Solving Equations Unit - · PDF fileSolving Equations Unit (Level IV Graduate Math) ... Moving from Balance Puzzles to Algebra ... 4 1 3 3 4 3 4 1 3 4 = = + − = −

Solving Equations Unit

(Level IV Graduate Math)

Draft

(NSSAL)

C. David Pilmer

©2008

(Last Updated: Dec 2011)

This resource is the intellectual property of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education. The following are permitted to use and reproduce this resource for classroom purposes.

• Nova Scotia instructors delivering the Nova Scotia Adult Learning Program • Canadian public school teachers delivering public school curriculum • Canadian nonprofit tuition-free adult basic education programs

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Individuals, not including teachers or instructors, are permitted to use this resource for their own learning. They are not permitted to make multiple copies of the resource for distribution. Nor are they permitted to use this resource under the direction of a teacher or instructor at a learning institution. Acknowledgments

The Adult Education Division would also like to thank the following NSCC instructors for piloting this resource and offering suggestions during its development. Eileen Burchill (IT Campus)

Elliott Churchill (Waterfront Campus)

Barbara Leck (Pictou Campus)

Suzette Lowe (Lunenburg Campus)

Floyd Porter (Strait Area Campus)

Brian Rhodenizer (Kingstec Campus)

Joan Ross (Annapolis Valley Campus)

Tanya Tuttle-Comeau (Cumberland Campus)

Jeff Vroom (Truro Campus)

NSSAL i Draft ©2008 C. D. Pilmer

Table of Contents Introduction…………………………………………………………………………………. ii Tracking Your Progress…………………………………………………………………….. ii The Big Picture……………………………………………………………………………... Course Timelines……………………………………………………………………………

iii iv

Solving Equations Balance Puzzles……………………………………………………………………………... 1 Revisiting the Balance Puzzles……………………………………………………………... 4 Moving from Balance Puzzles to Algebra………………………………………………….. 11 Solve and Check…………………………………………………………………………….. 19 Dealing with x and -x……………………………………………………………………….. 24 Formulas and Solving for an Unknown…………………………………………………….. 28 Equations with Decimals…………………………………………………………………… 31 Equations with Fractions……………………………………………………………………. 38 Equations in the Real World………………………………………………………………... 46 Equations Where a Variable is Squared or Cubed………………………………………….. 54 Post-Unit Reflections……………………………………………………………………….. 65 Answers……………………………………………………………………………………... 66 Online Support……………………………………………………………………………… 74

NSSAL ii Draft ©2008 C. D. Pilmer

Introduction In this unit you will learn to solve and check equations of the form DCxBAx +=+ ,

CBAx =+2 and CBAx =+3 . Examples of these types of equations are shown below.

• 13795 +=− xx • 8.41.02.16.0 −=+ xx

• 4311

43

+=− xx

• 47194 2 =−x • 4.732.36.5 3 =+ x

By the end of the unit, you will learn how to work with equations derived from a real world context. Tracking Your Progress Date

Started Date

Completed Balance Puzzles………………………………………….…… 1 Revisiting the Balance Puzzles……………………..………… 4 Moving from Balance Puzzles to Algebra……………….…… 11 Solve and Check……………………………………………… 19 Dealing with x and -x…………………………………...…….. 24 Formulas and Solving for an Unknown……………...……….. 28 Equations with Decimals………………………...…………… 31 Equations with Fractions……………………..………………. 38 Equations in the Real World……………...………………….. 46 Equations Where a Variable is Squared or Cubed…………… 54

NSSAL iii Draft ©2008 C. D. Pilmer

The Big Picture The following flow chart shows the five required units and the four optional units (choose two of the four) in Level IV Graduate Math. These have been presented in a suggested order.

Note: You are not permitted to complete four ALP Approved Projects and thus avoid selecting from the Linear Functions and Linear Systems Unit, Trigonometry Unit, or Statistics Unit.

Math in the Real World Unit (Required) • Fractions, decimals, percents, ratios, proportions, and

signed numbers in real world applications • Career Exploration and Math

Solving Equations Unit (Required) • Solve and check equations of the form DCxBAx +=+ , CBxA += 2

and CBxA += 3

Consumer Finance Unit (Required) • Simple Interest and Compound Interest • TVM Solver (Loans and Investments) • Credit and Credit Scores

Graphs and Functions Unit (Required) • Understanding Graphs • Linear Functions and Line of Best Fit

Measurement Unit (Required) • Imperial and Metric Measures • Precision and Accuracy • Perimeter, Area and Volume

Choose two of the four.

Linear Functions and Linear Systems

Unit

Trigonometry Unit

Statistics Unit ALP Approved Projects

(Complete 2 of the 5 projects.)

NSSAL iv Draft ©2008 C. D. Pilmer

Course Timelines Graduate Level IV Math is a two credit course within the Adult Learning Program. As a two credit course, learners are expected to complete 200 hours of course material. Since most ALP math classes meet for 6 hours each week, the course should be completed within 35 weeks. The curriculum developers have worked diligently to ensure that the course can be completed within this time span. Below you will find a chart containing the unit names and suggested completion times. The hours listed are classroom hours. Unit Name Minimum

Completion Time in Hours

Maximum Completion Time

in Hours Math in the Real World Unit 24 36 Solving Equations Unit 20 28 Consumer Finance Unit 18 24 Graphs and Functions Unit 28 34 Measurement Unit 24 30 Selected Unit #1 20 24 Selected Unit #2 20 24 Total: 154 hours Total: 200 hours As one can see, this course covers numerous topics and for this reason may seem daunting. You can complete this course in a timely manner if you manage your time wisely, remain focused, and seek assistance from your instructor when needed.

NSSAL 1 Draft ©2008 C. D. Pilmer

Balance Puzzles Example: In the diagram below you have objects on two sides of scale. The objects are perfectly balanced on the scale. The round objects are metal ball bearings. The rectangular objects are cardboard boxes that contain an unknown number of ball bearings. If more than one box is used on the same scale, you are to assume that all the boxes contain the same number of unknown ball bearings.

If the ball bearings are all of the same size and weight, and we assume that the cardboard used to make each box weighs virtually nothing, then determine how many ball bearings are hidden in each box. One hint is to remove similar items at the same time from both sides of the scale so that everything remains balanced.

Solution:

Step 1: Remove 1 box from both sides Step 2: Remove 3 ball bearings from both sides.

Step 3: If 2 boxes are equal to 10 ball bearings, then 1 box must be equal to 5 ball bearings.

There are 5 ball bearings in each box.


Questions: 1. For each of the diagrams below, determine how many ball bearings are in a box. Provide a

brief description to explain how you arrived at your answer. (a) (b)

(c) (d)

(e) (f)

(g) (h)


(i) (j)

(k) (l)

Scrambled Answer Key for Question 1

2 6 4 2 5 1 0 4 3 5 2 3 2. Create your own balance question and solve it.


Revisiting the Balance Puzzles Consider the balance puzzle question on the right. You have three boxes and two ball bearings on one side of a balanced scale. On the other side of the scale, you have one box and eight ball bearings. Each of the four boxes has the same, yet unknown, number of ball bearings in them. If all the ball bearings are of the same size and weight, determine how many ball bearings are in a box. Solution:

Step 1: Remove 1 box from both sides Step 2: Remove 2 ball bearings from both sides.

Step 3: If 2 boxes are equal to 6 ball bearings, then 1 box is equal to 3 ball bearings.

This process of determining the number of ball bearings in a box can be represented using algebra. Let n represent the unknown number of ball bearings in a box.

Algebraic Steps: Written Description: 8123 +=+ nn Original Question

822181123

=+−+=−+

nnnnn

Remove 1 box from both sides of the scale.

62

28222=

−=−+nn

Remove 2 ball bearings from both sides of the scale.

326

22

=

=

n

n

If 2 boxes are equal to 6 ball bearings, then 1 box is equal to 3 ball bearings. There are 3 ball bearings in each box.

There are 3 ball bearings in each box.


