CHAPTER 5 – Equations Instructor: Dr.Gehan Shanmuganathan.

CHAPTER 5 – Equations

Instructor:

Dr.Gehan Shanmuganathan

Business Math, Ninth EditionCheryl Cleaves, Margie Hobbs & Jeffrey Nobel

© 2012 Pearson Education, Inc.Upper Saddle River, NJ 07458 All Rights Reserved

Learning Outcomes

Solve equations usingmultiplication or division.

Solve equations usingaddition or subtraction.

Solve equations usingmore than one operation.

Solve equations containing multipleunknown terms.

Solve equations containing parentheses.

Solve equations that are proportions.

5-1



Equations

5-1-1Section 5-1

Solve equations usingmultiplication or division

An equation is a mathematical statement in which two quantities are equal.

Solving an equation means finding the valueof an unknown.

Example:

8x = 24To solve this equation, the

value of x must be discovered.

Division is used to solve this equation.



EquationsSection 5-1


Letters, such as (x,y,z) represent unknown amounts and are called

unknowns or variables.

4x = 16

The numbers are calledknown or given amounts.





Any operation performed on one side of the equation must be performed on the otherside of the equation as well.

– If you “multiply by 2” on one side, you must“multiply by 2” on the other side.

– If you “divide by 3” on one side, you must also“divide by 3” on the other side.





STEP 2Use division to divide both sides by 8.

STEP 3Simplify: x = 3

Isolate the unknown value and determineif multiplication or division is needed.

STEP 1

8x = 24

3 x 8 = 24

HOW TO:




Find the value of an unknownusing multiplication

Multiply both sides by 3 to isolate a.

The left side becomes 1a or a.

The right side becomes theproduct of 6 x 3, or 18.

a = 18

HOW TO:

= 63

aFind the value of a:




STEP 2Perform the same operation to both sides.

STEP 3Isolate the variable and solve.

Determine which operation is needed.STEP 1

2b = 40

Division

Divide both sides by 2.

40 = = 20

2b

An Example…




Solve equations usingaddition or subtraction5-1-2

Adding or subtracting any number from one side must be carried out on the other side as well.

– Subtract “the given amount” from both sides.

Would solving 4 + x = 16 require additionor subtraction of “4” from each side?

Subtraction




Solve equations usingaddition or subtraction

STEP 2Use subtraction to isolate x.

STEP 3Simplify: x = 6

Isolate the unknown value and determineif addition or subtraction is needed.

STEP 1

4 + x = 10

HOW TO:




STEP 2Perform the same operation to both sides.

STEP 3Isolate the variable and solve.

Determine which operation is needed.STEP 1

b - 12 = 8

Addition

Add 12 to both sides

An Example…

b = 8 + 12 = 20




Solve equations usingmore than one operation5-1-3

Isolate the unknown value.

– Add or subtract as necessary first.

– Multiply or divide as necessary second.

Identify the solution.

– The number on the side opposite the unknown.

Check the solution by “plugging in” thenumber using the original equation.



When two or more calculations are written symbolically, the operations are performed according to a specified order of operations.

– First — perform multiplication and division as theyappear from left to right.

– Second — perform addition and subtraction as theyappear from left to right.


Order of operations



To solve an equation, undo the operations, working in reverse order

– First — undo the addition or subtraction.

– Second — undo multiplication or division.


Order of operations




STEP 2Divide each side by 7.

STEP 3Verify by plugging in 5 in place of x .

Undo the addition by subtracting 4 from each side.STEP 1

7x + 4 = 39

7x = 35

An Example…

35= = 5

7 x

7 (5) + 4 = 39 35 + 4 = 39



In some equations, the unknown value may occur more than once.

The simplest instance is when the unknown value occurs in two addends, such as 3a + 2a = 25

– Add the numbers in each addend (2+3).

– Multiply the sum by the unknown (5a = 25).

– Solve for a (a = 5).


5-1-4Solve equations containing

multiple unknown terms




STEP 2Undo the subtraction.

STEP 4Check by replacing a with 7.

Combine the unknown value addends.STEP 1

Find a if: a + 4a – 5 = 39

a + 4a = 5a 5a – 5 = 30

An Example…

5a = 35

STEP 3Undo the multiplication.a = 7

7 + 4(7) = 35Correct!



