Chapter 3: Solving Equations and Problem Solving

Index

Section Page 3.1 Simplifying Algebraic Expressions 2 – 83.2 Solving Equations: The Addition Property 9 – 133.3 Solving Equations: The Mult. Property 14 – 21

Ch. 3 Integrated Review 22 – 263.4 Solving Linear Equations in 1 Variable 27 – 363.5 Linear Equations in 1 Var. & Problem Solving 37 – 42

Review 43 – 54Practice Test 55 – 59

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§3.1 Simplifying Algebraic Expressions

First, let’s review some vocabulary that we encountered in section 8 of chapter 1. This is vocabulary that we need in order to talk about simplification of algebraic expressions.

Variable – Any unknown value denoted by a letter of the alphabet (most commonly “x”).Example: a) x

b) z

Term – Number, variable, product of a number and a variable or a variable raised to a power.

Example: a) 5b) 5xc) xyd) x2

Numeric Coefficient – The numeric portion of a term with a variable.Example: What is the numeric coefficient?

a) 3x2

b) x/2

c) - 5x/2

d) – z

Constant – Any term that is a single number, which is not multiplied by a variable.Example: What is the constant in each expression?

a) 3x + 5b) 27y 1c) –72 + y

This next one is a new concept, but it follows directly from the information that we already have.

Like Term – Terms that have a variable, or combination of variables, that are raised to the exact same power.

Example: Are the following like terms?a) 7x 10xyb) - 15z 23zc) t 15tvd) 5 5we) xy 6xyf) xy - 2xyz

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Simplifying an algebraic expression by combining like terms means adding or subtracting terms in an algebraic expression that are alike.

Simplifying An Algebraic ExpressionStep 1: Change all subtraction to additionStep 2: Use the distributive property wherever necessaryStep 3: Group like termsStep 4: Add numeric coefficients of like termsStep 5: Don’t forget to separate each term in your simplified

expression by an addition symbol!!

Example: Simplify each

a) 15x + 13x

b) 8z 9z 35z

c) 2x + 4y + 5x 2y + 5

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d) 5xy + 2 + 7 2xy

e) 6(9 + 2xy)

f) 6(x + 7) 2x + 5

g) -9(2 + y) + 4y y

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h) x + 6(6 x) 4x

j) 5(x + 7) + 2(x 4)

k) –(x + y) + 2y

Note: This problem shows an important concept of how to distribute the negative sign when it is in front of parentheses.

l) 7 (x + 2)

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Note: This problem shows another important concept – dealing with the subtraction when there are parentheses following it.

The last thing that we need to discuss in this section is the use of algebraic expressions in defining areas and perimeters of figures. We will be using the area formulas that we have already discussed in chapter one and we will be using the knowledge that we have just acquired in simplifying algebraic expressions.

Example: Find the perimeter of the following figures.

a) The following is a regular octagon.

b) Ignore the fact that this figure looks regular.

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s = (z + 5) in.

S1= 12 metersS5= 25 meters

Example: Find the area of the following figures.

a)

b) The following is a compound shape. Find its area.

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S2= (x + 3)meters

S3=(5 + 2x) meters

S4=(3x + 10) meters

= 21 in.

w = (x + 21) in.

l1= 21 cm

w1 = (x + 2) cm

HW §3.1p. 177-180 #3-6(mental math) #2-10even,#12-20mult.of4,#24-74even,#78,80,#81-86all,#92

§ 3.2 The Addition Property of Equality

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l2 = (2x + 3) cm

w2 = 7 in.

Linear Equations in One Variable are equations that can be simplified to an equation with one term involving a variable raised to the first power which is added to a number and equivalent to a constant. Such an equation can be written as follows:

ax + b = ca, b & c are constantsa 0x is a variable

Our goal in this and the next section will be to rewrite a linear equation into an equivalent equation (an equation with the same value) in order to arrive at a solution. We will do this by forming the equivalent equation:

x = # or # = x

x is a variable # is any constant

Remember that only one value will make a linear equation true and that is the value that we desire. This also means that we can check our value and make sure that the original equation, when evaluated at the constant, makes a true statement.

