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If...X+5 = $200. What does X =?. Applications of Linear Equations. Chapter 5. Learning Objectives. After completing this chapter, you will be able to:. Solve two linear equations with two variables. LO 1. LO 2. Solve problems that require setting up linear equations with two variables.

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Applications of linear equations

If...X+5 = $200

What does X =?

Applications of Linear Equations

Chapter 5


Applications of linear equations

Learning Objectives

After completing this chapter, you will be able to:

Solvetwolinear equationswith two variables

LO 1.

LO 2.

Solveproblems that require setting up linear equations withtwo variables

Also


Applications of linear equations

Learning Objectives

Perform linearCost-Volume-Profit and break-even analysis employing:

LO 3.

- The contribution margin approach

A.

- The algebraic approach of solving the cost and revenue functions

B.


Applications of linear equations

Substitute y = 2

Multiply by 2

Subtract

Divide by -5

Solving Two Equations with Two Unknowns

LO 1.

Equations

2x – 3y = – 6

x + y = 2

(A) Solve for y

(B) Solve for x

(A) Solve for y

2x – 3y = – 6

x + y = 4

2x + 2y = 4

-5y = -10

y = 2

(B) Solve for x

2x – 3(2) = – 6

2x – 3y = – 6

2x – 6 = – 6

2x = + 6 – 6

Check…

x = 0


Applications of linear equations

2x – 3y = – 6

x + y = 2

Equations

=

LeftSide

RightSide

Show

Note

Substituting

= 2

= – 6

LS =RS

LS =RS

Solving Two Equations with Two Unknowns

You should always checkyour answer by substituting the valuesinto each of the equations!

x = 0y=2

Equation 1

Equation 2

LeftSide

RightSide

LeftSide

RightSide

= 2x – 3y

= x + y

= 2(0) – 3(2)

= 0 + 2

= – 6

= 2


Applications of linear equations

Setting up

linear equations

with Two Variables

LO 2.


Applications of linear equations

Setting up

linear equations

with Two Variables

Q

York Daycare purchases the same amount of

milkand orange juice each week. After price increases from $1.10 to $1.15 per litre for milk, andfrom $0.98 to$1.14 per canof frozen orange juice, the weekly bill rose from $84.40 to $91.70.

How manylitres of milkand cans of orange juiceare purchased each week?


Applications of linear equations

Setting up

linear equations

with Two Variables

Purchases

Let x = # litres of milk Let y = # cans of orange juice

Equations

After price increases from $1.10 to $1.15 perlitre of milk,

A.

Development of…

and from

$0.98 to$1.14 per canof frozen orange juice,

(1)

1.10x+

0.98y

= 84.40

B.

1.15x +

1.14y

= 91.70

(2)

the weekly bill rose from $84.40 to $91.70.

C.

Solving…


Applications of linear equations

Setting up

linear equations

with Two Variables

Equation

Equation

(1)

(2)

Eliminate x by Dividing by 1.10

Eliminate x by Dividing by 1.15

Let x = # litres of milk Let y = # cans of orange juice

1.10x+ 0.98y = 84.40

(1.10x+ 0.98y)/1.10 = 84.40/1.10

x+ 0.8909y = 76.73

1.15x + 1.14y = 91.70

(1.15x+ 1.14y)/1.15 = 91.70/1.15

x+ 0.9913y = 79.74

…continue


Applications of linear equations

Setting up

linear equations

with Two Variables

Equation

Equation

Equation

(1)

(2)

(1)

1.10x+ 0.98y = 84.40

Subtract

Substitute into

Proof

x+ 0.8909y = 76.73

x+ 0.9913y = 79.74

.1004y = 3.01

y = 29.98 i.e. 30 cans

1.10x+ 0.98(29.98) = 84.40

1.10x+ 29.38 = 84.40

1.10x= 84.40 -29.38

1.10x= 55.02

x= 50.02 i.e. 50 litres


Applications of linear equations

Setting up

linear equations

with Two Variables

Quantity

Price

$

= New Weekly Cost to Purchase

$91.70

Proof

Litres of Milk

50

$1.15

$57.50

Cans of Orange Juice

30

1.14

34.20


Applications of linear equations

LO 3.

