Applications of Linear Equations · Applications of Linear Equations Basic Problems Example: Basic Problem A health club charges members $10.00 per month plus $5.00 per hour to use

Applications of Linear Equations

Basic Problems Example: Basic Problem A health club charges members $10.00 per month plus $5.00 per hour to use the facilities. If a member was charged $55.00 this month, how many hours did he use the facilities?

Solution: Step 1: Analyze the Problem: Let x = the number of hours he uses the facilities. Step 2: Create an Equation: The charge will be the sum of the monthly rate and the hourly charge times the number of hours of use.

55105 =+x

Step 3: Solve the Equation:

9

455

55105

==

=+

x

x

x

Step 4: State the Conclusion: He uses the facilities for 9 hours this month.

/////// Example: Basic Problem The office manager of an election campaign office needs to print 5000 postcards. Suppose that it costs $0.02 to print a large postcard and $0.01 to print a small postcard. If $85.00 is allocated for printing postcards, how many of each type of postcard can be printed?

Solution: Step 1: Analyze the Problem: The total number of postcards will be 5,000. Let x = the number of $0.02 postcards. The 5000-x = the number of $0.01 postcards. Step 2: Create an Equation:

85)5000(01.002.0 =−+ xx


3500

3501.0

8501.05002.0

85)5000(01.002.0

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: 3,500 $0.02 postcards and 1,500 $0.01 postcards may be printed.

//////

Example: Basic Problem A parking garage charges $3.00 for the first hour and $2.00 for each additional hour or fraction thereof. If $9.00 was paid for parking, how many hours was the car parked in the garage?

Solution: Step 1: Analyze the Problem: Let x = number of hours car is parked. Step 2: Create an Equation: The charge for parking will be the sum of the charge for the 1st hour and the hourly rate times the number of hours.

932 =+x Step 3: Solve the Equation:

3

62

932

==

=+

x

x

x

Step 4: State the Conclusion: The car was parked for 3 hours.

////// Example: Basic Problem Two tanks hold a total of 45 gallons of a toxic solvent. One tank holds 6 gallons more than twice the amount in the other. How many gallons does each tank hold?

Solution: Step 1: Analyze the Problem 1st tank = x gallons 2nd tank = 2x+6 gallons. Step 2: Create an Equation The equation will be based on adding the volumes of the tanks.

45)62( =++ xx Step 3: Solve the Equation:

13

393

4563

45)62(

==

=+=++

x

x

x

xx

The 1st tank has 13 gallons, therefore the 2nd tank has

326266)13(262 =+=+=+x gallons. Step 4: State the Conclusion: The two tanks hold 13 and 32 gallons.

////

Example: Basic Problem A Toyota Tacoma is purchased for $12,000. The truck’s value depreciates $1100 in value per year for each of the first 6 years of ownership. At what point will the truck have a value of $6500?

Solution: Step 1: Analyze the Problem We are looking for the present value after x years of ownership, where x will be less than or equal to 6. Let x = the number of years of ownership. Step 2: Create an Equation The equation will be based on the fact that each years depreciation will be subtracted from the purchase price to find the present value.

65001100000,12 =− x Step 3: Solve the Equation:

5

55001100

65001100000,12

==

=−

x

x

x

Step 4: State the Conclusion: The truck will have a value of $1100 in 5 years.

//////// Example: Basic Problem High school students enrolling in a private tutoring program must first take a placement test at a cost of $25.00 before receiving tutoring at a cost of $18.75 per hour. If a family has set aside $400.00 to get their child extra help, how many hours of tutoring can they afford?

Solution: Step 1: Analyze the Problem: The total cost of tutoring will be based on the variable costs ($18.75/hour) and the fixed costs ($25.00). Let x = the number of hours of tutoring. Step 2: Create an Equation: The equation will be based on the formula:

Total Costs = Variable Costs + Fixed Costs

2575.18400 += x Step 3: Solve the Equation:

20

75.18375

2575.18400

==

+=

x

x

x

Step 4: State the Conclusion: The family can afford 20 hours of tutoring.

///////

Example: Basic Problem The owner of a small factory has 8 employees. Some of the employees make $10 per hour, whereas the others make $15 per hour. If the total payroll is $105 per hour, how many employees make the higher rate?