Eliminate the parentheses.

– Multiply the number just outside the parenthesesby each addend inside the parentheses.

– Show the resulting products as addition orsubtraction, as indicated

Solve the resulting equation.


5-1-5 Solve equations containing parentheses




STEP 2Show the resulting products.

STEP 3Check by replacing a with 7.

Multiply 6 by each addend.STEP 1

Solve: 6A + 2 = 24

6 multiplied by A + 6 multiplied by 2

An Example…

6A + 12 = 24

7 + 4(7) = 35



5x -10 = 45

5x = 55

x = 11


5 (x - 2) = 45

Remove the parentheses first.TIP:

An Example…



A proportion is based on two pairs of related quantities.

The most common way to write proportions isto use fraction notation—also called a ratio.

– When two ratios are equal, they form a proportion.


5-1-6 Solve equations that are proportions



A cross product is the product of the numerator of one fraction, times the denominator of another fraction.

– An important property of proportions is that the cross products are equal.


5-1-6 Solve equations that are proportions



STEP 2Multiply the denominator of the first fraction by the numerator of the second fraction.

Multiply the numerator from the first fraction by the denominator of the second fraction.

STEP 1

4 x 18 = 72

6 x 12 = 72


Verify that two fractions form a proportionHOW TO:

Do and form a proportion?4

12

6

18

Are they equal? Yes, they form a proportion.





Learning Outcome

Use the problem-solvingapproach to analyze andsolve word problems.

5-2



Five step problem solving approach:– What you know.

• Known or given facts.

– What you are looking for.• Unknown or missing amounts.

– Solution Plan.• Equation or relationship among known/unknown facts.

– Solution. • Solve the equation.

– Conclusion.• Solution interpreted within context of problem.

Using Equations to Solve Problems

5-2-1Section 5-2

Use the problem-solving approachto analyze and solve word problems.



Using Equations to Solve Problems

5-2-1Section 5-2

Use the problem-solving approachto analyze and solve word problems.

Key words in Table 5-1will guide you in using the problem-solving approach.

See page533

These words help you interpret the information andbegin to set up the equation to solve the problem.

“of” often implies multiplication.

“¼ of her salary” means “multiply her salary by ¼”Example:



Full time employees work morehours than part-time employees.

What are we looking for?Number of hours that FT employees work.

What do we know?

PT employees work 6 hours, and thedifference between FT and PT is 4 hours.

If the difference is four per day,and part-time employees worksix hours per day, how many

hours per day do full-timers work?

Use the solution planHOW TO:Using Equations to Solve ProblemsSection 5-2



Set up a solution plan.

FT – PT = 4FT = N [unknown] PT = 6 hours

N – 6 = 4

Solution plan: N = 4 + 6 = 10

What are we looking for?Number of hours that FT employees work.

What do we know?

PT employees work 6 hours, and thedifference between FT and PT is 4 hours.


Full time employees work morehours than part-time employees.

If the difference is four per day,and part-time employees worksix hours per day, how many

hours per day do full-timers work?We also know that

“difference” implies subtraction.

Conclusion:Full time employees

work 10 hours.



1. What are you looking for?

The number of cards that Jill has.

2. What do you know?

The relationship in the number of cards is 3:1; total is 200

3. Set up a solution plan.

x(Matt’s) + 3x(Jill’s) = 200

4. Solve it.

x + 3x = 200; 4x = 200; x = 50

5. Draw the conclusion.

Jill has 3x, or 150 cards

Jill has three times as many trading cards as Matt. If thetotal number both have is 200, how many does Jill have?




Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order?


MORE

How many humorous cards were ordered and how many nature cards were ordered—the total of H + N = 600 or N = 600 – H.

If we let H represent the humorous cards, Nature cards will be600 – H, which will simplify the solution process by using onlyone unknown: H.

What are you looking for?





MORE

A total of $950 was spent.

Two types of cards were ordered.

The total number of cards ordered was 600.

Humorous cards cost $1.75 each and nature cardscost $1.50 each.

What do you know?