Questions: 1. (a) On question 1 of a previous sheet titled “Balance Puzzles”, you determined how many

ball bearings were hidden in each of the cardboard boxes. Look back at that question and your answers, and match questions (a) through (l) to the appropriate algebraic solution. Three solutions are missing. This will be addressed in part (b) of this question.

(i) Matches with ______ (ii) Matches with ______ (iii) Matches with ______

661

39331931

1911329132

==

−=−+=+

−+=−++=+

nnnn

nnnnnn

236

33

6317113

713171114

7114

=

=

=−=−+

=+−+=−+

+=+

n

nnnn

nnnnnn

326

22

6217112

712

=

=

=−=−+

=+

n

nnnn

(iv) Matches with ______ (v) Matches with ______ (vi) Matches with ______

n

nn

nn

nnnnnn

=

=

=−+=−

+=−+=−+

+=+

22

224

2422226

226224262

2462

133

33

3325223

523

=

=

=−=−+

=+

n

nnnn

441

15111511

2522135213

==

−=−+=+

−+=−++=+

nnnn

nnnnnn

(vii) Matches with ______ (viii) Matches with ______ (ix) Matches with ______

331

14111411

3433144314

==

−=−+=+

−+=−++=+

nnnn

nnnnnn

n

nn

nn

nnnnnn

=

=

=−+=−

+=−+=−+

+=+

02

220

2033233

323133131

3331

52

102

2102

1101131013

=

=

=−+=−

+=

n

nn

nnnnnn


(b) You probably figured out that the solutions were not provided for questions (f), (k) and (l). Your mission is to complete the left column of the chart for each of these questions. You are required to fill in the algebraic steps that correspond to the written descriptions. If you are not sure how to complete this, look at the sample question that was completed on page 4.

Question (f)

Algebraic Steps: Written Descriptions:

Original Question




Question (k)


Original Question

Remove 2 boxes from both sides of the scale.




Question (l)


Original Question




2. (a) Complete the diagram that corresponds to the

equation 9234 +=+ nn . (b) Given the following algebraic steps, provide the written descriptions for a balance puzzle

question involving boxes and ball bearings.

Algebraic Steps: Written Descriptions: 9234 +=+ nn Original Question

932292234

=+−+=−+

nnnnn

62

39332=

−=−+nn

326

22

=

=

n

n


3. Given the following diagram showing boxes and ball bearings on each side of scale, determine how many ball bearings are hidden in the boxes. Remember that all the boxes contain the same number of ball bearings. You are required to solve the question using written descriptions as well as algebra.


Original Question

4. Given the following diagram showing boxes and ball

bearings on each side of scale, determine how many ball bearings are hidden in the boxes. Remember that all the boxes contain the same number of ball bearings. You are required to solve the question using written descriptions as well as algebra.


Original Question


5. Solve for n in the following algebraic equations. Remember to think about this process as

boxes and ball bearings balancing on a scale. You are always trying to do the same thing to both sides of the equation just like you were trying to do the same thing to both sides of the scale to maintain the balance.

(a) 10547 +=+ nn (b) 12126 +=+ nn (c) 16215 +=+ nn (d) 19234 +=+ nn (e) 17356 +=+ nn (f) 118172 +=+ nn Scrambled Answer Key for Question 5

2 8 5 3 1 4


Wrap-Up Statement: In the sections “Balance Puzzles” and “Revisiting the Balance Puzzles”, you learned that

balance puzzle problems can serve as an introduction to algebra and solving equations. In an attempt to find the number of ball bearings hidden in each of the boxes, you removed items of the same quantity from both sides of the scale until you could use logic to determine the answer. Each one of these steps can be expressed with a algebraic statement where n represents the number of ball bearings hidden in the each box.

Reflect Upon Your Learning Fill out this questionnaire after you have completed both sections (Balance Puzzles and

Revisiting the Balance Puzzles). Select your response to each statement. 1 - strongly disagree 2 - disagree 3 - neutral 4 - agree 5 - strongly agree (a) I understand all of the concepts covered in the section,

“Balance Puzzles.” 1 2 3 4 5

(b) I understand all of the concepts covered in the section, “Revisiting the Balance Puzzles.”

1 2 3 4 5

(c) I do not need any further assistance from the instructor on the material covered in this section.

1 2 3 4 5

(d) I do not need any more practice questions. 1 2 3 4 5 (e) I see how each of the steps to solve the Balance Puzzles

connects to an algebraic statement. 1 2 3 4 5

(f) I understand that to maintain the balance on the scale, I must remove an item of the same quantity from both sides of the scale. I can also see this process occurring when I look at the algebraic statements.

1 2 3 4 5


Moving From Balance Puzzles to Algebra Let’s look at some of balance puzzle equations you solved in the last section titled “Revisiting the Balance Puzzles.” We will look at the answers to questions 3 and 4.

3. 4.

Algebraic Steps:

339

33

93413443

13432132245

13245

=

=

=−=−+

=+−+=−+

+=+

n

nnnn

nnnnnn

Algebraic Steps:

25

105

5105

3133351335

113113613136

=

=

=−=−+

=+−+=−+

+=+

n

nnnn

nnnnnn

What are the three properties that we learned by using the balance puzzles as an introduction to solving equations using algebra.

• Whatever you do to one side of the equation must also be done to the other side of the equation to insure that the balance is maintained. For example in question 3, we subtracted 2n from both sides of the equation to insure that both sides of the equation remained equal to each other.

• We eventually want to collect all the variables terms (i.e. letters) on one side of the equation and the constants (i.e. numerical terms) of the other side. For example in question 3, we got down to the step with 93 =n . In question 4, we got down to the step with 105 =n . In both cases the variables are on the left and constants are on the right.

• The last thing we learned is a little more difficult to grasp and involves the term inverse operations.

An inverse operation "reverses" another operation. Addition and subtraction are inverses of each other because adding and subtracting the same number does not change the original number. For example, 7 - 6 + 6 = 7 and 13 + 11 - 11 = 13. Similarly, multiplication and division are inverses of each other because multiplying and dividing by the same number does not change the original number.

Consider question 3 above. In the second step we removed the +2n from the left hand side of the equation by -2n from both sides of the equation. Subtraction is the inverse of addition. (Subtraction reverses the work done by addition.) In the fifth step of question 3, we divided both sides of the equation by 3. That was done to change 3n (i.e. 3 multiplied by n) to simply n. Dividing by 3 reversed the work done by multiplying by 3. Division is the inverse of multiplication.


The following chart shows how these inverse operations can be used to solve simple equations. In the balance puzzle problems you only encountered questions that required that you use the inverse operations found in rows one and three of this chart. You will eventually learn how to use all of these inverse operations when solving equations.

Equation How to Solve for n. Solution Inverse Operation

75 =+n Subtract 5 from both sides of the equation. 2

5755=

−=−+nn

Subtraction is the inverse of addition.

43 =−n Add 3 to both sides of the equation 7

3433=

+=+−nn

Addition is the inverse of subtraction.

324 =n Divide both sides of the equation by 4

84

324

4

=

=

n

n Division is the inverse of

multiplication.

26=

n Multiply both sides of the equation by 6

12

266

6

=

×=×

n

n Multiplication is the inverse

of division.

Let’s see these three new properties in action. A series of equations has been solved and we have indicated if one or more of these properties has been used in each step of the solution.

Property 1: What is done to one side of the equation must be done to the other side.

Property 2: Eventually you want the variable terms (i.e letters) on one side of the equation and constants (i.e. numerical terms) on the other.

Property 3: Use inverse operations to eventually simplify equations. Example 1 Property 1 Property 2 Property 3

7214 +=+ nn nnnn 272214 −+=−+ Subtraction is the inverse of addition.

712 =+n 17112 −=−+n Subtraction is the inverse of addition.

62 =n

26

22

=n Division is the inverse of multiplication.

3=n Example 2 Property 1 Property 2 Property 3

10326 +=− nn nnnn 3103326 −+=−− Subtraction is the inverse of addition.

1023 =−n 210223 +=+−n Addition is the inverse of subtraction.