The Addition Property of Equality says that given an equation, if you add (subtract) the same thing to both sides of the equal sign, then the new equation is equivalent to the original.

a = b and a + c = b + care equivalent equations

Example: Form an equivalent equation by adding the opposite of the constant term on the left.

9 + x = 12

We’ll use the additive property of equality to move all variables to one side of the equation (for simplicity, in the beginning, we will move variables to the left) and all numbers to the other side (in the beginning to the right). This is known as isolating the variable.Isolating the variable is done by adding the opposite of the constant term to both sides of the equation (the expression on each side of the equal sign must first be simplified of

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course.) We add the opposite of the constant term in order to yield zero! This makes the constant disappear from the side of the equal sign with the variable! Don’t get to carried away and forget that in order to make it disappear, you must add the opposite to both sides of the equal sign!

Example: Solve by using the addition property of equality and the additive inverse. In order to solve these, you must look at the equation and ask yourself, “How will I make x stand alone?” This is where the additive inverse comes in -- You must add the opposite of the constant term in order to get the variable to “stand alone.”

a) x + 9 = 11

b) 15 = x + 2

c) x 3 = 10

d) 8 = x 3

e) x + 3 = 0

Yes, you can probably look at each of the above equations and tell me what the variable should be equivalent to, but the point here is to develop a method that will work when the equation is more complicated!!

How do we know that we got the right number for the above answers? We check them using evaluation!

Example: Check to problems from the previous example

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a) x = 2 so we replace x with 2 and get the following(2) + 9 = 11

11 = 11 true statement, therefore()this checks b)

c)

d)

e)

Not only can we move the constant from one side of the equation to the other, but we can also move the variable term from one side of the equation to the other.

Example: Solve each equation using the addition property of equality.a) 8 + 7y = 8y

b) 19z = 9 + 18z

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Of course, every problem will not be as simplistic as the ones used as examples so far, and sometimes we will have to simplify a problem before we can solve it. We will do this by treating each side of the equation as an algebraic or numeric expression (recall algebraic expressions involve terms with variables, but no equal sign) and use the skills developed in section 1 of chapter 3.

Example: Solve the following problems.a) 6a 2 5a = -9 + 1

b) 5a + 6 4a = 7a + 8 7a

c) 3(a + 7) 2a = 1

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d) - 2(2a 12) (-5a) = 4

e) 3(a + 6) = 4a + 1

Note: When our equations get complicated we must always remember that once we have simplified both sides of the equation that we must remember to keep the variable positive because at this time we do not know what to do if the variable is not positive!

HW §3.2p. 185-188 #4,8,12,#16-60mult.of4,#67-71all,#73,79,81§ 3.3 The Multiplication Property of Equality

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Let’s first introduce a concept that your book will not discuss until chapter 4, but one that I prefer to discuss when introducing the multiplication property of equality. The concept is that of a reciprocal. A reciprocal is any number which when multiplied by the number at hand yields the identity element of multiplication (recall this is one). The reciprocal can be thought of in several other ways also. The easiest way at this time will be to think of it as one over the number, so that it will look like a fraction, 1/x. Remember that whenever you multiply any number by its reciprocal, the product is one!! Let’s practice the reciprocal and then we will talk about one other important concept.

Example: Find the reciprocal of each number.a) 5

b) -7

c) 3

d) 0

Division is multiplication by the reciprocal of the divisor!!! I have mentioned this in passing before, but now I am stating it as a new way of defining division, and we will be using the new definition so we should get used to it!

Example: Dividea) 27(1/3)

b) 8(1/2)

c) 105(1/5)

Note: I prefer to introduce these concepts and teach the multiplication property of equality using them rather than learning it one way (the way the book shows, using division) and then having to go back and seemingly re-learn the whole process when we introduce numeric coefficients that are fractions. By learning the concept this way we can apply it to either numeric coefficients that are whole numbers or that are fractions!

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The Mulitiplication Property of Equality says that two equations are equivalent if the second is the same as the first, where both sides have been multiplied by a nonzero constant.

a = b is equivalent to ac = bc

We will be using this to solve such problems as:Example: 4x = 24

By inspection we see that if x = 6, both sides are equivalent! However, if we want to isolate x what must 4x be multiplied by? Think about it’s

reciprocal! (Remember that we just learned that if we multiply a number by its reciprocal we will always get 1!)

