Download Section 4.4 Problem Solving Using Systems of Equations

Math 123 - Section 4.4 - Problem Solving Using Systems of Equations - Part I - Page 1
Section 4.4
Problem Solving Using Systems of Equations - Part I
I.
In general
A.
Remember to
1.
Define the variables.
2.
Write a system of equations.
3.
Solve the system.
4.
Answer the question with a complete sentence.
B.
When you define your variables, remember to generalize. It is when you write the
system of equations that you will get specific.
II.
Number Problems
A.
Remember all the "buzz words" that we saw in Section 2.5.
B.
Examples - Let x represent one number and let y represent the other. Use the given
conditions to write a system of equations. Solve the system and find the numbers.
1.
The sum of three times a first number and twice a second number is 43. If
the second number is subtracted from twice the first number, the result is 4.
Find the numbers. (Page 319, #4)
We begin by defining our variables:
Let x = The first number, Let y = The second number
Let's work on the first equation.
"sum" means add; "times" means multiply; "twice" means multiply by 2; "is" means =
So our first equation is:
3x + 2y = 43
Now we can work on our second equation.
"subtracted from" means subtract, switch the order!; "twice" means multiply by 2; "is"
means =
So our second equation is:
2x  y = 4
Now we need to write a system of equations describing this situation:
E
:
 1 3 x  2y  43

E2 : 2 x  y  4
Use Elimination! Multiply E2 by 2 and add the equations.
3 x  2y
 4 x  2y
7x
x=5
3(5) + 2y = 43
15 + 2y = 43
2y = 28
y = 14
Answer:
 43
 8
 35
Divide both sides by 7 to isolate the variable.
Back-substitute into E1 to find y.
Multiply.
Subtract 15 from both sides to isolate the variable term.
Divide both sides by 2 to isolate the variable.
Answer the question.
The numbers are 5 and 14.
© Copyright 2012 by John Fetcho. All rights reserved
Math 123 - Section 4.4 - Problem Solving Using Systems of Equations - Part I - Page 2
2.
One Snickers bar and two Reese's Peanut Butter Cups contain 737 calories.
Two Snickers bars and one Reese's Peanut Butter Cup contain 778 calories.
Find the caloric content of each candy bar. (Page 320, #8)
First, we have to define our variables.
Let x = The number of calories in a Snickers bar.
Let y = The number of calories in a Reese's Peanut Butter Cup
Now let's work on the first equation, which comes from the first sentence.
x + 2y = 737
Next we work on the second equation, which comes from the second
sentence.
2x  y = 778
So our system is:
 x  2 y  737

2 x  y  778
Use Elimination! Multiply the first equation by 2.
 2 x  4 y

 2x  y
3y = 696
y = 232
  1474

778
x + 2(232) = 737
x + 464 = 737
x + 464 – 464 = 737 – 464
x = 273
Answer:
3.
Add the two equations.
Divide both sides by 3.
Substitute 232 in for y in one of the original
equations to find x. I choose the first equation.
Multiply 2 & 232.
Subtract 464 from both sides.
Simplify.
Answer the question.
A Snickers bar has 273 calories and a Reese's Peanut Cup has
232 calories.
Now you try one:
One Mr. Goodbar and two Mounds bars contain 780 calories. Two Mr. Goodbars
and one Mounds bar contain 786 calories. Find the caloric content of each item.
(Page 320, #7)
Answer:
A Mr. Goodbar has 264 calories and a Mounds bar has 258
calories.
III.
Money Problems
A.
General Principles
1.
Total Value = (Unit Value)(Number of Things)
2.
Remember that if I have 5-$20 bills, I have much more than $5! The value
and the number are two very different things!
© Copyright 2012 by John Fetcho. All rights reserved
Math 123 - Section 4.4 - Problem Solving Using Systems of Equations - Part I - Page 3
B.
Examples - Solve each of the following.
1.
A restaurant purchased eight tablecloths and five napkins for $106. A week
later, a tablecloth and six napkins were bought for $24. Find the cost of one
tablecloth and one napkin, assuming the same prices for both purchases.
(Page 320, #12)
Let's begin by defining our variables:
Let x = The price of a tablecloth
Let y = The price of a napkin
Let's write an equation for the first purchase, using this idea:
(Number Tablecloths)(Price Tablecloths) + (Number Napkins)(Price Napkins) = Total Spent
8x + 5y = 106
For the second purchase, we have:
x + 6y = 24
Now we need to set up a system of equations, using the idea of total value
outlined above:
 E1 : 8 x  5 y  106