Solution: Step 1: Analyze the Problem: Let x = the number of employees that earn $15 per hour. Then 8-x will represent the number of employees that earn $10 per hour. Step 2: Create an Equation:

105)8(1015 =−+ xx


5

255

105805

105108015

105)8(1015

==

=+=−+=−+

x

x

x

xx

xx

Step 4: State the Conclusion: Therefore, 5 employees earn $15 per hour.

/////

Integer Problems: Example: Integer Problem. Write an equation that represents three consecutive integers whose sum is 39. Find the 3 integers.

Solution: Step 1: Analyze the Problem: This is a consecutive integer problem. Every integer is one more than the preceding integer.

Let x = 1st integer Let x+1 = 2nd integer Let x+2 = 3rd integer.

Step 2: Create an Equation: The sum of the integers is 39.

39)2()1( =++++ xxx Step 3: Solve the Equation:

12

363

3933

39)2()1(

==

=+=++++

x

x

x

xxx

Step 4: State the Conclusion: The three integers are 12, 13, and 14.

//////

Example: Integer Problem Find four consecutive even integers whose sum is 340.

Solution: Step 1: Analyze the Problem: This is a consecutive integer problem. Every even integer is two more than the preceding even integer. Let x = 1st integer Let x+2 = 2nd integer Let x+4 = 3rd integer. Let x+6 = 4th integer. Step 2: Create an Equation: The sum of the integers is 340.

340)6()4()2( =++++++ xxxx


82

3284

340124

340)6()4()2(

==

=+=++++++

x

x

x

xxxx

Step 4: State the Conclusion: The integers are 82, 84, 86, and 88.

////// Example: Integer Problem Find two consecutive odd integers whose sum is 56.

Solution: Step 1: Analyze the Problem: This is an integer problem. Every odd integer is two more than the preceding odd integer. Let x = the 1st odd integer. Let x+2 = the 2nd odd integer. Step 2: Create an Equation: The sum of the integers is 56.

562 =++ xx Step 3: Solve the Equation:

292

27

542

5622

562

=+=

==+

=++

x

x

x

x

xx

Step 4: State the Conclusion: The 1st odd integer is 27 and the 2nd odd integer is 29.

//////

Example: Integer Problem Friends and Leave it to Beaver are two of the most popular television shows of all time. The number of episodes of each show are consecutive even integers whose sum is 470. If there are more episodes of Friends, how many episodes of each were there?

Solution: Step 1: Analyze the Problem: This is a consecutive integer problem. Every even consecutive integer is two more than the even integer before it. 1st Integer = x 2nd Integer = x+2 Step 2: Create an Equation: The sum of the two even integers is 470.

470)2( =++ xx Step 3: Solve the Equation:

234

4682

47022

470)2(

==

=+=++

x

x

x

xx

Therefore, 23622342 =+=+x Step 4: State the Conclusion: There were 234 episodes of Leave it to Beaver and 236 episodes of Friends.

//////

Geometry Problems Example: Geometry Problem Find an algebraic equation for the perimeter of a rectangle, given that the length is x yards and the width is 10 yards shorter. Find the length and width of the rectangle if the perimeter is 60.

Solution: Step 1: Analyze the Problem: Start by defining the length and width in terms of x.

Let x = the length Let x-10 = the width

Step 2: Create an Equation: The formula for the perimeter of a rectangle is Perimeter = 2L +2W

)10(22

22

−+=+=

xxP

WLP

Step 3: Solve the Equation: Now substitute in the value of P and solve for x.

20

804

20460

)10(2260

==

−=−+=

x

x

x

xx

Step 4: State the Conclusion:

Therefore, the length is 20 and the width is 10.

/////

Example: Geometry Problem The perimeter of a regulation singles tennis court is 210 feet and the length is 3 feet less than three times the width. What are the dimensions of the court?

Solution: Step 1: Analyze the Problem: This is a geometry problem. The formula for the perimeter of a rectangle is LWP 22 += . Let the width = x. Let the length = 3x-3 Step 2: Create an Equation: Like all geometry problems, the equation will be based on the formula.

)33(22210 −+= xx Step 3: Solve the Equation:

27

8216

68210

662210

)33(22210

==

−=−+=

−+=

x

x

x

xx

xx

Therefore, 783813)27(333 =−=−=−x Step 4: State the Conclusion: The width of the tennis court is 27 feet and the length is 78 feet.