Cost

Analysis


Applications of linear equations

Terminology

Fixed Costs

Business Costs

Business Expenses

or

Variable Costs

…do NOT change if sales increase or decrease

e.g. rent, property taxes, some forms of depreciation

…do change in direct proportion to sales volume e.g. material costs, direct labour costs


Applications of linear equations

Terminology

BreakEvenPoint

… is the point at which neither a Profit or Loss is made


Applications of linear equations

Terminology

A Contribution Margin statement

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


Applications of linear equations

Terminology

A Contribution Margin Statement

$ %

NetSales(Price * # Units Sold) x100

Less: Variable Costs x x

Contribution Margin x x

Less: Fixed Costs x x

Net Income x x


Applications of linear equations

Scenario 1

Q

uestion:

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

How much does the sale of an additional unit of a firm’s product contributetowards increasing its net income?


Applications of linear equations

Formulae

Applying

Formulae

- To Find -

Contribution Margin

CM = S - VC

Contribution Rate

CR=CM/S* 100%

*BreakEven Point:

x =FC / CM

...in Units (x)

$x =(FC / CM)*S

...in Sales $

...in % of Capacity

BEPin Units/PC*100

* At BreakEven,NetProfit or Loss = 0


Applications of linear equations

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 $30per unit, and theProductionCapacity is 900 units per period.

CM = S - VC

= $50 - $30 = $20

CR=CM/S* 100%

= $20/$50 * 100 = 40%

BreakEven Point:

Unitsx =FC / CM

= $8640/$20 = 432 Units

In $ x =(FC / CM)*S

= ($8640/$20)* $50 = $21,600

BEPin unitsPC*100

= 432/ 900*100 = 48% of Capacity


Applications of linear equations

Scenario 2

Q

uestion:

The Lighting Division of Seneca Electric Co. plans to introduce a new street light based on the following accounting information:

FC= $3136VC= $157.S= $185 Capacity = 320 units

Calculate the breakeven point (BEP)

…in units

…in dollars

…as a percent of capacity


Applications of linear equations

Scenario 2

S – VC = CM

$185 – 157 = $28

FC= $3136VC= $157.S= $185 Capacity = 320 units

Break Even Point

…in units

=FC / CM

= $3136/

28 =

112 Units

…in dollars

=(FC / CM)*S

= ($3136/

28) *

$185 = $20720

…as a percent of capacity

= BEPin units/PC*100

= 112/320 * 100 = 35% of Capacity


Applications of linear equations

Scenario 2 -1

FC= $3136VC= $157.S= $185Capacity = 320units

$2688

Determine the BEP as a % of capacity if FC are reduced to $2688.

Formula

=BEPin units/PC*100

Step 3… Find

% of Capacity

Step 2… FindBEP in units

Step 1… FindCM

=BEPin units /PC*100

S = $185

= FC/CM

VC= - 157

CM $28

= $2688/$28

= 96/320*100

= 96 Units

= 30% of Capacity

=


Applications of linear equations

Scenario 2 -2

$4588

Formula

= BEPin units /PC*100

FC= $3136VC= $157S= $185Capacity = 320units

$148

VC =S*80% = $148

Determine the BEP as a % of capacity if FC are increased to $4588, and VC reduced to80%ofS.

Step 3… Find

% of Capacity

Step 2… FindBEP in units

Step 1… FindCM

=BEPin units /PC*100

S = $185

= FC/CM

VC= - 148

CM $ 37

= $4588/$37

= 124/320*100

= 124 Units

= 39% of Capacity


Applications of linear equations

Scenario 2 -3

Formula

= BEPin units /PC*100

FC= $3136VC= $157S= $185Capacity = 320units

$171

Determine the BEP as a % of capacity if S is reduced to$171.