MOREVolumeunknowns

$1.75(H) + $1.50 (600 – H) = $950.00

Unit prices

Totalspent

Solution PlanSet up the equation by multiplying the unitprice of each by the volume, represented bythe unknowns equaling the total amount spent.





MORE

$1.75H + $1.50(600 - H) = $950.00

$1.75H + $900.00 - $1.50H = $950.00

$0.25H + $900.00 = $950.00

$0.25H = $50.00

H = 200

Solution





The number of humorous cards ordered is 200.

Since nature cards are 600 – H, we can conclude that 400 nature cards were ordered.

Using “200” and “400” in the original equation proves that the volume amounts are correct.

Conclusion



Denise ordered 75 dinners for the awards banquet. Fish dinners cost $11.75 and chicken dinners cost $9.25 each.If she spent a total of $756.25, how many of each type of dinner did she order?


$11.75(F) + $9.25(75 - F) = $756.25

$11.75F + $693.75 - $9.25F = $756.25

$2.50F + $693.75 = $756.25

$2.50F = $62.50

F = 25

Conclusion:

25 fish dinners and 50 chicken dinners were ordered.



The relationship between two factors is often described in proportions.

– You can use proportions to solve for unknowns.

ProportionsUsing Equations to Solve ProblemsSection 5-2

The label on a container of weed killer gives directions to mix three ounces of weed killer with every two gallons of water. For five gallons of water, how many ounces of weed killer should you use?

Example:



1. What are you looking for?

Number of ounces of weed killer needed for 5 gallons of water.

2. What do you know?

For every 2 gallons of water, you need 3 oz. of weed killer.

The relationship between two factors is often described in proportions.

– You can use proportions to solve for unknowns.


Example:

The label on a container of weed killer gives directions to mix three ounces of weed killer with every two gallons of water. For five gallons of water, how many ounces of weed killer should you use?

Set up a solution plan.2 5

= 3 x

Solve it.

Cross multiply: 2x = 15; x = 7.5

Conclusion.

You need 7.5 oz of weedkiller for 5 gal of water.



Many business-related problems that involve pairs of numbers that are proportional involve direct proportions.

– An increase (or decrease) in one amount causes an increase(or decrease) in the number that pairs with it.




Using Equations to Solve ProblemsSection 5-2

In this example, an increase in the amount of gas would directly and proportionately increase the mileage yielded.

Your car gets 23 miles to the gallon.

How far can you go on 16 gallons of gas?

Cross multiply: 1x = 368 miles

Conclusion:

You can travel 368 miles on 16 gallons of gas.

1 gallon 16 gallons =

23 miles milesx

An Example…





Learning Outcomes

Evaluate a formula.

Find an equivalent formulaby rearranging the formula.

5-3



Evaluate the formula

Write the formula. Rewrite the formula substituting known

valuesfor the letters of the formula.

Solve the equation for the unknown letter or perform the indicated operations, applying the order of operations.

Interpret the solution within the context of the formula.

Formulas

5-3-1Section 5-3



FormulasSection 5-3

A plasma TV that costs $2,145 is marked up $854. What is the selling price of the TV?

Use the formula S = C + M where S is the selling price, C is the cost, and M is Markup.

S = $2,145 + $854

S or Selling Price = $2,999

An Example…



Determine which variable of the formula is to be isolated (solved for).

Highlight or mentally locate all instances of the variable to be isolated.

Treat all other variables of the formula as you would treat numbers in an equation, and perform normal steps for solving an equation.

If the isolated variable is on the right side of the equation, interchange the sides so that it appears on the left side.

Find an Equivalent Formulaby Rearranging the Formula

FormulasSection 5-3

5-3-2



FormulasSection 5-3

An Example…

The formula forSquare Footage = Length x Width or S = L x W.

Solve the formula for W or width.Isolate W by dividing both sides by L.

The new formula is:

= S

WL





EXERCISE SET A



EXERCISE SET A



EXERCISE SET A



EXERCISE SET A



EXERCISE SET A



EXERCISE SET A





PRACTICE TEST



PRACTICE TEST



PRACTICE TEST



PRACTICE TEST



PRACTICE TEST



PRACTICE TEST



PRACTICE TEST



PRACTICE TEST



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CHAPTER 5 – Equations Instructor: Dr.Gehan Shanmuganathan.

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