Using the idea of isolating the variable and the multiplication property of equality, we can arrive at a solution for x!

Example: Solve using the multiplicative property of equalitya) 16x = 48

b) 14x = 28

c) 2x = 24

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Now let’s combine what we have learned in the last section (3.2) with what we have learned in this section. We will use the addition property first to move all variables to one side and number to the other and then we will use the multiplication property to allow the variable to stand alone, thus arriving at our solution. Here are the steps that we’ll follow:

Solving Algebraic EquationsStep 1: Simplify the expressions on either side of the equal sign as much as possible

by combining all like terms in each expression.Step 2: Move all variables to one side of the equal sign using the addition property of

equality. Try to keep the variable positive.Step 3: Move all the numbers to the opposite side of the equal sign from the variables

using the addition property of equality.Step 4: Remove the numeric coefficient by multiplying both sides of the equation by

its reciprocal (divide by numeric coefficient; multiplication property of equality).

Step 5: Check your solution by evaluating the original equation at the answer.

Example: Solve the following and checka) 2x + 3 = 13 4

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b) 7x 2 + 2x = 25

c) 2 17 8 = -2 + 5x 2x

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The answer to an equation can be zero. (We will find as we progress that some equations also have no solution, but at this point we will have no problems that have no solution.) The answer will be zero because we will have multiplication by zero. Do not stop solving an equation, until you have the variable standing alone, even if the equation is something equals zero, be sure that you multiply both sides by the reciprocal of the numeric coefficient and obtain the answer.

Example: Solve 7x + 2x 9 = -9

The next concept that has great importance in this chapter is the translation of algebraic expressions. This is just like the translation of algebraic expressions that we covered in section 1.8, but they may be a little more complicated. Here are the phrasings again.

Words and Phrasing for Translation Problems, by OperationNote: Let any unknown be the variable x.

AdditionWord Phrasing Algebraic Expression

sum The sum of a number and 2 x + 2more than 5 more than some number x + 5added to Some number added to 10 10 + xgreater than 7 greater than some number x + 7increased by Some number increased by 20 x + 20years older than

15 years older than John x + 15

Note: Because addition is commutative, each expression can be written equivalently in reverse, i.e. x + 2 = 2 + x

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SubtractionWord Phrasing Algebraic Expression

difference of The difference of some number and 2The difference of 2 and some number

x 22 x

years younger than

Sam's age if he is 3 years younger than John

x 3

diminished by

15 diminished by some numberSome number diminished by 15

15 xx 15

less than 17 less than some numberSome number less than 17

x 1717 x

decreased by Some number decreased by 1515 decreased by some number

x 1515 x

subtract from Subtract some number from 51Subtract 51 from some number

51 xx 51

Note: Because subtraction is not commutative, x 2 2 x

MultiplicationWord Phrasing Algebraic Expression

product The product of 6 and some number 6xtimes 24 times some number 24xtwice Twice some number

Twice 242x2(24)

multiplied by 8 multiplied by some number 8xat Some number of items at $5 a piece $5x"fractional part" of

A quarter of some number ¼ x or x/4 .

"Amount" of "$" or "¢"

Amount of money in some number of dimes (nickels, quarters, pennies, etc.)

0.1x (dollars) or 10x (cents)

percent of 3 percent of some number 0.03xNote: Because multiplication is commutative all of the above algebraic expressions can be written equivalently in reverse, i.e. 6x = x6. It is standard practice to write the numeric coefficient and then the variable, however.

DivisionWord Phrasing Algebraic Expression

quotient The quotient of 6 and some numberThe quotient of some number and 6

6 xx 6

divided by Some number divided by 2020 divided by some number

x 2020 x

ratio of The ratio of some number to 8The ratio of 8 to some number

x 88 x

Note: Division can also be written in the following equivalent ways, i.e. x 6 = x/6 = 6x = x

6Exponents

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Words Phrasing Algebraic Expressionsquared Some number squared x2

square of The square of some number x2

cubed Some number cubed x3

cube of The cube of some number x3

(raised) to the power of

Some number (raised) to the power of 6

x6

Recall that when you see the phrasing: twice the sum(difference) of there will be parentheses around the sum(difference).