E2 : x  6 y  24
Use Elimination! Multiply E2 by 8 and add:
 106
8x 
5y
 8 x  48 y  192
Divide both sides by 43 to isolate the variable.
43 y  86
y=2
Substitute into E2 to find x.
x + 6(2) = 24
Multiply to simplify the left-hand side.
x + 12 = 24
Subtract 12 from both sides to isolate the variable.
x = 12
Answer the question.
Answer:
2.
The tablecloths cost $12 each. The napkins cost $2 each.
You are choosing between two plans at a discount warehouse. Plan A offers
an annual membership fee of $300 and you pay 70% of the manufacturer's
recommended list price. Plan B offers an annual membership fee of $40 and
you pay 90% of the manufacturer's recommended list price. How many dollars
of merchandise would you have to purchase in a year to pay the same amount
under both plans? What will be the cost for each plan? (Page 321, #20)
This problem employs the idea that
Total Cost = Fixed Price + (Unit Cost)(Number Bought)
In each of these cases, the fixed cost will be the yearly membership fee. The
unit cost will be the percentage. The number bought we don't know and will
have to find out.
Defining our variables:
Let T = The total cost of each plan.
Plan A
T = 300 + 0.70p
Let p = The manufacturer's list price
Plan B
T = 40 + 0.90p
© Copyright 2012 by John Fetcho. All rights reserved
Math 123 - Section 4.4 - Problem Solving Using Systems of Equations - Part I - Page 4
Writing our system of equations:
 E1 : T  300  0.70 p

E2 : T  40  0.90 p
Use Substitution! Set E1 = E2 and solve for p.
300 + 0.70p = 40 + 0.90p
Subtract 0.70p from both sides to get the variable
terms on one side.
300 = 40 + 0.20p
Subtract 40 from both sides to isolate the variable
term.
260 = 0.20p
Divide both sides by 0.20 to isolate the variable.
1300 = p
Substitute into E1 (or E2) to find the total cost.
T = 300 + 0.70(1300)
T = 300 + 910 = 1210
Answer the question.
Answer:
You would have to buy $1300 in merchandise per year for the
plans to cost the same. That cost would be $1210 per year.
3.
Now you try one: You are choosing between two long-distance telephone
plans. Plan A has a monthly fee of $20 with a charge of $0.05 per minute for
all long-distance calls. Plan B has a monthly fee of $5 with a charge of $0.10
per minute for all long-distance calls. (Page 320, #17)
a.
For how many minutes of long-distance calls will the costs for the two plans be
the same? What will be the cost for each plan?
Answer:
b.
If you make approximately 10 long-distance calls per month, each averaging
20 minutes, which plan should you select? Explain your answer.
Answer:
4.
After 300 minutes, both plans will cost $35.
Choose Plan B since 10(20) = 200 minutes, and Plan B would cost
less than Plan A.
On a special day, tickets for a minor league baseball game cost $5 for adults
and $1 for students. The attendance that day was 1281 and $3425 was
collected. Find the number of each type of ticket sold. (Page 321, #22)
First, the problem is talking about two different things: the total number of
people attending and the total amount of money collected. So our two
equations will have to reflect that fact. One equation will look at the number
of people in attendance; the other will look at the amount of money collected.
So let's define our variables, using the number of each group in attendance as
our guide.
Let x = The number of adults
Let y = The number of students
Number of adults + Number of students = Total Attendance
x + y = 1281
Remember that Total Cost = (Cost of a Ticket)(Number Sold)
(Cost of Adult)(Number of Adults) + (Cost of Student)(Number of Students) = Total
© Copyright 2012 by John Fetcho. All rights reserved
Math 123 - Section 4.4 - Problem Solving Using Systems of Equations - Part I - Page 5
5x + 1y = 3425
So putting these together as a system, we have:
x  y

5 x  y
 1281
 3425
We will use elimination and I think that eliminating y is the
way to go. So I want to multiply the first equation by 1.
 x   y   1281

 5 x  y  3425
4x = 2144
x = 536
536 + y = 1281
y = 745
Now add the two equations.
Divide both sides of the equation by 4.
Now substitute into the first equation to find y.
Subtract 536 from both sides of the equation.
Now answer the question.
Answer: There were 536 adults and 745 students.
© Copyright 2012 by John Fetcho. All rights reserved