1. Linear
5-1
Equations
Apps.
Applications of Linear Equations
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2. 5-2
Linear
Equations
Apps.
Learning Objectives
After completing this chapter, you will be able to:
LO 1
Graph a linear equation
in two variables
LO 2
Express a linear equation in the
slope-intercept form
LO 3
Solve two equations in two unknowns
by the graphical method
Also
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3. 5-3
Linear
Learning Objectives
Equations
Apps.
Distinguish between fixed costs and
variable costs
Perform linear Cost-Volume-Profit and
LO 4
LO 5
B
break-even analysis employing:
- The algebraic approach of solving
the cost and revenue functions
- A breakeven chart
C
- The contribution margin approach
A
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4. 5-4
Linear
Equations
Apps.
LO 1
Step 1
Step 2
Step 3
Graphing a Linear Equation
Graph the equation 3x – 2y = 8
over the range x=0 to x=8
Rearrange the equation to solve for y
Substitute a few values for x, and
calculate the corresponding values for y
Plot and connect the points on the
xy graph
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3x – 2y =8
– 2y = 8 - 3x
y = 8 - 3x
-2
y = 1.5x - 4
x y
0 -4
3 0.5
6
5
…continue

5. Linear
5-5
Graphing a Linear Equation
y = 1.5x - 4
x y
0 -4
3
0.5
6
5
Equations
Apps.
Step 3
7
6
5
4
3
Y 2
1
0
-1
0
-2
-3
-4
Plot and connect the points on the
xy graph
y = 1.5x - 4
1
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2
3
4
5
X
6
7
8
9
10

6. 5-6
Linear
Graphing a Linear Equation
Equations
Apps.
y = 1.5x - 4
LO 2
This is called the slope–intercept form
of this linear equation
Where 1.5 is the slope, and
-4 is the y-intercept
1
0
-1
Y -2
-3
-4
0
1
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2
3
4
5
6
7
8
y = 1.5x - 4
X
9
10

7. Linear
Equations
Apps.
LO 3
Step 1
5-7
Solving 2 equations in 2 unknowns
graphically
Find the point of intersection for the
equations 3x – 2y = 8 and 4x+2y = 6
Rearrange each equation to solve for y
y = 1.5x - 4
y = -2x + 3
Step 2
Step 3
4x+2y = 6
2y = 6- 4x
y = 6 – 4x
2
Substitute a few values for x, and
calculate the corresponding values for y
in each equation
Plot and connect the points on the xy
graph – the point where the lines cross in
the point of intersection, your solution
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y = -2x + 3
x y
0 3
3 -3
2 -1

8. Linear
Equations
Apps.
5-8
Solving 2 equations in 2 unknowns
graphically
Find the point of intersection for the equations
3x – 2y = 8 and 4x+2y = 6
Y
7
6
5
4
3
2
1
0
-1
-2
-3
-4
(2, -1)
0
1
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2
3x-2y=8
or
y= 1.5x - 4
y = -2x + 3
x y
0 3
3 4 5 6 7 8 9 10
3 -3
4x+2y=6
or y = -2x + 3
2 -1

9. 5-9
Linear
Equations
Apps.
Cost
Analysis
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10. 5 - 10
Linear
Equations
Apps.
Terminology
LO 4
Fixed Costs
Business Costs
Business Expenses
…do NOT change if
sales increase or decrease
e.g. rent, property taxes, some
forms of depreciation
Variable Costs
…do change in direct
proportion to sales volume
e.g. material costs, direct
labour costs
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11. 5 - 11
Linear
Equations
Apps.
Terminology
Break Even Point
… is the point at which
neither
a Profit or Loss is made
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12. 5 - 12
Linear
Equations
Apps.
LO 5C
Terminology
Contribution Margin
…is the dollar amount that is found by
deducting ALL Variable Costs
from Net Sales
and ‘contributes’ to meeting
Fixed Costs and making a ‘Net Profit’.
Contribution Rate
…is the dollar amount expressed as
a percent (%) of Net Sales
A Contribution Margin statement
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13. 5 - 13
Linear
Equations
Apps.
Terminology
A Contribution Margin Statement
Net Sales(Price * # Units Sold)
Less: Variable Costs
Contribution Margin
Less: Fixed Costs
Net Income
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$
x
x
x
x
x
%
100
x
x
x
x