Example: Translate the following algebraic expressions.a) Twice the sum of 15 and some number.

b) Twice the difference of a number and 20.

The final concept in this chapter that I wish to discuss is the use of a formula called the distance formula. This formula says that if you multiply rate (of speed) and time you will get the distance. This is a physics formula and it is used all the time in algebra problems. Using variables the equation is written as follows:

Distance Formula

d = r t where d = distancer = rate of speedt = time

Example: A car is traveling at 60 mph. It is traveling from LA to SF which

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is a distance of approximately 300 miles.

Example: A train travels from Portland to Los Molinos a distance of approximately 450 miles in 9 hours. What is the train’s rate of speed? Use the distance formula to set up the equation and solve the problem.

HW §3.3p. 195-198 #4-20mult.of4,#22-32even,#36-68mult.of4,#72-90mult.of3,#100,106,107&110

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Ch. 3 Integrated Review

Evaluation1. Place parentheses where variables are to be replaced2. Use order of operations to evaluate and achieve a single numeric answer

Example: Evaluate the following when x = 5, y = 7 & z = -7a) 7x 2 + 3z

b) 2x2 + 2yz

c) 2(y + z)2

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Simplifying Algebraic Expressions1. Change all subtraction to addition2. Use the distributive property wherever necessary3. Group like terms4. Add numeric coefficients of like terms5. Don’t forget to separate each term in your simplified expression by an addition symbol!!

Example: Simplify each of the followinga) 2x + 7 7x + -3

b) 2(x + 7) 9

c) 5(2 7x) 3(x + 4)

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Solving Algebraic Equations1. Simplify the expressions on either side of the equal sign as much as possible by combining all like terms in each expression.2. Move all variables to one side of the equal sign using the addition property of equality. Try to keep the variable positive.3. Move all the constants to the opposite side of the equal sign from the variables using the addition property of equality.4. Remove the numeric coefficient by multiplying both sides of the equation by its reciprocal (divide by numeric coefficient; multiplication property of equality).5. Check your solution by evaluating the original equation at the answer.

Example: Solve each equation & checka) a + 17 = 59

b) 9x = 108

c) 9z + 7 = 8z 2

d) 3(2z + 2) 2z = -2

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Area Using Algebraic ExpressionsArea of Rectangle = Length Width Area of Square = side2

Area of Triangle = ½ Base HeightArea of Trapezoid = ½ Height (Big Base + Small Base)Area of Parallelogram = Height Length In finding the area of any figure that involves an algebraic expression, the answer will more than likely be achieved by using the distributive property and will definitely involve simplifying an algebraic expression. Always remember that the area of any figure has square units, that the formula used to arrive at the area needs to be shown and that the variables used in the formula (equation) must be defined and specified.

Example: Find the area of the following figures.

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(2x + 5) meters

4 meters

15 cm

(x 3) cm

Translation1. Review the words and phrases for operators on pages 127-1292. Remember that the phrasing “twice the sum (difference) of,” indicates 2(x + y)3. Work slowly and circle operators and underline numbers. Don’t forget to find the “and” that goes with phrasings such as “the sum of ‘this’ and ‘that’”.4. Don’t forget to define your variable!

Example: Translate each of the following into algebraic expressions

a) The difference of 54 and some number

b) Twice the sum of 15 and some number

c) Fifteen subtracted from the quotient of 64 and a number

d) A number added to the product of the number and 8

HW Integrated Reviewp. 199-200 all

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§3.4 Solving Linear Equations in One Variable

Linear Equations in One Variable are equations that can be simplified into an equation with one term involving a variable raised to the first power, which is added a second term which is a constant and the sum of those two terms is equivalent to a constant. Such an equation can be written as follows:

ax + b = ca, b & c are constantsa 0x is a variable

Note: All the problems for which we have been finding solutions in the last two sections have been linear equations in one variable. Even though an equation does not have a “b” term, it is always there, it is just sometimes zero, and therefore we do not have to write it out!

Let’s review the steps for solving an algebraic equation, which can now be referred to as a linear equation in one variable.

