations and
the inequalities. Once the students are finished ask them if they need to go over any of
the answers. Once the students are done tell them that they are using the same types of
equations, but they are now in the form of word problems. So you will need to read the
problem and figure out what type of equation you will need to solve the problem.
Developmental Activity: Students will be given a worksheet called equations or
inequalities with different types of word problems. The first few problems will be ones
that I make-up. They will include equations and inequalities. The inequalities will need
to be done with a calculator. I will make up four problems which will include two of
each type. The last few problems will be from different Math A exams. If there is not a
lot of time left in the class then have the students do one equation and one inequalities
problem. Then have them work on the math A questions. At this time ask students if
they have any questions. If the students have questions try to answer them still allowing
time for the closure. This activity should take about 25 minutes.
Closure: Give each student an index card. On the card have the question what did you
most like about today’s class and what was the most important thing that you have
learned. If the students so not mention the things that they should have learned then that
means you will need to re teach the lesson again.
Assessment: Give students the exit slip. There will be one question that the students will
have to answer. The question will be:
The piano teacher bought 14 pieces of music and 7 beginner books. The total cost
was $28.00. The trumpet teacher bought 10 pieces of music and 5 beginner books. The
total cost was $20.50. How much did each piece of music and each beginner book cost?
What’s the Answer?
Name: _________________________
Directions: Solve each problem for x and y.
1.)
x  y  20
x  5y  15
2.)
5x  3y  20
3x  2 y  15
For these two problems solve for x.
3.)
3x  7  2
4.)
2  5x  3x  14
Equations or Inequalities
Name: _______________________________
Directions: For each problem read the question, pick which method you want to use and
solve the problem.
1.) Joe purchased 12 pens and 14 notebooks for $20. Jen bought 7 pens and 4
notebooks for $7.50. Find the price of one pen and the price of one notebook.
2.) Kim a nurse bought 5 uniform tops and 3 bottoms which cost $20.00. Adam a
cook bought 3 uniform tops and 2 bottoms which cost $15.00. How much did the
tops cost and how much did the bottoms cost.
3.) Herman decides to take up golf. His golf club membership will cost $450 for the
season and he will be charged $18 for each round of golf that he plays. Herman has
decided not to spend more than $1000 on golf for the season.
a.) Write an inequality that describes the relationship between the maximum
amount Herman wants to spend and the total golf costs for the season.
b.) Solve the inequality to determine the maximum number of rounds of golf he
can play yet not exceed his $1000 limit.
4.) Pizza Palace is running a promotional sale on cinnamon sticks. The hope is to
attract more customers into the shop so that they will also buy a pizza with two toppings
at the regular price. The Pizza Palace will lose $0.78 on very cinnamon stick order. The
profit, however, on each pizza will be $1.32.
a.) "Breaking even" is the worst the Pizza Palace is willing to accept. They want the
losses from the cinnamon sticks to be less than or equal to the profits from the pizzas.
Write an inequality expression for this situation. Let c represent the number of cinnamon
stick orders sold and p the number of pizzas.
b.) Graph the inequality plotting the number of cinnamon stick orders on the horizontal
axis and the number of pizzas on the vertical axis. Indicate the region where the Pizza
Palace will profit from the promotion.
5.) Solve the following system of equations algebraically or graphically:
x 2  y 2  25
3y  4 x  0
6.) Alexandra purchases two doughnuts and three cookies at a doughnut
shop and is charged $3.30. Briana purchases five doughnuts and two
cookies at the same shop for $4.95. All the doughnuts have the same
price and all the cookies have the same price. Find the cost of one
doughnut and find the cost of one cookie.
Answer sheet
Anticipatory Set:
1.)
x  y  20
x  5 y  15
Use the first equation and subtract x from both sides. This will give you
y  20  x .
Substitute this equation into y for the second equation and then solve for x.
x  5 20  x   15
x  2125
.
Then plug x into the second equation and solve for y.
2125
.  5y  15
y  125
.
2.)
5x  3 y  20
3x  2 y  15
Multiply the first equation by 3 and the second by -5.
35x  3y  20
 53x  2 y  15
The x terms will cancel out and then that means you can solve for y.
 1y   15
y  15
Then plug 15 in for y in the first equation and solve for x.
5x  3 * 15  20
x  5
3.) 3x  7  2
3x  9
x 3
4.) 2  5x  3x  14
Subtract 2 from both sides.
 5x  3x  16
Subtract 3x from both sides.
 8x   16
Divide by -8 and remember to flip the sign because you are dividing by a
negative number.
x 2
Developmental Activity:
1.) The two equations are:
12 p  14n  20
7 p  4n  7.50
Multiply the first equation by 7 and the second equation by -12.
712 p  14n  20
 127 p  4n  7.50
The p terms will cancel out and then you can solve for the n term.
50n  50
n1
Now that you have the value for n, plug the value for n into the first equation and
solve for p.
12 p  14 * 1  20
p  .5
So the pens cost fifty cents each and the notebooks cost a dollar each.
2.) The answer will be the same as in question 2 from the “What’s the answer”
worksheet.
3.) a.) Let x = rounds of golf
450 + 18x < 1000
b.) 450 + 18x < 1000
18x < 550
x < 30.55555556
Herman can play at most 30 rounds of golf for the season. Playing a
fractional part of a round of golf is not allowed.
4.) a.) Let c = cinnamon stick orders
p = pizza orders
Losses < Profits
0.78c < 1.32p
c < (1.32/.78)p
c < 1.692p
or
0.78c < 1.32p
(0.78/1.32)c < p
0.591c < p
p > 0.591c
b.) The shaded region above the line identifies where the profits will be
greater than the losses.
5.) x 2  y 2  25
3y  4 x  0
Solve the second equation for y.
y
4x
3
Substitute this value of y into the first equation where y is. Then solve for
x.
2
 4x 
x     25
 3
2
16 x 2
x 
 25
9
2
25x 2
 25
9
x 3
Using the second equation plug 3 in for x and solve for y.
3y  4 * 3  0
y 4
6.) The two equations are:
2d  3c  3.30
5d  2c  4.95
Multiply the first equation by -2 and the second equation by 3.
 2 2d  3c  3.30
3 5d  2c  4.95
The c terms will cancel out and that means you can solve fore the d term.
11d  8.25
d  .75
Now that you have the value for d you can plug it into the first equation and solve
for c.
2*.75  3c  3.30
c  .60
Therefore the cost of one doughnut is 75 cents and the cost of one cookie is 60 cents.