123 =n

312

33

=n Division is the inverse of multiplication.

4=n


Additional examples with explanations have been provided. Notice that the three properties listed on the previous pages reappear in these examples. Example 3 Explanation:

18237 −=− nn

nnnn 2182237 −−=−− • To get rid of the +2n on the right hand side of the equation, subtract 2n from both sides of the equation.

1835 −=−n • Simplify: nnn 527 =− and 022 =− nn

318335 +−=+−n • To get rid of the subtract 3 from the left hand side of the equation, add 3 to both sides of the equation.

155 −=n • Simplify: 033 =+− and 15318 −=+−

515

55 −

=n • To get rid of the 5 that is being multiplied by the n, divide both

sides of the equation by 5. 3−=n

Example 4 Explanation:

11573 +=+ xx

xxxx 5115573 −+=−+ • To get rid of the +5x on the right hand side of the equation, subtract 5x from both sides of the equation.

1172 =+− x • Simplify: xxx 253 −=− and 055 =− xx

711772 −=−+− x • To get rid of the add 7 from the left hand side of the equation, subtract 7 from both sides of the equation.

42 =− x • Simplify: 077 =− and 4711 =−

24

22

−=

−− x • To get rid of the -2 that is being multiplied by the x, divide both

sides of the equation by -2. 2−=x


aa 33142 −=−

aaaa 3331342 +−=+− • To get rid of the -3a on the right hand side of the equation, add 3a to both sides of the equation.

3145 =−a • Simplify: aaa 532 =+ and 033 =+− aa

431445 +=+−a • To get rid of the subtract 4 from the left hand side of the equation, add 4 to both sides of the equation.

355 =a • Simplify: 044 =+− and 35431 =+

535

55

=a • To get rid of the 5 that is being multiplied by the a, divide both

sides of the equation by 5. 7=a



kk 22763 −=− kkkk 2227263 +−=+−

• To get rid of the -2k on the right hand side of the equation, add

2k to both sides of the equation. 2743 =− k • Simplify: kkk 426 −=+− and 022 =+− kk

327343 −=−− k • To get rid of the +3 from the left hand side of the equation, subtract 3 from both sides of the equation.

244 =− k • Simplify: 033 =− and 24327 =−

424

44

−=

−− k • To get rid of the -4 that is being multiplied by the k, divide both

sides of the equation by -4. 6−=k

Questions: 1. Complete the following statements. A sample question has been provided. Sample Question: To solve the equation 37 =−x you must add 7 to both sides of the equation. When you

do this, you would find that x equals 10.

(a) To solve the equation 98 =+n you must _____________________________________

___________________________. When you do this, you would find that n equals ____.

(b) To solve the equation 213 =p you must ______________________________________

___________________________. When you do this, you would find that p equals ____.

(c) To solve the equation 24 =−d you must _____________________________________

___________________________. When you do this, you would find that d equals ____.

(d) To solve the equation 43=

x you must ________________________________________

___________________________. When you do this, you would find that x equals ____.

(e) To solve the equation 84 =n , you must _______________________________________

___________________________. When you do this, you would find that n equals ____.

(f) To solve the equation 149 =+ g , you must ____________________________________

___________________________. When you do this, you would find that g equals ____.


2. The partial solutions to three equations have been provided. Fill in the boxes with the appropriate information.

(a) 5476 −=+ nn nnnn 454476 −−=−+ • To get rid of the +4n on the right hand side of the

equation, subtract 4n from both sides of the equation.

• Simplify

75772 −−=−+n • To get rid of the +7 from the left hand side of the equation, subtract 7 from both sides of the equation.

• Simplify

212

22 −

=n • To get rid of the 2 that is being multiplied by the n,

divide both sides of the equation by 2.

(b) 23357 +=− xx • To get rid of the +3x on the right hand side of the

equation, subtract 3x from both sides of the equation.

2354 =−x

• Simplify

• To get rid of the -5 from the left hand side of the equation, add 5 to both sides of the equation.

284 =x

• Simplify

• To get rid of the 4 that is being multiplied by the x, divide both sides of the equation by 4.

7=x

(c) pp 41312 −=+ • To get rid of the -4p on the right hand side of the

equation, add 4p both sides of the equation.

• Simplify

• To get rid of the +1 from the left hand side of the equation, subtract 1 from both sides of the equation.

126 =p

• Simplify

6

126

6=

p • To get rid of the 6 that is being multiplied by the p, divide both sides of the equation by 6.

2=p


3. Look back at your answer to question 2(c). Would the final answer ( 2=p ) be different if you started by getting rid of the +1 from the left hand side of the equation, and then getting rid of the -4p from the right side of the equation? Explain.

4. The partial solutions to six equations have been provided. Fill in the boxes with the

appropriate information.

(a) 26486 +=+ nn (b) xx 31157 +=− nnnn 4264486 −+=−+ xxxx 3311357 −+=−− 826882 −=−+n 511554 +=+−x 182 =n

416

44

=x

9=n (c) pp 22373 −=− (d) aa 21352 −=+ pppp 2223273 +−=+− 1354 =+a 305 =p 84 =a

6=p 2=a (e) 25973 −=− xx (f) nn 23316 −=+ 2576 −=−− x 3336 =+ n 186 −=− x

327

33

=n

(g) 1156 −=−p (h) 1937 =− x

66

66 −

=p

3

123

3−

=−− x


5. An equation has been solved. Provide written explanations for each of the indicated steps.

xx 44753 −=+ Written Explanations xxxx 4447453 +−=++

4757 =+x

547557 −=−+x

427 =x

742

77

=x

6=x 6. Solve the following equations. Show all your work. (a) 3496 −=+ xx (b) 14275 +=− nn (c) pp 61372 +=− (d) mm 21974 −=+ (e) 1351 −=+ xx (f) 36234 −=− nn Scrambled Answer Key for Question 6

7 3 2 -6 -5 8


Wrap-Up Statement: In the section “Moving From Balance Puzzles to Algebra”, you examined the solutions to

some of the balance puzzle problems in order to create some rules that you could use to solve a variety of equations. You learned that you use inverse operations to collect the variable on one side of the equation and the number on the other side. You had to ensure that you used the same operation on both sides of the equation in order to maintain the balance.

Reflect Upon Your Learning Fill out this questionnaire after you have completed questions 1 to 6. Select your response to

each statement. 1 - strongly disagree 2 - disagree 3 - neutral 4 - agree 5 - strongly agree (a) I understand all of the concepts covered in the section,

“Moving From Balance Puzzles to Algebra.” 1 2 3 4 5

(b) I do not need any further assistance from the instructor on the material covered in this section.

1 2 3 4 5

(c) I do not need any more practice questions. 1 2 3 4 5 (d) I understand the term, inverse operations. 1 2 3 4 5 (e) When I solve an equation, I can explain what is occurring in

each step of my solution. 1 2 3 4 5


Solve and Check If you are not supplied with an answer key, how do you know if you solved an equation correctly? It is simply a matter of substituting your answer back into the original equation and see if that answer satisfies the equation (i.e. make sure that both sides of the equation are equal to each other). Example 1

7214

22

14268662

862585567

8567

−=

−=

−=−−=−+

−=+−−=−+

−=+

x

xxxx

xxxxxx

Check:

( ) ( )835649

8756778567

−−=+−−−=+−

−=+ xx

4343 −=−

7−=x is the correct solution.

Example 2

34

124

4124

8208842084

222028222082

=

=

=−=−+

=++−=++

−=+

n

nnnn

nnnnnn

Check:

( ) ( )62086

322083222082

−=+−=+−=+ nn

1414 =

3=n is the correct solution.

Example 3

4520

55

20511911115

9115

=

=

=+=+−

=−

p

pppp

Check:

( )91120

911459115

=−=−=−p

99 =

4=p is the correct solution. Questions:

1. Kiana was given the equation 13315 +=− xx . Her answer was 7=x . Without solving the equation yourself, determine if Kiana’s answer is correct.


2. Andrea was given the equation 9214 +=+ xx . Her answer was 5=x . Without solving the equation yourself, determine if Andrea’s answer is correct.