Solving Linear Equations in One Variable1. Simplify the expressions on either side of the equal sign as much as possible by combining all like terms in each expression.2. Move all variables to one side of the equal sign using the addition property of equality. Try to keep the variable positive.3. Move all the constants to the opposite side of the equal sign from the variables using the addition property of equality.4. Remove the numeric coefficient by multiplying both sides of the equation by its reciprocal (divide by numeric coefficient; multiplication property of equality).5. Check your solution by evaluating the original equation at the answer.

Although we have been solving linear equations in one variable for the last two sections they will now get a little more complicated. We have all the tools that we need to solve any problem; we just have to apply them in the correct order!

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Example: Solve 12x 5 = 23 2xStep 1: Move all variables to the one sideStep 2: Move all constants to other sideStep 3: Isolate the variable(use multiplicative prop. of equality)**Steps 1 & 2 are interchangeable. These steps both use the additive property of equality.

Example: Solve 2x + 10 + 14x = 2x 18Step 1: Simplify each side of the equationStep 2: Move all variables to one sideStep 3: Move all constants to the other sideStep 4: Isolate x by multiplying by numeric coefficient’s reciprocal

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Example: Solve 5(2x 1) 6x = 7Step 1: Simplify each side of the equation (this includes

distributive property)Step 2: Move all variables to one sideStep 3: Move all constants to the other sideStep 4: Isolate x by multiplying by the reciprocal of the numeric

coefficient

Note: Step 1 should simplify the problem! You should, 1. use the distributive property2. combine all like terms on each side of the equation

Now, here are some more to practice in class. Raise your hand if you especially need help. We will do these together in a few minutes, but you are expected to try them on your own, while I circulate.

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Your TurnExample: Solve each of the following linear equations in one variable

a) -7 (10x + 3) = 5x (x 18)

b) z + 7 18 = 3z + 1

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c) 3(z + 2) = 6z 4(z + 1)

f) 8(c + 9) 7 = 3 + 5(c 2)

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Now, for some more translation problems, but this time we will be translating linear equations in one variable. In this section the author only discusses translation of a mathematical equation, but I will also discuss it from the viewpoint of the next section. As a result I will not be covering the work from the next section involving translation of a linear equation in one variable and its solution. Here are some words that we will be seeing that mean equals.

EqualsLet x=# for all of the following.

Words Phrasing Algebraic Equationis The sum of some # and 4 is 9. x + 4 = 9will be 12 decreased by a # will be 8. 12 x = 8was The quotient of some # and 6 was 2. x 6 = 2equals The difference of 3 and some # equals 2 3 x = 2is equal to Some # minus 3 is equal to 7 x 3 = 7yields The product of some # and 2 yields 26 2x = 26gives A # subtracted from 2 gives –1 2 x = -1amounts to -5 and a # amounts to 15 -5 + x = 15is the same as 4 is the same as a # minus –15 4 = x (-15)Note: Any form of the word “is” can be used to mean equal.

Example: Write a mathematical equation.a) The sum of 27 and negative 3 is 24.

b) The difference of –18 and 12 yields -30.

c) Five times the difference of 3 and –2 equals 25.

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Remember that defining your variable, the linear equation, solving the equation, and the answer are all part of the work and the credit you will receive. They must all be there!

Example: Write a linear equation in one variable and solve it.a) Five more than twice a number is 37. What is the number?

b) Negative 5 added to twice some number yields 27. Find the number.

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c) Twice the sum of a number and –5 is the same as 48.

d) Twice a number, subtracted from 28, gives 14. Find the number.

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e) Fourteen less half of 24 amounts to the sum of a number and 10. What is the number?

e) Twenty less a number is 3 times the sum of the number and 8.

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The last thing that we need to do in this section is review some complex evaluation problems. Remember that evaluation is very important because this is what we use to check our answers to linear equations in one variable.

Example: Evaluate (-y)3 5xy when x = 2 and y = -1

Example: Evaluate y3 5xy when x = 2 and y = -1

Note: Always make your variable a parenthesis before you substitute!!

HW §3.4p.205-207 #4-60mult.of4,#61-68all,#74-82even,#85&86

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§3.5 Linear Equations in 1 Variable & Problem Solving

In the last section we went over many translation of algebraic equation problems. This section continues to beat this concept into the ground. I am not going to continue the discussion, as I feel that although you need more practice, that practice is best done on your own. Thus we will skip to the very important portion of this chapter, word problems involving algebraic equations.

