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Activity 3 - 1
Business Checking Account
Objectives
• Solve a system of two linear equations numerically
• Solve a system of two linear equations graphically
• Solve a system of two linear equations using the
substitution method
• Recognize the connections between the three
methods of solution
• Interpret the solution to a system of two linear
equations in terms of the problem’s content
Vocabulary
• System of linear equations – two equations that relate the same
two variables
• Numerical method – using a table of values to see with input
results in the same output for the two equations
• Graphical method – graph the functions and determine the
coordinates of the point of intersection
• Substitution method – an algebraic method using substitution
to reduce the problem to one variable
• Consistent – has exactly one solution (the graphs of the lines
intersect)
• Inconsistent – has no solutions (the graphs of the lines are
parallel)
Solving a System of 2 Linear Equations
• Numerically – by completing a table and noting which
x-value gives you the same y-value
• Graphically – by graphing the equations and finding
their point of intersection
• Algebraically – by using properties of equality to solve
the equations for one variable and then the other
– Substitution method: (also known as elimination)
– Addition method (the emphasis in lesson 3.3)
Business Checking Account
In setting up your part-time business, you have
two choices for a checking account at the local
bank.
MONTHLY FEE
TRANSACTION FEE
REGULAR
$11.00
$0.17 for each transaction
BASIC
$8.50
$0.22 for each transaction in excess of
20
If you anticipate making about 50 transactions each
month, which checking account will be more economical?
Basic
Business Checking Account
MONTHLY FEE
TRANSACTION FEE
REGULAR
$11.00
$0.17 for each transaction
BASIC
$8.50
$0.22 for each transaction in excess of
20
2. Let x represent the number of transactions. Write an equation that
expresses the total monthly cost, C, for the regular account.
C = $11 + $0.17t
3. Let x represent the number of transactions. Write an equation that
expresses the total monthly cost, C, for the basic account. This is
a more complicated equation because the transaction fee does
not apply to the first twenty checks.
C = $8.50
for t ≤ 20
C = $8.50 + $0.22(t – 20) = $0.22t + $4.10 for t > 20
Fill in the Table
• Use the two equations from the previous slide
C = 11 + 0.17x
and
Number of
Transactions
20
50
100
Cost of Regular ($) 14.40 19.50 28.00
Cost of Basic ($)
8.50 15.10 26.10
C = 4.10 + 0.22x
150
36.50
200
45.00
250
53.50
300
62.00
37.10
48.10
59.10
70.10
TI-83 Table Feature
• Use the table feature of your calculator to
determine the x-value that produces two
identical y-values
• TABLE (2nd Graph)
– Put in x values of interest (independent) in table
– Set Y1 = to equation of interest
– Go back to table to read off dependent values
TI-83 Graph
• Use your calculator to graph the functions
y
window:
Xmin: 0
Xmax: 300
Xscl: 10
Ymin = 0
Ymax = 80
Yscl = 10
Xres = 1
x
Graphical Method
• Step 1: Solve both equations for y = …
• Step 2: Put into your calculator (y1 = for one and y2
= for the other) and graph (or graph by hand)
• Step 3: If the lines intersect, then the intersection
point is the solution; if the lines are parallel, then
there is no solution; and if the lines are the same,
then there are an infinite number of solutions
• Step 4: Write the solution (intersection point) (use
TRACE on your calculator to find it)
Graphical Method - Solutions
• Consistent
Inconsistent
One Solution
y
 Solutions
No Solution
y
x
y
x
x
Substitution Method
• Step 1: Solve one or both equations for a variable
(both x = … or both y = …)
• Step 2: Substitute the expression that represents
the variable in one equation for that variable in the
other equation
• Step 3: Solve the resulting equation for the
remaining variable
• Step 4: Substitute the value from step 3 into one of
the original equations and solve for the other
variable
Substitution Example
Given:
y + x = 9 and y = 3x – 3
• Step 1: y = 9 – x
and y = 3x – 3
• Step 2: 9 – x = 3x – 3
• Step 3:
9
+3
12
3
+x
+x
= 4x – 3
+3
= 4x
= x
• Step 4: y + 3 = 9
y=6
or
or
y = 3(3) – 3
y=9–3=6
Problem 1
Given: y = 3x – 10 and y = 5x + 14
Solve using substitution
• Step 1: y = 3x – 10
and y = 5x + 14
• Step 2: 3x – 10 = 5x + 14
• Step 3:
+ 10
+ 10
3x = 5x + 24
- 3x
- 3x
0 = 2x + 24
-24 = 2x
-12 = x
• Step 4: y = 3(-12) – 10 = -46
Problem 2
Use the substitution method to sol the following
system of checking account cost functions:
C = 0.17x + 11
and
C = 0.22x + 4.10
• Step 1: C = 0.17x + 11
and y = 0.22x + 4.10
• Step 2: 0.17x + 11 = 0.22x + 4.10
• Step 3: -.17x
-.17x
11 = 0.05x + 4.10
- 4.1
- 4.1
6.9 = 0.05x
138 = x
• Step 4: 0.17(138) + 11 = 23.46 + 11 = 34.46
Summary and Homework
• Summary
– The solution of a system of equations is the set of all
ordered pairs that satisfy both equations.
– The three standard methods for solving a system of
equations are the Numerical method, Graphical
method and Substitution method.
– A linear system is consistent if there is at least one
solution, the point of intersection of the graphs.
– A linear system is inconsistent if there is no solution -that is, the lines are parallel.
• Homework
– 1- 4, 7, 8