Step 3… Find

% of Capacity

Step 2… FindBEP in units

Step 1… FindCM

=BEPin units /PC*100

S = $ 171

= FC/CM

VC= -157

CM $ 14

= $3136/$14

= 224/320*100

= 224 Units

= 70% of Capacity


Applications of linear equations

Scenario 2 -4

Formula

NI = #Units aboveBEP*CM

FC= $3136VC= $157S= $185Capacity = 320units

Determine the NI if 134 units are sold!

Units

Sold 134

BEP 112

OverBEP 22

Step 2… FindBEP in units

Step 1… FindCM

S = $185

= FC/CM

VC= - 157

CM $ 28

= $3136/$28

= 112 Units

CM of $28 per unit

Company had a NI of 22

* $28 = $616.


Applications of linear equations

Scenario 2 -5

Formula

#Units aboveBEP =NI/CM

Step 2… FindBEP in units

Step 1… FindCM

S = $185

= FC/CM

VC= - 157

CM $ 28

= $3136/$28

= 112 Units

FC= $3136VC= $157S= $185 Capacity = 320units

What unit sales will generate NI of $2000?

NI/CM

= $2000/$28per Unit

= 72 Units above Break Even

CM of $28 per unit

72 Units + 112 BEP Units

= Total Sales Units = 184


Applications of linear equations

Scenario 2 -6

Formula

# Units belowBEP =(NI)/CM

Step 2… FindBEP in units

Step 1… FindCM

S = $185

= FC/CM

VC= - 157

CM $ 28

= $3136/$28

= 112 Units

FC= $3136VC= $157S= $185Capacity = 320units

What are the unit sales if there is a NetLoss of $336?

(NI)/CM

= ($336)/$28per Unit

= 12 Units below Break Even

CM of $28 per unit

112 BEP- 12 Units Below

= Total Sales Units = 100


Applications of linear equations

Scenario 2 -7

Formula

# units above BEP *CM = NI

Step 2… FindBEP in units

Step 1… FindCM

S = $185

= FC/CM

VC= - 157

CM $ 28

= $3136/$28

= 112 Units

FC= $3136VC= $157S= $185Capacity = 320units

272

The company operates at 85% capacity. Find the Profit or Loss.

320*.85

= 272

Units

Production 272

BEP 112

OverBEP 160

CM of $28 per unit

160 Units * $28 = Profit $4480


Applications of linear equations

Case

The Marconi Co. year end operating results were as follows:

Total Sales of $375000

Operated at 75% of capacity

TotalVariable Costs were $150000

TotalFixed Costs were $180000

What was Marconi’s BEP expressed in dollars of sales?


Applications of linear equations

Case

The Marconi Co. year end operating results were as follows:

Total Sales of $375000

Operated at 75% of capacity

TotalVariable Costs were $150000

TotalFixed 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

2. VCper Unit

3. CM

4. Total Costs

5. BEP in $


Applications of linear equations

Case

The Marconi Co. year end operating results were as follows:

Total Sales of $375000

Operated at 75% of capacity

TotalVariable Costs were $150000

TotalFixed Costs were $180000

What was Marchoni’s BEP expressed in dollars of sales?

1. Number of Units sold

Let S = $1 and X be the Number of $1 Units sold

Sales of$375 000 = 375000 Total Units sold

$150000

375000

2. VCper Unit

Total VC

Total Unit Sales

= $0.40pu

=

S $1.00

VC .40

CM $ .60

3. CM


Applications of linear equations

Case

The Marconi Co. year end operating results were as follows:

Total Sales of $375000

Operated at 75% of capacity

TotalVariable Costs were $150000

TotalFixed Costs were $180000

What was Marchoni’s BEP expressed in dollars of sales?

# Of Units

= $300000

$BEP

4. Total Costs

TC = FC + VC

= $180 000 + 0.40X

$BEP =(FC/CM)*S

5. BEP in $

= ($180000/0.60)*$1.00

= (300000)*$1.00


Applications of linear equations

This completes Chapter 5


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