//// Example: Geometry Problem Andrew is constructing two adjacent rectangular dog pens. Each pen will be three times as long as it is wide, and the pens will share a common long side. If Andrew has 65 ft of fencing, what are the dimensions of each pen?

Solution: Step 1: Analyze the Problem: Although not shown in this example, it will be beneficial to draw a diagram. This is usually a good idea for any type of geometry problem. Begin by letting x equal the width of one pen. Then the length will be 3x. Step 2: Create an Equation: Adding up all the lengths and widths will give us the following equation to solve.

65333 =++++++ xxxxxxx


5

6513

65333

==

=++++++

x

x

xxxxxxx

Step 4: State the Conclusion: Therefore, the width of each pen is 5 ft. and the length is 15 ft.

///////

Example: Geometry Problem Given a triangle, one angle is 15 degrees less than twice the smallest angle. The third angle is 35 degrees more than the smallest angle. Find these three angle measures.

Solution: Step 1: Analyze the Problem: The sum of the angles of a triangle is always equal to 180 degrees. Let x = the measure of the smallest angle. Then the other angles are 2x-15 and x+35. Step 2: Create an Equation:

180)35()152( =++−+ xxx Step 3: Solve the Equation:

40

1604

180204

180)35()152(

==

=+=++−+

x

x

x

xxx

Step 4: State the Conclusion: The angles are 40, 65, and 75 degrees.

///// Example: Geometry Problem Find the measure of two complimentary angles if one angle is 10 degrees more than three times the other angle.

Solution: Step 1: Analyze the Problem: Two angles are called complimentary angles when the sum of their measures is 90 degrees. Let x = the measure of the smaller angle. Then 3x+10 will be the measure of the other angle. Step 2: Create an Equation: The equation will be based on the sum of the two angles being equal to 90.

90103 =++ xx Step 3: Solve the Equation:

20

804

90104

90103

==

=+=++

x

x

x

xx

Step 4: State the Conclusion: The angles are 20 degrees and 70 degrees.

////////

Uniform Motion Problems

Example: Motion Problem Twenty minutes after a father left for work on the bus, he noticed that he left his briefcase at home. His son left home, driving 36 mph, to catch the bus that was travelling at 24 mph. How long did it take the son to catch the bus?

Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula Distance = Rate x Time or D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Father 24 x 24x Son 36 x-20 36(x-20)

Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal.

)20(3624 −= xx Step 3: Solve the Equation:

60

72012

7203624

)20(3624

==

−=−=

x

x

xx

xx

Step 4: State the Conclusion: The father was travelling for 60 minutes so it took the son 60-20 or 40 minutes to catch the bus.

/////// Example: Motion Problem If a student drives from college to home at 40 mph, then he is 15 minutes late. However, if he makes the same trip at 50 mph, he is 12 minutes early. What is the distance between his college and home?

Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula Distance = Rate x Time or D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Trip 1 40 x 40x Trip 2 50 x-27 50(x-27)

Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal.

)27(5040 −= xx


135

135010

13505040

)27(5040

==

−=−=

x

x

xx

xx

135 minutes is 2.25 hours. Substituting this into 40x gives a distance of

miles90)25.2(40 = Step 4: State the Conclusion: The student lives 90 miles from campus.

////// Example: Motion Problem John rowed his boat from his camp on the bayou to his crab traps. Going down the bayou he caught a falling tide that increased his normal speed by 2 mph, but coming back it decreased his normal speed by 2 mph. When going with the tide, the trip took 10 minutes and going against the tide, the trip took 30 minutes. How far is it from John’s camp to the crab traps?

Solution: Step 1: Analyze the Problem: This is a motion problem that involves use of the formula rtD = . As with most motion problems, it is helpful to set up a table to organize the information.

Trip Rate Time Distance To Traps x+2 10=1/6 1/6(x+2) To Camp x-2 30=1/2 1/2(x-2)

Step 2: Create an Equation: From this table we can create the following equation which is based on the distance in both directions being equal.

)2(2

1)2(

6

1 −=+ xx


4

82

632

)2(2

1)2(

6

1

==

−=+

−=+

x

x

xx

xx

Step 4: State the Conclusion: The distance to the traps is 60 miles.