14. 5 - 14
Linear
Scenario 1
Equations
Apps.
Market research for a new product indicates that the product
can be sold at $50 per unit.
Cost analysis provides the following information:
Fixed Costs per period = $8640
Variable Costs = $30 per unit.
Production Capacity per period = 900 units
uestion: How much does the sale of an additional
unit of a firm’s product contribute
towards increasing its net income?
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15. 5 - 15
Linear
Equations
Apps.
Formulae
- To Find -
Contribution Margin
Contribution Rate
*Break Even Point:
...in Units (x)
...in Sales $
...in % of Capacity
CM = SP - VC
CR = CM/SP * 100%
x = FC / CM
$x = (FC / CM)* SP
BEPin Units/PC*100
* At Break Even, Net Profit or Loss = 0
Applying
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Formulae

16. Linear
Equations
Apps.
5 - 16
Applying the Formulae
As in the previous scenario, the new product can be sold at $50
per unit. Costs are as follows: Fixed Costs are $8640 for the
period , Variable Costs are $30 per unit, and the Production
Capacity is 900 units per period.
CM = SP - VC
CR = CM/SP * 100%
Break Even Point:
Units x = FC / CM
In $ x = (FC / CM)* SP
BEPin units / PC*100
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= $50 - $30 = $20
= $20/$50 * 100 = 40%
= $8640/$20 = 432 Units
= ($8640/$20)* $50 = $21,600
= 432/ 900*100 = 48% of
Capacity

17. 5 - 17
Linear
Equations
Apps.
Scenario 2
The Lighting Division of Seneca Electric Co. plans to
introduce a new street light based on
the following accounting information:
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
uestion: Calculate the breakeven point (BEP)
…in units
…in dollars
…as a percent of capacity
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18. 5 - 18
Linear
Equations
Apps.
Scenario 2
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
…in units
Break Even Point
= FC / CM
= $3136/ 28 = 112 Units
SP – VC = CM
$185 – 157 = $28
…in dollars
= (FC / CM)* SP
= ($3136/ 28) * $185 = $20720
…as a percent of capacity = BEPin units/PC*100
= 112/320 * 100 = 35% of Capacity
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19. 5 - 19
Linear
Scenario 2 -1
Equations
Apps.
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
$2688
Determine the BEP as a % of capacity if
FC are reduced to $2688.
Formula
Step 1… Find CM
SP = $185
VC = - 157
CM $ 28
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=BEPin units/PC*100
Step 2… Find
Step 3… Find
BEP in units
= FC/CM
= $2688/ $28
= 96 Units
% of Capacity
=BEPin units /PC*100
= 96/320*100
= 30% of Capacity

20. 5 - 20
Linear
Scenario 2 -2
Equations
Apps.
FC = $4588 VC = $157 SP= $185 Capacity = 320 units
$3136
$148
VC =SP*80% = $148
Determine the BEP as a % of capacity if
FC are increased to $4588, and VC reduced to 80% of SP.
Formula
Step 1… Find CM
SP = $185
VC = - 148
CM $ 37
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= BEPin units /PC*100
Step 2… Find
Step 3… Find
BEP in units
= FC/CM
= $4588/ $37
= 124 Units
% of Capacity
=BEPin units /PC*100
= 124/320*100
= 39% of Capacity

21. 5 - 21
Linear
Scenario 2 -3
Equations
Apps.
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
$171
Determine the BEP as a % of capacity if
SP is reduced to $171.
Formula
Step 1… Find CM
SP = $ 171
VC = -157
CM $ 14
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= BEPin units /PC*100
Step 2… Find
BEP in units
= FC/CM
= $3136/ $14
= 224 Units
Step 3… Find
% of Capacity
=BEPin units /PC*100
= 224/320*100
= 70% of Capacity

22. 5 - 22
Linear
Scenario 2 -4
Equations
Apps.
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
Determine the NI if 134 units are sold!
Formula
Step 1… Find CM
SP = $185
VC = - 157
CM $ 28
CM of $28 per unit
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NI = #Units above BEP*CM
Step 2… Find
BEP in units
= FC/CM
= $3136/$28
= 112 Units
Units
Sold
134
BEP
112
Over BEP 22
Company had a NI of 22* $28 = $616.