Almost all of the word problems in this section follow the following pattern:

First Information = Second + Some Number (Redefine with a sum S + #1)Second Information = Unknown (Define with variable; S)Total = First + Second = #2 (Substitute variable

expressions to create an equation; (S+#1) + S = #2)

Following are some problems that follow this pattern, some with a little more complex nature, but nonetheless, all the same pattern and pattern is what we want to get used to searching for.

Example: Juan and his brother collect baseball cards. Juan has seventeen more cards than his brother. Together they have 267 cards. How many cards does Juan have?

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Example: During the Student Government Association elections, one candidate received 38 more votes for Senator than the other candidate. If the total number of votes cast for the two candidates was 164, how many votes did each candidate get?

Example: A mountain climber was at 632 feet elevation He descended to an elevation of 430 feet. How many feet did he descend?

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Example: The length of a rectangle is twice the width. The perimeter is 60 meters. Find the width.

One other important bit of information that we should mention is the wording:Something has 5 times as (many, more, fast) as Another Thing 5 times Another Thing

Thus if Another Thing = x then Something = 5x

Example: A gambler lost some money on cards on Friday. On Saturday, he lost twice as much money. After the two nights of cards, he has lost $81. How much did he lose on Friday?

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Example: A bicyclist travels twice as fast as someone who is walking. The combined speed is 18 miles per hour. Find the speed of the bicyclist.

Example: A washing machine with installation costs $420. If the washing machine costs 5 times as much as the installation, find the cost of installation.

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There are few problems of the following type:

First Information = Second #1 = #2 (Redefine with S #1 = #2)Second Information = Unknown (Redefine with variable; S)

The equation in this type of problem comes from the sum (difference) in the first information.

Example: Ida has $57 less in her checking account than Ray. Ida has $279. How much does Ray have?

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Example: While playing volleyball the Wildcats scored 8 points less than the Bulldogs. The Wildcats scored a total of 45 points. How many points did the Bulldogs score?

§3.5 HWp. 213-217 #1-8all,#10-28even,#30-54even

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Chapter 3 Review

Simplifying by Combining Like Terms

Example 1: 7x + 2x

Example 2: 5x 15x + 17 9

Example 3: 4(8 + 2z)

Example 4: 7(z 4) + z 1

Example 5: 2v 4(8v + 3) + 3

Example 6: (5 2x) + 9

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Finding Perimeter & Area

Example 7: Find the perimeter of a rectangle with a width of (2x + 5) inches and a length of (x + 7) inches.

Example 8: Find the perimeter of the following:

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(5x + 5) cm

(x + 3) cm

(2x + 3) cm

5 cm

Example 9: Find the area of a rectangle whose length is 5 km and whose width is(5x + 3) km.

Example 10: Find the area of the figure below.

Addition Property of Equality

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(x 5) m

7 m

(4x + 12) m

5 m

Example 11: x + 5 = 29

Example 12: x 5 = 13

Example 13: 12 x = 0

Example 14: 7x + 13 6x = 3

Example 15: 7x 5 = 6x + 3

Example 16: 7(x 3) + -2x = 4x 1

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Checking Solutions

Example 17: Check the solution to example 11.

Example 18: Check the solution to example 16.

Equations vs. Expressions

Example 19: What is the difference between an expression and an equation?

Example 20: T or F An expression has a solution.

Example 21: T or F An expression can be evaluated.

Example 22: T or F An equation’s solution can be checked by evaluation at the

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solution?

Reciprocals & Division as Multiplication

Example 23: What is the reciprocal of 15?

Example 24: What is the reciprocal of –2?

Example 25: T or F Multiplication by the reciprocal of the divisor is the same as division?

Multiplication Property of Equality

Example 26: Solve 8x = 72

Example 27: Solve 27x = 0

Example 28: Solve -2x = 28

Example 29: Solve -3z = -27

Combining All Steps In Solving Equations

Example 30: Solve 2y + 1 = 5(y 4)

Example 31: Solve 3 5 + 2x = x 6

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Example 32: Solve 9 + 2(7y 4) = -27

Example 33: Solve 3(5x 7) + 12 = 7 -2(7x 3) + 5

Translation Problems

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Example 34: Twice the sum of a number and six is 86. Find the number.