//////

Example: Motion Problem A cyclist leaves his training base for a morning workout, riding at the rate of 18 mph. One and one half hours later, his support staff leaves the base in a car going 45 mph in the same direction. How long will it take the support staff to catch up with the cyclist?

Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Cyclist 18 x 18x

Car 45 x-1.5 45(x-1.5) Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal.

)5.1(4518 −= xx Step 3: Solve the Equation:

5.2

5.6727

5.674518

)5.1(4518

==

−=−=

x

x

xx

xx

Step 4: State the Conclusion: The cyclist has been travelling for 2.5 hours; therefore it will take the support car 1 hour to catch up.

/////// Example: Motion Problem Two trains are 330 miles apart, and their speeds differ by 20 mph. Find the speed of each train if they are travelling toward each other and will meet in 3 hours.

Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Train 1 x 3 3x Train 2 x+20 3 3(x+20)

Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal to 330.

330)20(33 =++ xx Step 3: Solve the Equation:

5.52

2706

330606

3306033

330)20(33

==

=+=++

=++

x

x

x

xx

xx

Step 4: State the Conclusion: The trains are travelling at 52.5 mph and 72.5 mph.

/////// Example: Motion Problem A cyclist rode uphill and then turned around and made the same trip downhill. The downhill speed was twice the uphill speed. If the trip each way was 16 miles and the entire trip lasted 4 hours, at what speed was the cyclist going uphill?

Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Let x = the rate of going uphill and fill in the known information.

To determine the time we need to rewrite the formula asr

Dt = .

Direction D r t

Uphill 16 x 16/x Downhill 16 2x 16/2x = 8/x

Step 2: Create an Equation: The equation will be based on total time. To solve the equation begin by multiplying both sides of the equation by the LCD which is x.

xx

tttT

8164

21

+=

+=


6

816)4(

8164

21

=

+=

+=

+=

x

xxxx

xx

tttT

Step 4: State the Conclusion: The cyclist was going uphill at a speed of 6 mph.

///////

Formula Problems Example: Formula Problem In nursing, Clark’s Rule for medication expresses a relationship between the recommended dosages for a child and for an adult. The rule states that a child’s dosage C equals the product of the weight W of the child in pounds divided by 150 and the adult’s dosage A. Determine the formula for adult dosage and then determine the adult dosage if a child weighing 60 lbs requires a dosage of 125 mg.

Solution: Step 1: Analyze the Problem: First, we need to solve this formula for A, the adult dosage. Then, we can substitute in the given values to find the indicated dosage. Step 2: Create an Equation:

AW

C ×=150

Step 3: Solve the Equation: Solve for A

AW

C

WAC

AW

C

=

=

×=

150

150150

Solve for the indicated dosage.

5.31260

)125(150

150

=

=

=

A

A

AW

C

Step 4: State the Conclusion: Therefore, the adult dosage is 312.5 mg.

////// Example: Formula Problem Cowling’s rule is another method for determining the dosage of a drug to prescribe to a child. For this rule, the formula

24

)1( += aDd

Gives the child’s dosage d, where D is the adult dosage and a is the age of the child in years. If the adult dosage of a drug is 600 mg. and a doctor uses this formula to determine that a child’s dosage is 200 mg, then how old is the child?

Solution: Step 1: Analyze the Problem: The formula is presently solved for d, the childs dosage. We need to solve for a, the age of the child and then substitute in the given values to find the indicated age. Step 2: Create an Equation: The formula is already given, we only need to solve it for a.

24

)1( += aDd

Step 3: Solve the Equation: Solve for a.

124

124

)1(2424

)1(

−=

+=

+=

+=

D

da

aD

d

aDd

aDd

Determine the indicated age.

7

1600

)200(24

124

=

−=

−=

a

a

D

da

Step 4: State the Conclusion: The child is 7 years old.

///////////

Basic Percent Problems Example: Percent Problem Last year at Grand Canyon University, 70% of the awarded degrees were to undergraduates. If 2,450 degrees were awarded last year, how many were to undergraduates?

Solution: This problem may be restated as an IS-OF statement, then converted into mathematical symbols and solved accordingly.

70% of 2,450 is what number?