23. 5 - 23
Linear
Scenario 2 -5
Equations
Apps.
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
What unit sales will generate NI of $2000?
Formula
Step 1… Find CM
SP = $185
VC = - 157
CM $ 28
CM of $28 per unit
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#Units above BEP = NI/CM
Step 2… Find
BEP in units
= FC/CM
= $3136/$28
= 112 Units
NI/CM
= $2000/$28 per Unit
= 72 Units above
Break Even
72 Units + 112 BEP Units
= Total Sales Units = 184

24. 5 - 24
Linear
Scenario 2 -6
Equations
Apps.
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
What are the unit sales if there is a Net Loss of $336?
Formula
Step 1… Find CM
SP = $185
VC = - 157
CM $ 28
CM of $28 per unit
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# Units below BEP = (NI)/CM
Step 2… Find
BEP in units
= FC/CM
= $3136/$28
= 112 Units
(NI)/CM
= ($336)/$28 per Unit
= 12 Units below
Break Even
112 BEP - 12 Units Below
= Total Sales Units = 100

25. 5 - 25
Linear
Scenario 2 -7
Equations
Apps.
FC = $3136 VC = $157 SP= $185 Capacity = 320 units
The company operates at 85% capacity. Find the Profit or Loss.
Step 1… Find CM
SP = $185
VC = - 157
CM $ 28
Step 2… Find
BEP in units
= FC/CM
= $3136/$28
= 112 Units
CM of $28 per unit
Formula
320*.85
= 272
Units
Production 272
BEP
112
Over BEP 160
# units above BEP *CM = NI
160 Units * $28 = Profit $4480
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26. 5 - 26
Linear
Case
Equations
Apps.
The Marconi Co. year end operating results were as follows:
Total Sales of $375000
Operated at 75% of capacity
Total Variable Costs were $150000
Total Fixed Costs were $180000
What was Marconi’s BEP expressed in dollars of sales?
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27. 5 - 27
Linear
Equations
Apps.
Case
The Marconi Co. year end operating results were as follows:
Total Sales of $375000
Operated at 75% of capacity
Total Variable Costs were $150000
Total Fixed Costs were $180000
What was Marconi’s BEP expressed in dollars of sales?
What information is needed to calculate the $BEP?
1. Number
of Units sold
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2. VC
per Unit
3. CM
4. BEP
in $

28. 5 - 28
Linear
Case
Equations
Apps.
The Marconi Co. year end operating results were as follows:
Total Sales of $375000
Operated at 75% of capacity
Total Variable Costs were $150000
Total Fixed Costs were $180000
What was Marchoni’s BEP expressed in dollars of sales?
1. Number
of Units sold
Let SP = $1 and X be the Number of $1 Units sold
Sales of $375 000 = 375000 Total Units sold
2. VC
per Unit
Total VC
=
Total Unit Sales
3. CM
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SP $1.00
VC
.40
CM $ .60
$150000
375000
= $0.40perunit

29. 5 - 29
Linear
Case
Equations
Apps.
The Marconi Co. year end operating results were as follows:
Total Sales of $375000
Operated at 75% of capacity
Total Variable Costs were $150000
Total Fixed Costs were $180000
What was Marchoni’s BEP expressed in dollars of sales?
4. BEP
in $
$BEP = (FC/CM)*SP
= ($180000/0.60)*$1.00
# Of Units
= (300000)*1.00
= $300 000 $BEP
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30. 5 - 30
Linear
Equations
Apps.
This completes Chapter 5
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