Example 35: Solve: Twice the sum of a number and ten is equal to six.

Example 36: Solve: Three times a number less forty-five is equal to twice the same number.

Distance Formula

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Example 37: It is 250 miles from point A to point B. If a car travels at 50 miles per hour, how long will it take for the car to arrive at point B, after leaving point A?

Example 38: It took Tim 3 hours to travel from St. Paul to Dakota. If the distance between St. Paul and Dakota is 216 miles, what was Tim’s average speed?

Complex Evaluation

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Example 39: Evaluate x3 + 4vw when x = -2, v = -3, & w = -1

Example 40: Evaluate -(x y)2 when x = 15 & y = -5

Example 41: Evaluate 2x3 + 7xy when x = -2 & y = 3

Word Problems

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Example 42: During the student government association elections, candidate A received 28 more votes for senator than candidate B. If the total number of votes cast for the two candidates was 254, how many votes did each candidate receive?

Example 43: Find the length of a rectangle if it is four less than twice the width and the perimeter is 40 cm.

Example 44: A book costs three times as much as a CD. If the cost of the book and the

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CD together is $72, find the cost of each.

Example 45: Ivan has 191 CD’s in his collection. This is 54 more than Yvonne has. How many CD’s does Yvonne have?

HW Ch. 3 Test (from book) p. 227 – 228 all

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Ch. 3 Practice Test

The following are a compilation of problems from past tests that are relevant to this class. Note that some of the solutions to linear equations may be fractions which we have not covered in our work, but the steps to solving such problems do not vary, so they are relevant. (Just be aware!)

1. Simplify 2x + 5x 4 6

2. Simplify 8(5y + 1) + 6y

3. Evaluate -4x + 7 when x = 5

4. Solve 10 6 = 3x 2x + 4

5. Solve 4(x 1) = 4

6. Solve 8(x 7) = 2(x 4)

7. The sum of a number and 5 is 8. Find the number.

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8. If F = 9 C + 32, what is the Celsius temperature in a room that 5

is 25F?

9. Add 3x2 + 3x + 5 + 3x2 + 5x + 5

10. Find the value of 3x2 + 2x + 3 when x = 1

11. Simplify ( 9 + 9x ) + 4

12. Simplify:9 ( c + 2 ) = ____________________

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13. Simplify (combine all like terms)5 + x + 2 + 3x

14. Simplify. 2 x + 5 ( 2 x - 2 ) - 9 x

15. Solve.3 + x = 5

16. Solve.2 ( x - 1 ) + 5 = 13

17. Solve.8 + 3 ( x - 2 ) = 9 ( x + 8 ) - 2 - 2 x

18. Solve x - 6 = 9

19. Solve 2 x + 3 = 15

20. Solve 2 - 5 x = 11 - 8 x

21. Solve 5 ( x + 3 ) = 3 + x

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22. Solve 2 ( x - 3 ) + 4 = 7 ( 3 + 2x ) + 1

23. Solve x - 9 = 3

24. Solve 4 + 6x = 28

25. Solve 49 - 7x = 56

26. Solve 16x + 6 = 9 + 13x

27. Solve 8 + 4 ( 3x + 1 ) = 36

28. Solve 21 - 2 ( 3x - 1 ) = 2x - 1

29. Solve 9 ( x + 2 ) + 4 = 5 ( x + 6 ) - 4

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Instructions: Solve the following word problems using your algebraic skills.

30. John is 5 years older than twice Eva's age. If John is 29, how old is Eva?

31. The price of the book is three times the sum of the calculator and the pencil. The book is sold for $48 and the pencil for $4. What is the price of the calculator?

32. I am 28. Melissa is 7 years younger than twice my age. How old is Melissa?

33. The perimeter of my bedroom is 48 feet. The bedroom is twice as long as it is wide. What is the equation for the perimeter of my bedroom in terms of its width?

34. Evaluate [ a + b ] 2 - n 3 if a = 3 , b = -1 , and n = 3

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