1715

)2450(7.0

==

N

N

/// Example: Percent Problem If a bank account pays $120.00 in simple interest on an initial balance of $2,000, what is the annual interest rate?

Solution: This problem may be restated as an IS-OF statement, then converted into mathematical symbols and solved accordingly.

120 is N% of 2,000

%6

06.0%2000

120%

)2000%(120

==

=

=

N

N

N

N

//////

Example: Percent Problem Gwen sold her car on consignment. The saleswoman’s commission was 10% of the selling price and Gwen received $6,570. Find the selling price of the car.

Solution: This problem may be restated as an IS-OF statement, then converted into mathematical symbols and solved accordingly. If the commission is 10% of selling price then Gwen will receive 90% of the selling price. Let the selling price equal N.

6570 is 90% of N

.300,79.0

6570

90.06570

=

=

=

N

N

N

Therefore the selling price of the car is $7,300.

/////// Example: Percent Problem After getting a 15% discount on the price of a new Chrysler Sebring convertible, Helen paid $27,000. What was the original price of the convertible to the nearest dollar?

Solution: This problem may be restated as an IS-OF statement, then converted into mathematical symbols and solved accordingly. If the discount was 15%, then Helen paid 85% of the original price.

27,000 is 85% of N.

70.764,3185.0

27000

85.027000

=

=

=

N

N

N

The original price was $31,764.70.

/////// Example: Percent Problem A cattle rancher is going to sell one of his prize bulls at an auction and would like to make $45,500.00 after paying a 9% commission to the auctioneer. For what selling price will the rancher make this amount of money?

Solution: Step 1: Analyze the Problem: If the commission is 9% that means that the rancher takes home 91%. Let x = the selling price. Then the IS-OF statement is $45,500 is 91% of what number (selling price.) Step 2: Create an Equation: The equation is based on the IS-OF statement.

x91.0500,45 = Step 3: Solve the Equation:

000,50$

91.0500,45

==

x

x

Step 4: State the Conclusion: The selling price must be $50,000 in order for the rancher to take home $45,500.

/////

Example: Percent Problem Gwen bought a new Chevy Prius. The selling price plus the 8% state sales tax was $15,714. What was the selling price?

Solution: This problem may be restated as an IS-OF statement, then converted into mathematical symbols and solved accordingly. Let the selling price equal N.

15,714 is 108% of N

550,1408.1

714,15

08.1714,15

=

=

=

N

N

N

The selling price was $14,550.

//////// Example: Percent Increase Between the years 2000 and 2006, the average cost for auto insurance nationwide grew 27%, to $867.00 What was the average cost in 2000?

Solution: Step 1: Analyze the Problem: This is a percent increase problem. The formula for percent increase is always the amount of increase divided by the original number times 100. Let x = the average cost in 2000. Then we can create the following equation based on percent increase. Step 2: Create an Equation:

%27)100(867 =

−x

x

Step 3: Solve the Equation: This is a rational equation. To solve, multiply both sides of the equation by the LCD, which is x.

[ ]

67.682

27.1867

27.0867

27.0867

==

=−

=

−

x

x

xx

xx

xx

Step 4: State the Conclusion: The average cost of auto insurance in 2000 was $682.67.

//////

Example: Percent Problem What was the MSRP for a Hummer H1 that sold for $107,272 after an 8% discount?

Solution: Step 1: Analyze the Problem: This is a percent problem that may be put into the form of an IS-OF Statement. If the discount is 8% that means that the selling price is 92% of the MSRP. Let x = MSRP. Then,

$107,272 is 92% of what number (MSRP)

Step 2: Create an Equation:

x92.0272,107 = Step 3: Solve the Equation:

600,116$92.0

272,107

272,10792.0

=

=

=

x

x

x

Step 4: State the Conclusion: The MSRP is $116,600.

///////

Mixture Problems Example: Mixture Problem The owner of a health foods store sells dried apples for $1.20 per quarter pound, and dried apricots for $1.80 per quarter pound. How many pounds of each must he mix together to get 20 pounds of a mixture that sells for $1.68 per quarter pound?

Solution: Step 1: Analyze the Problem: This is a mixture problem. A table is helpful for organizing the information. Note that I have converted the prices per quarter lb to prices per lb.