3. Solve each of the following equations and then check your answer by substituting it back into

the original equation. Show all your work. (a) 13256 +=+ xx Check: (b) 5257 −=+ pp Check: (c) 9573 +=− aa Check:


(d) 5229 +=− nn Check: (e) aa 32654 −=+ Check: (f) dd 31782 −=− Check: (g) 17261 +=− xx Check:


(h) 41324 −=− pp Check: (i) 31632 +=+ nn Check: (j) pp 51023 −−=− Check: (k) mm 4123 =− Check:


(l) xx 2184 −=+ Check: Scrambled Answer Key for Question 3

-8 -3 2 9 3 -12 -4 -7 -1 5 1 -2


Dealing with x and -x Up to this point you have been dealing with equations with terms like 2x, 3p, -4m. In all of these cases, it is easy to identify the coefficient of the term. For example, the coefficient of 2x is 2. The coefficient of 3p is 3. The coefficient of -4m is -4. The coefficient is the numerical multiplier of the variable.

Initially when learners see terms like x and -x, they are uncertain what these terms mean because the coefficient doesn’t seem to be present. The term x can also be written as 1x. The coefficient is 1. The term -x can also be written as -1x. The coefficient in this case is -1.

Understanding the meaning of x and -x allows you to solve a wider variety of equations. Example 1

574 −=+ xx Example 2

pp −=− 5102 Example 3

3574 −=+ dd Answer:

4312

33

12375773

573151174

5174574

−=

−=

−=−−=−+

−=+−−=−+

−=+−=+

x

xxxx

xxxxxx

xx

Check:

( )

999716

54744574

−=−−=+−

−−=+−−=+ xx

Answer:

53

153

3153

105101035103

115110215102

5102

=

=

=+=+−

=−+−=+−

−=−−=−

p

pppp

pppppp

pp

Check:

( )

0001010

5510525102

==−

−=−−=− pp

Answer:

101

101

1101

73771371

5355743574

=−−

=−−

−=−−−=−+−

−=+−−−=−+

−=+

d

dddd

dddddd

Check:

( ) ( )

4747350740

310571043574

=−=+

−=+−=+ dd

Questions: Solve each equation and check your answer. Show all your work. 1. 9116 +=− xx

Check:


2. 1647 +=+ xx

Check:

3. 937 +=− xx

Check:

4. xx −=+ 36123

Check:

5. xx −=+ 2205

Check:


6. 2632 −=− xx

Check:

7. xx −=+ 75

Check:

8. 1695 −=+ xx

Check:

9. 127 −=− xx

Check:


10. 1311 −=− xx

Check:

Scrambled Answer Key:

10 -8 -1 1 4 6 12 7 2 -6 -3

Wrap-Up Statement: In the sections “Solve and Check” and “Dealing with x and -x”, you learned two different

concepts. In “Solve and Check” section you discovered that you could check your answer by substituting the answer back into the original equation and seeing if both sides of the equation were still equal to each other. If they were equal, then you knew that your answer was correct. In the “Dealing with x and -x” section, you discovered that x can be written as 1x. You also found out that -x can be written as -1x.

Reflect Upon Your Learning Fill out this questionnaire after you have completed both sections (Solve and Check, and

Dealing with x and -x). Select your response to each statement. 1 - strongly disagree 2 - disagree 3 - neutral 4 - agree 5 - strongly agree (a) I understand all of the concepts covered in the section, “Solve

and Check.” 1 2 3 4 5

(b) I understand all of the concepts covered in the section, “Dealing with x and -x.”

1 2 3 4 5


1 2 3 4 5

(d) I do not need any more practice questions. 1 2 3 4 5 (e) I find that it’s taking me less time to solve equations compared

to the previous sections. 1 2 3 4 5

(f) I feel confident about the math that I have learned and used in these two sections.

1 2 3 4 5


Formulas and Solving for an Unknown As you continue your studies, you will have to work with formulas that may require you to use some algebra skills to solve for an unknown. For example, you may know the perimeter and length of a rectangle, but don’t know its width. You could use the length, perimeter, and perimeter formula to solve for the width.

Example: The perimeter of a rectangle is 62 cm. The length of the rectangle is 24 cm. Determine the

width of the rectangle.

Answer:

72

22

14214

48248486224862

2)24(26222

=

=

=−+=−

+=+=

+=

w

ww

ww

wwlP

We start with the formula for the perimeter of a rectangle, wlP 22 += . Substitute the known values (62 and 24) into the appropriate spots in the formula. All we have left to do is solve for the unknown, w.

At this point, we are not going to be solving word problems like the ones above. For now, you’re going to be supplied with a formula and numerical values for all but one of the variables. You’ll substitute those numerical values into the equation and then solve for the unknown variable.

Example 1: Example 2: Example 3: 1 ,4 ;72 ==+=+ cbcba 9 ;253 ==− xyx 6 ,2 ;61034 =−=−=− nmpnm

Answer:

224

22

4248442

8421742

72

=

=

=−=−+

=++=++=+

a

aaaaa

cba

Check: cba +=+ 72 ( ) 17422 +=+ 88 =

Answer:

( )

5525

55

25527227527

25272593

253

=−−

=−−

−=−−=−−

=−=−=−

y

yy

yyy

yx

Check: 253 =− yx ( ) ( ) 25593 =− 22527 =− 22 =

Answer:

( ) ( )

p

pp

pp

pp

pnm

=−

=−

=−+−=+−

−=−−=−−

−=−−−=−

210

101020

10206610626

61026610188

610632461034

Check: 61034 −=− pnm

( ) ( ) ( )620188

62106324−−=−−

−−=−−

2626 −=−


Questions: Solve for the missing variable.

1. 1 ,423 −==+ xxy

2. 10 ,364 ==+ ffe

3. 4 ,2 ,6426 ==+=+ dgdhg

4. 3 ,5 ,4532 ==+=− nmntm

5. 2 ,1 ,2335 −=−=−=+ yxzyx

6. 4 ,3 ,5417 ==−=− rsrst


7. 7 ,4 ,32413 −==−=− rqrqp

8. 12 ,5 ,11 ,627 ===−=− zxwywzx

Scrambled Answer Key

-3 0 2 -1 5 8 6 -4 7 Wrap-Up Statement: In the section “Formulas and Solving for an Unknown”, you learned that equations can have

more than one variable in them. If a numerical value is given for all but one of the variables, you can solve for the unknown variable. This is done by substituting the numerical values into their appropriate positions in the equation, and then using algebra to rearrange the equation and solve for the unknown variable.



“Formulas and Solving for an Unknown.” 1 2 3 4 5


1 2 3 4 5

(c) I do not need any more practice questions. 1 2 3 4 5 (d) I feel confident about the math that I have learned and used in

this section. 1 2 3 4 5


Equations with Decimals What happens when your equation has decimal numbers and/or decimal coefficients? Do you have to do anything differently? No, the rules for solving equations remain the same when dealing with equations where decimals are used. The only difference is that you may want to have a calculator to make life a little easier when you’re adding, subtracting and dividing with decimals. Example 1:

85.04

5.05.0

45.031335.0

135.02.012.02.037.0

12.037.0

−=

−=

−=−−=−+

−=+−−=−+

−=+

x

xxxx

xxxxxx

Example 2:

57.05.3

7.07.0

5.37.02.17.42.12.17.0

7.42.17.03.03.07.43.02.14.0

3.07.42.14.0

=

=

=−=−+

=++−=++

−=+

p

pppp

pppppp

Example 3:

4006.04.2

06.006.0

4.206.01.23.01.21.206.0

3.01.206.04.03.04.04.01.246.0

3.04.01.246.0

=

=

=+=+−

=−−+=−−

+=−

d

dddd

dddddd

Example 4:

7.08.0

56.08.0

8.056.08.0

44.0144.08.044.018.044.0

5.05.015.03.044.05.013.044.0

−=−

=−−

=−−=−−

=−−+=−−

+=−

x

xx

xx

xxxxxx

Example 5:

6.95.08.4

5.05.0

8.45.02.02.08.42.03.0

2.08.43.0

=

=

=+−=+

−=

a

aa

aaaaaa

Example 6:

208.0

168.0

8.0168.0

116112.01612.0

162.0

=−−

=−−

−=−−−=−

−=−=

y

yy

yyyyyy

yy


Questions: Solve each of the following equations. Show all your work. 1. 63.025.0 +=+ xx

2. 51.044.0 +=− dd

3. 6.43.02.07.0 +=+ xx


4. 1.11.04.03.0 −=− yy

5. 22.03.04 −=− pp

6. 6.52.04.2 −=+ xx


7. mm 3.04.02.42 −=−

8. 4.21.02 −=− rr

9. 102.0207.0 −=+ xx


10. cc 01.05203.0 −=−

11. 1605.024.0 +=+ ff

12. pp 02.0573.0 −=+


13. 075.405.021.0 +=−− xx

14. 81.006.024.0 +=− dd

15. 16.05.09.0 −= xx


16. kk 2.0318.04.0 −=

Scrambled Answer Key:

-6.25 20 0.53 2 -48.5 4 11 40 -10 -0.4 -3.5 175 30 -40.5 -60 12


Equations with Fractions A lot of learners find solving equations challenging enough without including those dreaded fractions. There are two ways to handle equations with fractions where one method is far easier than the other. Example 1:

Solve the equation 1219

34

+=− xx .

Method 1 This method is far more challenging because you are forced to work with the fractions. Most learners will probably choose to use the other method. Do not spend a lot of time looking at this.

1

219

34

+=− xx

xxxx211

21

219

34

−+=−− • To get rid of the x

21 from the right hand side of the

equation, subtract x21 from both sides of the equation.

1

639

68

=−− xx • To subtract x21 from x

34 , you need to create a common

denominator of 6.

1965

=−x • x68 subtract x

63 gives you x

65 .

9199

65

+=+−x • To get rid of the -9 from the left hand side of the equation, add 9 to both sides of the equation,

10

65

=x • Add 1 and 9.

65

10

65

65

=x

• To get rid of the

65 which is being multiplied by x, divide

both sides of the equation by 65 .

5610×=x • Dividing by a fraction is the same as multiplying by the

fractions reciprocal. 12=x • Multiply 10 by 6 and then divide by 5.

Your head is probably spinning after looking at this method. Most people, including many math instructors, would not use this method. The other method eliminates all of the fractions in the first two steps and is much easier.


Method 2 We will eliminate the fractions in the equation by multiplying everything on both sides of the equation by the least common multiple of the numbers in the denominators of the fractions.

( ) ( )

125

605

5605

546545456545

363354863548

62654

324

1621696

346

1219

34

=

=

=+=+−

=−−+=−−

+=−

+=−

+

=−

+=−

x

xxxx

xxxxxx

xx

xx

xx

• The fractions we are dealing with have the

denominators 3 and 2. The least common multiple of 3 and 2 is 6. Multiply everything on both sides of the equation by 6.

• 324

346 =× and

26

216 =×

• The fractions are gone so we can solve the equation using the procedures we learned previously in this unit.

For the remaining examples, we are going to use method 2 to solve the equations.

Example 2:


54

+=+ xx .

( ) ( )

103

303

3303

30603030360303

56055308605308

602

1030540

6102110310

5410

6213

54

=

=

=−=−+

=+−+=−+

+=+

+=+

+

=+

+=+

x

xxxx

xxxxxx

xx

xx

xx



• 540

5410 =× and

210

2110 =×



Example 3:


31

+=− nn .

( ) ( )

245

1205

51205

48724848572485

97299484729484

724

36483

12

6124312412

3112

6434

31

−=−

=−−

=−+=+−−

=−−−+=−−

+=−

+=−

+

=−

+=−

n

nnnn

nnnnnn

nn

nn

nn



• 3

123112 =× and

436

4312 =×


Example 4:

Solve the equation xx43810

83

−=− .

( ) ( )

169

1449

91449

80648080964809

66646803664803

4246480

824

43888108

838

43810

83

=

=

=+=+−

=−+−=+−

−=−

−=−

−=−

−=−

x

xxxx

xxxxxx

xx

xx

xx



• 824

838 =× and

424

438 =×



Example 5:

Solve the equation 74332 +=− pp .

( ) ( ) ( )

8540

55

405122812125

2812532833128

283128

284

12128

744343424

74332

=

=

=+=+−

=−−+=−−

+=−

+=−

+

=−

+=−

p

pppp

pppppp

pp

pp

pp

• The only fraction we are dealing with has the

denominator 4. To get rid of this fraction multiply everything on both sides of the equation by 4.

• 4

12434 =×

• The fraction is gone so we can solve the equation using the procedures we learned previously in this unit.

Questions: 1. Fill in the blanks.

(a) 214 =× (b)

656 =× (c)

3112 =×

(d) 438 =× (e)

5410 =× (f)

236 =×

(g) 2310 =× (h)

14114 =× (i)

458 =×

(j) 232 =× (k)

5420 =× (l)

4316 =×


8 16 2 10 1 5 11 15 9 3 4 12 6 2. Fill in the blanks.

(a) To get rid of the fraction in the equation 1532

=+x , you would multiply everything on

both sides of the equation by _____.


(b) To get rid of the fraction in the equation 4537 =− x , you would multiply everything on

both sides of the equation by _____.

(c) To get rid of the fractions in the equation 5734

21

−=− tt , you would multiply

everything on both sides of the equation by _____.

(d) To get rid of the fractions in the equation xx5215

43

−=+ , you would multiply


(e) To get rid of the fractions in the equation 9321

65

−=+ mm , you would multiply


(f) To get rid of the fractions in the equation dd4342

61

−=− , you would multiply

everything on both sides of the equation by _____. Scrambled Answer Key for Question 2

6 14 3 15 20 12 5 3. Complete the following questions. (a)

( ) ( )7313323

7132

=+

=+

x

x

(correct answer: x = 9)

(b)

( ) ( )1123212312

4312

1323

43

−

=−

−=−

pp

pp

(correct answer: p = 24)


4. Solve each of the following equations. Show all your work. (a)

10143

=+x

(b) 46

52

=+d

(c) 2

2311 =− n

(d) 1

212

31

−=+ xx

(e) 3

214

56

−=+ hh

(f) 3

541

32

−=+ pp


(g) xx

311

413 −=−

(h) 3

61

32

−= tt

(i) 1

411

83

−=+ pp

(j) xx

6552

41

+=−

(k) xx

762 =−

(l) 1

323

94

+=+ xx


30 -5 -16 -10 14 -6 6 20 -24 12 9 -12 18


Wrap-Up Statement: In the sections “Equations with Decimals” and “Equations with Fractions”, you learned how

to solve equations with decimals or fractions. When an equation has decimals, you rearrange the equation using the same procedure you used with integers (…, 3,-2, -1, 0, 1, 2, 3,…). The only difference is that you may want to use a calculator so that you can quickly add, subtract, and divide the decimal numbers. When an equation has fractions, it is far easier to get rid of the fractions in the first few steps. This is done by multiplying everything on both sides of the equation by the least common multiple of the numbers in the denominators. Once the fractions are gone, you following the procedure that you have been using in the previous sections of this unit.

Reflect Upon Your Learning Fill out this questionnaire after you have completed the two sections (Equations with

Decimals and Equations with Fractions). Select your response to each statement. 1 - strongly disagree 2 - disagree 3 - neutral 4 - agree 5 - strongly agree (a) I understand all of the concepts covered in the section,

“Equations with Decimals.” 1 2 3 4 5

(b) I understand all of the concepts covered in the section, “Equations with Fractions.”

1 2 3 4 5


1 2 3 4 5

(d) I do not need any more practice questions. 1 2 3 4 5 (e) I can successfully get rid of the fractions in an equation by

multiplying both sides of the equation by the least common multiple of the numbers in the denominators.