Fruit # of lbs. Rate/lb. Total amount apples x 4.80 4.8x

apricots 20-x 7.20 7.2(20-x) mixture 20 6.72 6.72(20)

Step 2: Create an Equation: From the table we can create the equation which is based on adding the dried apple and apricots together to equal the mixture.

)20(72.6)20(2.78.4 =−+ xx


4

6.94.2

4.1342.71448.4

)20(72.6)20(2.78.4

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: The owner must mix 4 lbs of dried apples and 16 lbs of dried apricots.

/////// Example: Mixture Problem How many gallons of a 3% salt solution must be mixed with 50 gallons of a 7% solution to obtain a 5% solution?

Solution: Step 1: Analyze the Problem: This is a mixture problem. A table is helpful for organizing the information.

Solution % Salt Amount Salt 1 3% x 0.03x 2 7% 50 0.07(50)

Final 5% x+50 0.05(x+50) Step 2: Create an Equation: The equation will be based on the fact that the sum of the salt in solution 1 and solution 2 must be equal to the amount of salt in the final solution.

)50(05.0)50(07.003.0 +=+ xx Step 3: Solve the Equation:

50

102.0

5.205.05.303.0

)50(05.0)50(07.003.0

==

+=++=+

x

x

xx

xx

Step 4: State the Conclusion: 50 gallons of the 3% solution must be mixed.

/////// Example: Mixture Problem A printer has ink that is 8% cobalt blue color and ink that is 22% cobalt blue color. How many ounces of each ink are needed to make 1 gallon of ink that is 15% cobalt blue?


Solution % Cobalt Amount Cobalt 1 8% x 0.08x 2 22% 1-x 0.22(1-x)

Final 15% 1 0.15 Step 2: Create an Equation: The equation will be based on the fact that the sum of the cobalt in solution 1 and solution 2 must be equal to the amount of cobalt in the final solution.

15.0)1(22.008.0 =−+ xx Step 3: Solve the Equation:

5.0

07.014.0

15.022.022.008.0

15.0)1(22.008.0

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: 0.5 gallons of each solution must be mixed.

///////

Example: Mixture Problem A nurse wants to add water to 30 ounces of a 10% solution of benzalkonium chloride to dilute it to an 8% solution. How much water must she add?


Solution % BChloride Amount BChloride 1 10% 30 0.1(30) 2 0% x 0

Final 8% x+30 0.08(x+30) Step 2: Create an Equation: The equation will be based on the fact that the sum of the benzalkonium chloride in solution 1 and solution 2 must be equal to the amount of benzalkonium chloride in the final solution.

)30(08.0)30(1.0 += x Step 3: Solve the Equation:

5.7

6.008.0

4.208.03

)30(08.0)30(1.0

==

+=+=

x

x

x

x

Step 4: State the Conclusion: 7.5 gallons of water must be mixed.

///////////

Investment Problems Example: Investment Problem A professor wants to supplement his pension with investment interest. If he invests $28,000 at 6% interest how much does he have to invest at 7% interest to achieve a yield of $3,500 per year in investment interest.

Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information.

Investment Interest Rate Amount Yield 1 6% 28,000 0.06(28,000) 2 7% x 0.07x

Step 2: Create an Equation: The equation is based on the fact that the sum of the yields from the two investments must be equal to his desired yield of $3,500.

350007.0)000,28(06.0 =+ x Step 3: Solve the Equation:

000,26

182007.0

350007.01680

350007.0)000,28(06.0

==

=+=+

x

x

x

x

Step 4: State the Conclusion: $26,000 must be invested at 7%.

///// Example: Investment Problem A salesperson used his $3,500 year-end bonus to purchase some old coins, with hopes of earning 15% annual interest on the gold coins and 12% annual interest on the silver coins. If he saw a return on his investment of $480.00 the first year, how much did he invest in each type of coin?


Coins Earning Rate Amount Return Gold 15% x 0.15x Silver 12% 3500-x 0.12(3500-x)

Step 2: Create an Equation: The equation is based on the fact that the sum of the yields from the two investments must be equal to his total return of $480.00

480)3500(12.015.0 =−+ xx Step 3: Solve the Equation:

000,2$

6003.0

48012.042015.0

480)3500(12.015.0

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: $2,000 must be invested in gold and $1500 in silver.

/////// Example: Investment Problem Equal amounts are invested in each of three accounts paying 7%, 8%, and 10.5% interest annually. If one year’s combined interest income is $1,249.50, how much is invested in each account?

Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information. We can let x = the amount of each investment because we are told the amounts are equal.

Investment Interest Rate Amount Income 1 7% x 0.07x 2 8% x 0.08x 3 10.5% x 0.105x

Step 2: Create an Equation: The equation is based on the fact that the sum of the yields from the three investments must be equal to the total income of $1,249.5.

5.1249105.008.007.0 =++ xxx Step 3: Solve the Equation:

900,4$

5.1249255.0

5.1249105.008.007.0

==

=++

x

x

xxx

Step 4: State the Conclusion: $4,900 must be invested in each account.

/////// Example: Investment Problem Of the $50,000 that Natasha pocketed on her last real estate deal, $20,00 went to charity. She invested part of the remainder in Dreyfus New Leaders Fund with an annual yield of 16% and the rest in Templeton Growth Fund with an annual yield of 25%. If she made $6,060.00 on these investments in 1 year, then how much did she invest in each fund?


Investment Interest Rate Amount Yield Dreyfus 16% x 0.16x

Templeton 25% 30,000-x 0.25(30,000-x)

Step 2: Create an Equation: The equation is based on the fact that the sum of the yields from the two investments must be equal to the total yield of $6,060.00.

060,6)000,30(25.016.0 =−+ xx Step 3: Solve the Equation:

000,14000,30

000,16$09.0

1440

144009.0

606025.0500,716.0

6060)000,30(25.016.0

=−=

−−=

−=−=−+

=−+

x

x

x

x

xx

xx

Step 4: State the Conclusion: $16,000 was invested in Dreyfus New Leaders Fund and $14,000 was invested in Templeton Growth Fund.

////////// Example: Investment Problem Coughlin construction contracted Amherst High and Memorial stadium for a total cost of $4.7 million. Because the construction was not completed on time, Coughlin paid 5% of the amount of the high school contract and 4% of the stadium contract in penalties. If the penalty was $223,000 then what was the amount of each contract?


Contract Penalty Rate Amount Penalty School 5% x 0.05x Stadium 4% 4,700,000-x 0.04(4,700,000-x)

Step 2: Create an Equation: The equation is based on the fact that the sum of the penalties from the two contracts is equal to the total penalty of $223,000

0000,223)000,700,4(04.005.0 =−+ xx Step 3: Solve the Equation:

000,500,3

000,3501.0

000,223000,18801.0

000,22304.0000,18805.0

000,223)000,700,4(04.005.0

==

=+=−+

=−+

x

x

x

xx

xx

Step 4: State the Conclusion: The high school contract is $3,500,000 and the stadium contract is $1,200,000.

////////// Example: Investment Problem Norma invested the startup capital for her internet business in two hedge funds. After one year, one of the funds returned 5% and the other returned 6% for a total return of $5880. If the amount on which she made 6% was $10,000 larger than the amount on which she made 5%, then what was the original amount of her startup capital?

Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information. The table is based on the formula prtI = . Since the time is one year we may omit this column.

Investment Principle Rate Interest Fund 1 x 0.05 0.05(x) Fund 2 x+10,000 0.06 0.06(x+10,000)

Step 2: Create an Equation: From the table we can create the following equation and solve for x.

5880)000,10(06.005.0 =++ xx


000,48

528011.0

588060006.005.0

5880)000,10(06.005.0

==

=++=++

x

x

xx

xx

Step 4: State the Conclusion: Therefore the principle invested in hedge fund 1 is $48,000 and the amount invested in hedge fund 2 is $58,000. The total invested is $106,000.

///////

Example: Investment Problem Eric paid one-half of his game-show winnings to the government for taxes. He invested one-third of his winnings in Jeff’s coffee shop at 14% interest and one-sixth of his winnings in Zack’s German Bakery at 12% interest. If he earned a total of $4,000 on the investments in one year, then how much did he win on the game show?

Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information. Let x = Eric’s after tax winnings. Step 2: Create an Equation: Then, we can create the following equation and solve.