1 2 3 4 5


Equations in the Real World In the real world we often use equations to describe relationships between two variables. If you know the value of one of the variables, we can use the equation to solve for the other variable. Example 1: The volume of fluid in an intravenous (I.V.) bag changes with time. The equation 1000100 +−= tv describes the relationship between the volume, v, in terms of time, t. The volume is measured in milliliters. The time is measured in hours. (a) How much fluid is in the bag at 2=t hours. (b) At what time will there be 600 ml of fluid in the bag? (c) At what time will there be 300 ml of fluid in the bag? Answers: (a) We know t and have to solve for v.

( )1000200

100021001000100

+−=+−=

+−=

vv

tv

800=v There is 800 ml of fluid in the bag at 2=t hours. (b) We know v and have to solve for t.

100100

100400

100400100010001001000600

10001006001000100

−−

=−−

−=−−+−=−

+−=+−=

tt

tt

tv

t=4 At 4=t hours there will be 600 ml of fluid in the I.V. bag (c) We know v and have to solve for t.

100100

100700

100700100010001001000300

10001003001000100

−−

=−−

−=−−+−=−

+−=+−=

tt

tt

tv

t=7 At 7=t hours there will be 300 ml of fluid in the I.V. bag


Example 2: A cylindrical above ground pool is being drained using a pump. The depth of the water in the pool decreases with time. The equation 2.14.0 +−= td describes the depth, d, of the water in terms of the time, t. The depth is measured in metres. The time is measured in hours. (a) At what time is the depth 0.6 metres? (b) What is the depth at 2.5 hours?

Answers: (a) We know d and have to solve for t.

t

tt

tt

td

=−−

=−−

−=−−+−=−

+−=+−=

5.14.04.0

4.06.0

4.06.02.12.14.02.16.0

2.14.06.02.14.0

The depth is 0.6 metres at t = 1.5 hours.

(b) We know t and have to solve for d.

( )

2.02.10.1

2.15.24.02.14.0

=+−=

+−=+−=

ddd

td

At t = 2.5 hours, the depth of the water is 0.2 m.

Example 3: Angela is talking to a friend when she abruptly turns and walks away. The distance between

Angela and her friend increases with time. The equation 123

+= td describes the distance, d,

between Angela and her friend in terms of time, t. The distance is measured in metres. The time is measured in seconds. (a) How far is Angela from her friend at t = 8 seconds? (b) At what time will the distance between Angela and her friend be 19 metres?

Answers: (a)

( )

13112

12

24

1823

123

=+=

+=

+=

+=

dd

d

d

td

At t = 8 seconds, Angela is 13 metres from her friend.

(b)

( ) ( )

t

tt

tt

t

t

td

=

=

=−+=−

+=

+

=

+=

+=

1233

336

336223238

2338

12232192

12319

123

The distance is 19 m at t = 12 seconds.


Questions: Show all your work. 1. Tylena is having gravel delivered to her property. The more gravel she has delivered, the

greater the cost. The equation 4035 += tc describes the cost, c, of the gravel in terms of number of tonnes, t, that are delivered. The cost is measured in dollars.

(a) If she orders 8 tonnes of gravel, how much will it cost? (b) If the cost is $250, how many tonnes of gravel did she have delivered? (c) How much will it cost to order 10 tonnes of gravel? (d) How many tonnes of gravel can Tylena have delivered for $460.


2. Jun borrowed money from his parents and agreed to pay it back with weekly payments. The equation 48040 +−= pm describes the amount of money, m, Jun stills owes in terms of the number of weekly payments, p, he makes.

(a) How many payments will Jun have made if he stills owes his parents $320? (b) How much does Jun still owe his parents if he has made 10 payments? (c) If Jun still owes his parents $160, how many payments has he made? (d) If Jun has made 5 payments, how much does he still owe his parents?


3. An automobile gas tank is partially full when Adam decides to fill it up completely at a local gas station. During the filling process, the volume of gasoline in the tank increases with time. The equation 142.1 += tv describes the volume, v, of gasoline in the tank in terms of time, t. The volume is measured in litres. The time is measured in seconds.

(a) How much gas will be in the tank at t = 6 seconds? (b) At what time will there be 26 litres of gas in the tank? (c) How much gasoline is in the tank when Adam has been pumping for 17 seconds? (d) When will there be 32 litres of gasoline in the tank?


4. A car tire drives over a nail and springs a slow leak. The air pressure in the tire decreases slowly with time. The equation 2407.0 +−= tp describes the tire pressure, p, in terms of time, t. The pressure is measured in kilopascals. The time is measured in minutes.

(a) At what time will the tire’s pressure be 219 kilopascals? (b) At 12 minutes, what will the tire pressure be? (c) What will the tire pressure be if the tire has been leaking for 60 minutes? (d) If the tire’s pressure is 220.4 kilopascals, then how long has the tire been leaking air?


5. Micheline’s basement is flooding. When she wakes up, the basement floor is already

covered by water and the water level continues to rise. The equation 15320

+= td describes

the depth, d, of the water in her basement in terms of the time, t, since she woke up and discovered the flooding. The depth is measured in centimeters. The time is measured in hours.

(a) What is the depth of the water 9 hours after Micheline discovered the flooding? (b) When will the depth of the water be 135 cm? (c) If the depth of the water is 35 cm, how long has it been since Micheline discovered the

flooding problem? (d) If it’s been 24 hours since Micheline discovered the flooding, then what should the depth

of the water be? Scrambled Answer Key for Questions 1 to 5

198 4 15 175 390 18 8 6 10 30 21.2 75 280 12 28 231.6 34.4 80 3 320


Wrap-Up Statement: In the section “Equations in the Real World”, you were supplied with an equation that

showed the relationship between two real world variables. You were supplied with a numerical value for one variable and then had to solve for the other variable. Understanding the real world situation and the related equation was important when you worked on these types of questions.



“Equations in the Real World.” 1 2 3 4 5


1 2 3 4 5



(e) With each question, I can figure out what variable has been provided, and what variable I have to solve for.

1 2 3 4 5


Equations Where a Variable is Squared or Cubed Up to this point you have been solving equations of the form DCxBAx +=+ where you are asked to solve for the variable x. As you work through this course and other math courses, you will encounter equations that have a variable that is squared (i.e. 2x , 2a , 2r ) or a variable that is cubed (i.e. 3x , 3r ). These equations will be of the form CBAx =+2 or CBAx =+3 . How do we solve these types of equations? It all comes down to our understanding of inverse operations that we previously discussed in this unit. We learned that the inverse operation of addition was subtraction (and visa versa), and the inverse operation of multiplication was division (and visa versa). We used these inverse operations to solve a variety of equations. 1973 =+x 719773 −=−+x The inverse operation of adding 7 is subtracting 7. 123 =x

312

33

=x The inverse operation of multiplying by 3 is dividing by 3.

4=x So how can we use our understanding of inverse operations to answer slightly more complicated equations? It comes down to us being able to answer the following two questions. So what is the inverse operation of squaring? Answer: Square Rooting

So what is the inverse operation of cubing? Answer: Cube Rooting We can now use these two new inverse operations to solve a wider range of equations. Example 1: Solve the equation 1062 =−x .

Answer: 1062 =−x 610662 +=+−x The inverse operation of subtracting 6 is adding 6. 162 =x 162 =x The inverse operation of squaring is square rooting. 4or 4 −=x

You may be wondering why we have two answers for x. Even though your calculator tells that the square root of 16 is 4, you have to consider the third step. You are asked to find number that when squared are equal to 16. When you square 4 or -4, you get 16. You need to write down both answers. This issue only arises with questions involving squaring.


Example 2: Solve the equation 2923 =+x .