000,46

112.0

3

114.0 =

+

xx


5.701,59

4000067.0

400002.0047.0

000,46

112.0

3

114.0

==

=+

=

+

x

x

xx

xx

Step 4: State the Conclusion: Since this represents Eric’s after tax winnings, and he paid half his winnings in taxes, the total amount won on the game show is $119,403.00.

////////

Number-Value Problems Example: Number-Value Problem When a child emptied his coin bank, he had a collection of pennies, nickels, and dimes. There were 20 more pennies than dimes and the number of nickels were triple the number of dimes. If the coins had a value of $5.40, how many of each type coin were in the bank?

Solution: Step 1: Analyze the Problem: Number-value problems are also a type of mixture problem and a table is recommended to organize the information.

Coins Value Amount Total Pennies 0.01 x+20 0.01(x+20) Nickels 0.05 3x 0.05(3x) Dimes 0.1 x 0.1x

Step 2: Create an Equation: The equation is based on the fact that the total value of the pennies, nickels, and dimes is $5.40.

40.51.0)3(05.0)20(01.0 =+++ xxx Step 3: Solve the Equation:

20

52026

5402026

540101520

40.51.015.02.001.0

40.51.0)3(05.0)20(01.0

==

=+=+++

=+++=+++

x

x

x

xxx

xxx

xxx

Step 4: State the Conclusion: There are 20 dimes, 40 pennies, and 60 nickels.

//////// Example: Number-Value Problem The admission prices for the Coca-Cola Museum in Atlanta are $9.00 for adults, $8.00 for seniors and $5.00 for children. A family purchased 3 more children’s tickets than adults’ tickets, and one less senior ticket than adult tickets. The total cost of the tickets was $73.00. How many of each type of ticket did they purchase?

Solution: Step 1: Analyze the Problem: Number-value problems are also a type of mixture problem and a table is recommended to organize the information.

Coins Price Amount Total Value Adults 9.00 x 9x

Children 5.00 x+3 5(x+3) Seniors 8.00 x-1 8(x-1)

Step 2: Create an Equation: The equation is based on the fact that the total value of the adults tickets, children’s tickets and seniors tickets is $73.00

73)1(8)3(59 =−+++ xxx Step 3: Solve the Equation:

3

6622

73722

73881559

73)1(8)3(59

==

=+=−+++

=−+++

x

x

x

xxx

xxx

Step 4: State the Conclusion: 3 adult, 6 children and 2 senior tickets were purchased.

///////

Work Problems Example: Work Problem With the old combine Rick’s entire wheat crop can be harvested in 72 hours but a new combine can do the same job in 48 hours. How many hours would it take to harvest the crop with both combines operating?

Solution: Step 1: Analyze the Problem: This is a work problem. Once again a table is helpful for organizing the information.

Combine Work /hour hours working work old 1/72 x 1/72(x) new 1/48 x 1/48(x)

Step 2: Create an Equation: From the table we can create the following equation which is based on the old and new combines working together to complete 1 job.

14872

=+ xx


8.28

3456120

34567248

14872

==

=+

=+

x

x

xx

xx

Step 4: State the Conclusion: The time it takes the combines working together to harvest the crop (complete 1 job) is 28 hours and 48 minutes.

///////

Example: Work Problem Tom can paint a fence in 5 hours. Huckleberry can paint the same fence in 7 hours. If Tom gets up early and starts painting a 8:30 AM and Huckleberry joins him 2.5 hours later, at what time will they finish painting the fence?

Solution: Step 1: Analyze the Problem: This is a work problem. Once again a table is helpful for organizing the information.

Combine Work /hour hours working work Tom 1/5 x 1/5(x)

Huckleberry 1/7 x-2.5 1/7(x-2.5)

Step 2: Create an Equation: From the table we can create the following equation which is based on the old and new combines working together to complete 1 job.

17

5.2

5=−+ xx


95.3

5.4712

355.1257

17

5.2

5

==

=−+

=−+

x

x

xx

xx

Step 4: State the Conclusion: Tom will work for 3 hours and 57 minutes and Huckleberry will work for 1 hour and 27 minutes. The fence will be completed at 12:27 PM.

///////

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Applications of Linear Equations · Applications of Linear Equations Basic Problems Example: Basic Problem A health club charges members $10.00 per month plus $5.00 per hour to use