Answer: 2923 =+x 229223 −=−+x The inverse operation of adding 2 is subtracting 2. 273 =x 33 3 27=x The inverse operation of cubing is cube rooting. 3=x With the first two examples we ended up with numbers where we can figure out the square root or cube root in our head. Here are some other numbers. Square Roots:

11 = 24 = 39 = 416 = 525 =

636 = 749 = 864 = 981 = 10100 =

11121 = 12144 = Cube Roots

113 = 283 = 3273 = 4643 = 51253 = In most cases, however, we can not work out a square root or a cube root in our heads; we need to use a calculator. Square Roots on the TI Graphing Calculator

> Enter the number. > Close bracket with ) > ENTER Cube Roots on the TI Graphing Calculator MATH > 3 > Enter Number > Close bracket with ) > ENTER


Example 3: Example 4: Solve the equation 3835 2 =+x . Solve the equation 8.274.27.5 3 =+ x

Answer: Answer:

2.65-or 65.27

7535

55

355338335

3835

2

2

2

2

2

2

≈=

=

=

=

−=−+

=+

xx

x

xxxx

10.221.9

21.94.21.22

4.24.2

1.224.27.58.274.27.57.5

8.274.27.5

33 3

3

3

3

3

3

≈=

=

=

=

−=+−

=+

xx

x

xx

xx

You may have noticed that we have been using the symbol ≈ (approximately equal to). Since we are rounding our answers off, it is more appropriate to use the ≈ symbol, rather than the = symbol. Your answers will not be marked wrong if you don’t use this symbol. You may be wondering why we are looking at these slightly more complicated equations. As you work through the Measurement Unit and possibly through the optional Trigonometry Unit, you will encounter formulas that will require you to use these skills. Examples involving these formulas are shown below. Normally when we solve equations where the variable is squared, we end up with two answers (positive and negative number). In the context of these formulas, we will only be using the positive solution. This will make more sense when you get to the Measurement Unit and Trigonometry Unit. Example 5: Solve for b given that 222 cba =+ , a = 6, and c = 9.

Answer: 222 cba =+ 222 96 =+ b Substitute the known values in the equation. 8136 2 =+ b Evaluate 26 and 29 . 36813636 2 −=+− b Subtract 36 from both sides of the equation. 452 =b Simplify. 452 =b Square root both sides of the equation. 7.6or −≈ 6.7b Ignore the -6.7. We will only be dealing with the positive

answers when working with this formula and other formulas in this section of the unit.


Example 6:

Solve for r given that 3 34 rV π= , V = 48, and π = 3.14.

Answer: 3

34 rV π=

( ) 3 14.33448 r= Substitute the known values in the equation.

319.448 r= Using a calculator, multiply 34 by 3.14.

19.4

19.419.4

48 3r= Divide both sides of the equation by 4.19

349.11 r= 33 49.11 r= Cube root both sides of the equation. r≈26.2 Using a calculator, you learn that r is approximately equal

to 2.26. Example 7:

Solve for r given that hrV 2 31π= , V = 268, π = 3.14, and h = 15.2.

Answer:

hrV 2 31π=

( ) ( )2.15 14.331268 2r= Substitute the known values in the equation.

2 91.15268 r= Using a calculator, multiply 31 , 3.14 and 15.2.

91.15 91.15

91.15268 2r

= Divide both sides of the equation by 15.91.

284.16 r= 284.16 r= Square root both sides of the equation. r≈ 4.10-or 4.10 Ignore the negative answer.

Questions:

1. Solve the following equations. Show all your work. State all possible answers.

(a) 492 =x (b) 1253 =x


(c) 1452 =+x (d) 20123 =+x (e) 828 2 −= a (f) 16113 =−p (g) 197 2 =+ x (h) 31217 d+= (i) 2862 2 =−x (j) 2237 2 =+ t (k) 7419 3 += p (l) 351916 x+−=


(m) 8.71.24.32 2 −= h (n) 32.19.23.38 k+= (o) 89.37.05.32 3 =+− x (p) 2.4334.685.0 2 =+t Scrambled Answer Key for Question 1

3 or -3 3.73 3.46 or -3.46 3.09 5 1.71 1.44 7 or -7 1.91 2 4.12 or -4.12 6 or -6

4.38 or -4.38 3 6.59 or -6.59 2.24 or -2.24 2. Solve for the unknown variable. You will be dealing with trigonometry formulas, area

formulas, and volume formulas. For this reason, only state positive solutions.

(a) Solve for a given that 222 cba =+ , b = 7, and c = 11.


(b) Solve for s given that 26sA = and A = 42. (c) Solve for s given that hsV 2= , V = 60, and h = 6. (d) Solve for b given that 222 cba =+ , c = 12.5, and a = 7.2


(e) Solve for s given that hsV 2= , h = 8.4, and V = 93.2. (f) Solve for r given that 2 rA π= , π = 3.14 and A = 42 (g) Solve for r given that hrV 2 π= , h = 4, V = 83, and π = 3.14


(h) Solve for r given that 2 rA π= , A = 123.7 and π = 3.14.

(i) Solve for r given that 3 34 rV π= , π = 3.14 and V = 325.

(j) Solve for r given that 2 4 rA π= , π = 3.14 and A = 367.


(k) Solve for r given that hrV 2 31π= , π = 3.14, h = 14.5 and V = 423.

(l) Solve for r given that 3 34 rV π= , π = 3.14 and V = 237.5.


3.33 5.41 8.49 4.27 10.22 3.84 6.28 3.16 5.28 3.66 2.57 2.65


Wrap-Up Statement: In the section “Equations Where a Variable is Squared or Cubed”, you learned that the

inverse operation of squaring was square rooting, and that the inverse operation of cubing was cube rooting. You applied these inverse operations to solve equations of the form

CBAx =+2 and CBAx =+3 . Reflect Upon Your Learning Fill out this questionnaire after you have completed questions 1 and 2. Select your response

to each statement. 1 - strongly disagree 2 - disagree 3 - neutral 4 - agree 5 - strongly agree (a) I understand all of the concepts covered in the section,

“Equations Where a Variable is Squared or Cubed.” 1 2 3 4 5


1 2 3 4 5



For Your Information (FYI) Several formulas were used in question 2. Although you do not need to know where these formulas are used at this time, it is still nice to what they are called. 222 cba =+ Pythagorean Theorem 26sA = Surface Area of a Cube Formula hsV 2= Volume of a Square-based Prism Formula 2 rA π= Area of a Circle Formula hrV 2 π= Volume of a Cylinder Formula

3 34 rV π= Volume of a Sphere Formula

2 4 rA π= Surface Area of a Sphere Formula

hrV 2 31π= Volume of a Cone Formula


Post-Unit Reflections What is the most valuable or important thing you learned in this unit?

What part did you find most interesting or enjoyable?

What was the most challenging part, and how did you respond to this challenge?

How did you feel about this math topic when you started this unit?

How do you feel about this math topic now?

Of the skills you used in this unit, which is your strongest skill?

What skill(s) do you feel you need to improve, and how will you improve them?

How does what you learned in this unit fit with your personal goals?


Answers: Balance Puzzles (pages 1 to 3) 1. (a) - Remove 1 ball from each side. - If 2 boxes equal 6 balls, then 1 box must be equal to 3 balls. (b) - Remove 2 balls from both sides. - If 3 boxes equal 3 balls, then 1 box must be equal to 1 ball. (c) - Remove 1 box from each side. - If 2 boxes equal 10 balls, then 1 box must be equal to 5 balls. (d) - Remove 1 box from each side. - Remove 3 balls from each side. - That leaves us with 6 balls in 1 box. (e) - Remove 2 boxes from each side. - Remove 1 ball from each side. - That leaves us with 4 balls in 1 box. (f) - Remove 1 box from each side. - Remove 3 balls from each side. - If 2 boxes equal 4 balls, then 1 box must be equal to 2 balls. (g) - Remove 1 box from each side. - Remove 1 ball from each side. - If 3 boxes equal 6 balls, then 1 box must equal 2 balls. (h) - Remove 1 box from each side. - Remove 3 balls from each side. - If 2 boxes equal 0 balls, then 1 box must equal 0 balls. (i) - Remove 3 boxes from each side. - Remove 1 ball from each side. - That leaves us with 3 balls in 1 box. (j) .- Remove 2 boxes from each side. - Remove 2 balls from each side. - If 2 boxes equal 4 balls, then 1 box must equal 2 balls. (k) - Remove 2 boxes from each side. - Remove 4 balls from each side. - If 2 boxes equal 10 balls, then 1 box must equal 5 balls.


(l) - Remove 1 box from each side. - Remove 2 balls from each side. - If 3 boxes equal 12 balls, then 1 box must equal 4 balls. Revisiting the Balance Puzzles (pages 4 to 10) 1. (a) (i) matches with d (ii) matches with g (iii) matches with a (iv) matches with j (v) matches with b (vi) matches with e (vii) matches with i (viii) matches with h (ix) matches with c (b) Question (f) Algebraic Steps 3371 +=+ nn

327133171

+=−+=−+

nnnnn

n

n24

33237=

−+=−

n

n

=

=

22

224

Question (k) Algebraic Steps 14244 +=+ nn

14422142244

=+−+=−+

nnnnn

102

414442=

−=−+nn

52

102

2

=

=

n

n


Question (l) Algebraic Steps 24141 +=+ nn

23141241141

+=−+=−+

nnnnn

n

n312

223214=

−+=−

n

n

=

=

43

33

12

2. (a)

(b) Remove 2 boxes from both sides of the scale. Remove 3 balls from both sides of the scale. If 2 boxes are equal to 6 balls, then 1 box equals 3 balls. There are 3

balls in each box. 3. Algebraic Steps Written Descriptions 13245 +=+ nn Original Question

13432132245

=+−+=−+

nnnnn

Remove 2 boxes from both sides of the scale.

93

413443=

−=−+nn

Remove 4 balls from both sides of the scale.

339

33

=

=

n

n

If 3 boxes are equal to 9 balls, then 1 box equals 3 balls. There are 3 balls in each box.

4. Algebraic Steps Written Descriptions 13136 +=+ nn Original Question

13351131136

=+−+=−+

nnnnn


105

313335=

−=−+nn

Remove 3 balls from both sides of the scale.

25

105

5

=

=

n

n

If 5 boxes are equal to 10 balls, then 1 box equals 2 balls. There are 2 balls in each box.


5. (a) 3 (b) 2 (c) 5 (d) 8 (e) 4 (f) 1 Moving From Balance Puzzles to Algebra (pages 11 to 18) 1. (a) To solve the equation 98 =+n you must subtract 8 from both sides of the equation..

When you do this, you would find that n equals 1.

(b) To solve the equation 213 =p you must divide both sides of the equation by 3. When

you do this, you would find that p equals 7.

(c) To solve the equation 24 =−d you must add 4 to both sides of the equation. When you

do this, you would find that d equals 6.

(d) To solve the equation 43=

x you must multiply both sides of the equation by 3. When

you do this, you would find that x equals 12.

(e) To solve the equation 84 =n , you must divide both sides of the equation by 4. When you

do this, you would find that n equals 2.

(f) To solve the equation 149 =+ g , you must subtract 9 from both sides of the equation.

When you do this, you would find that g equals 5.

2. (a) 5476 −=+ nn nnnn 454476 −−=−+ 572 −=+n 75772 −−=−+n 122 −=n

212

22 −

=n

6−=n


(b) 23357 +=− xx xxxx 3233357 −+=−− 2354 =−x 523554 +=+−x 284 =x

428

44

=x

7=x (c) pp 41312 −=+ pppp 4413412 +−=++ 1316 =+p 113116 −=−+p 126 =p

612

66

=p

2=p 3. It doesn’t make any difference. 4. (a) 26486 +=+ nn (b) xx 31157 +=− nnnn 4264486 −+=−+ xxxx 3311357 −+=−− 2682 =+n 1154 =−x 826882 −=−+n 511554 +=+−x 182 =n 164 =x

218

22

=n

4

164

4=

x

9=n 4=x

(c) pp 22373 −=− (d) aa 21352 −=+ pppp 2223273 +−=+− aaaa 2213252 +−=++ 2375 =−p 1354 =+a 723775 +=+−p 513554 −=−+a 305 =p 84 =a

530

55

=p

48

44

=a

6=p 2=a


(e) 25973 −=− xx (f) nn 23316 −=+ 259973 −=−− xxx nnnn 2233216 +−=++ 2576 −=−− x 3336 =+ n 725776 +−=+−− x 633636 −=−+ n 186 −=− x 273 =n

618

66

−−

=−− x

3

273

3=

n

3=x 9=n

(g) 1156 −=−p (h) 1937 =− x 511556 +−=+−p 719737 −=−− x 66 −=p 123 =− x

66

66 −

=p

3

123

3−

=−− x

1−=p 4−=x 5. xx 44753 −=+ Written Explanations xxxx 4447453 +−=++ To get rid of the -4x on the right hand side of the equation,

add 4x to both sides of the equation. 4757 =+x Simplify 547557 −=−+x To get rid of the +5 on the left hand side of the equation,

subtract 5 from both sides of the equation. 427 =x Simplify

742

77

=x

To get rid of the 7 that is multiplied by the x, divide both sides of the equation by 7.

6=x 6. (a) -6 (b) 7 (c) -5 (d) 2 (e) 3 (f) 8 Solve and Check (pages 19 to 23) 1. The answer is correct. 2. The answer is incorrect. 3. (a) 2 (b) -4 (c) -8 (d) 1 (e) 3 (f) 5


(g) -2 (h) 9 (i) -7 (j) -1 (k) -12 (l) -3 Dealing with x and -x (pages 24 to 27) 1. 4 2. 2 3. -8 4. 6 5. -3 6. 7 7. 1 8. 10 9. -6 10. 12 Formulas and Solving for an Unknown (pages 28 to 30) 1. 2 2. 6 3. 5 4. -3 5. 7 6. -1 7. -4 8. 0 Equations with Decimals (pages 31 to 37) 1. 20 2. 30 3. 11 4. -3.5 5. 12 6. -10 7. 2 8. 4 9. -60 10. 175 11. 40 12. -6.25 13. -40.5 14. -48.5 15. -0.4 16. 0.53 Equations with Fractions (pages 38 to 45) 1. (a) 2 (b) 5 (c) 4 (d) 6 (e) 8 (f) 9 (g) 15 (h) 1 (i) 10 (j) 3 (k) 16 (l) 12 2. (a) 3 (b) 5 (c) 14 (d) 20 (e) 6 (f) 12 3 (a) 9 (b) 24 4. (a) 12 (b) -5 (c) 6 (d) 18 (e) -10 (f) 30 (g) -24 (h) -6 (i) -16


(j) -12 (k) 14 (l) 9 Equations in the Real World (pages 46 to 53) 1. (a) $320 (b) 6 tonnes (c) $390 (d) 12 tonnes 2. (a) 4 payments (b) $80 (c) 8 payments (d) $280 3. (a) 21.2 litres (b) 10 seconds (c) 34.4 litres (d) 15 seconds 4. (a) 30 minutes (b) 231.6 kilopascals (c) 198 kilopascals (d) 28 minutes 5. (a) 75 cm (b) 18 hours (c) 3 hours (d) 175 cm Equations Where a Variable is Squared or Cubed (pages 54 to 64) 1. (a) x = 7 or -7 (b) x = 5 (c) x = 3 or -3 (d) x = 2 (e) a = 6 or -6 (f) p = 3 (g) x = 3.46 or -3.46 (h) d = 1.71 (i) x = 4.12 or -4.12 (j) t = 2.24 or -2.24 (k) p = 1.44 (l) x = 1.91 (m) h = 4.38 or -4.38 (n) k = 3.09 (o) x = 3.73 (p) t = 6.59 or -6.59 2. (a) a = 8.49 (b) s = 2.65 (c) s = 3.16 (d) b = 10.22 (e) s = 3.33 (f) r = 3.66 (g) r = 2.57 (h) r = 6.28 (i) r = 4.27 (j) r = 5.41 (k) r = 5.28 (l) r = 3.84


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Solving Equations Unit - · PDF fileSolving Equations Unit (Level IV Graduate Math) ... Moving from Balance Puzzles to Algebra ... 4 1 3 3 4 3 4 1 3 4 = = + − = −