Ch. 2 Linear Equations and Inequalities in One Variable

Index

Section Pages2.1 The Addition Property of Equality 2 - 102.2 The Multiplication Property of Equality 11 - 122.3 More on Solving Linear Equations 13 - 202.4 An Intro to Applications of Linear Equations 21 - 282.5 Formulas and Applications from Geometry 29 - 352.6 Ratios and Proportions 36 - 462.7 More about Problem Solving 47 - 592.8 Solving Linear Inequalities 60 - 70

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§2.1 The Addition Property of Equality

Outline

Definitions Linear Equation in One Variable Equivalent Equations Solution Set Addition Property of Equality Check Isolating the Variable (Therefore) Simplification StepSolving Algebraic Equations Solution Set – i.e. Checking (is it a true statement)

Addition Property of Equality – Add the opposite to both sidesSimple – Just add the opposite2 Step – Simplify (including distributive property), then add oppositeComplex – Add the opposite of variable term and constant terms

Difference between Expression & EquationSolve Equation – One Step is Simplify, but continue to solveSimplify Expression – Don’t Place Equal Sign, DO NOT SOLVE

Homework p. 98-99 #2, 3, 4, #6-36mult.of3, #37, 38, #42-57mult.of3, #61-64all

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 solution set) in order to derive a solution set (the answer; note that I will simply call it a solution at this point). We will do this by forming the equivalent equation:

x = # or # = x

x is a variable # is any constant

Remember that only 1 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.

Recall the Fundamental Theorem of Fractions:a c = a (multiply/divide the numerator & denominator by the same number b c b and the resulting fraction is equivalent)

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The method for which we will be creating equivalent equations is very similar to the Fundamental Theorem of Fractions.

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 is the same as A + C = B + C (since c is added to both A & B)They are equivalent equations

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

9 + x = 12

Note: 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 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. Note: 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) x 3 = 10

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c) x 7/8 = 3/8

d) x + 3 = 0

e) -7 = y + 3

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 solution for the above answers? We check them using evaluation! When you are asked to see if a number is a solution to an equation, this is the process you take.

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Example: Check the solutions to a), c) & d) from the previous example

a) {2} so x = 2 so we replace x with 2 and get the following

(2) + 9 = 11

11 = 11 true statement, therefore( ) this checks

c)

d)

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. This means that will have to combine all like terms on both the left and the right sides of the equation

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before we can apply the addition property of equality. Sometimes this will also require the use of the distributive property before we can combine like terms. Do not neglect the simplification step!!

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

a) 5 + 6 = 9a + 8 8a

b) 3(a + 2) 2a = 1

c) 5/3 a 2(1/5 + 1/3 a) = 4/5

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d) 4.6a + 1.1 3.6a = 3.2 5.7

Not only do we have to move constants from one side of the equal sign to the other, some times we will also have to move the variable terms from one side of the equal sign to the other. We’ll start by practicing with the variable term only then we’ll do both variable and constant term! Remember that the idea is always to get the variables all on one side (usually the left at this point) and the constants on the other side (usually the right at this point)!

Example: Solve.a) 5y = 4y 2

b) 5y + 6 = 4y 2

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c) 7z + ½ = 8z 7/2

Note: The variable term has to move to the right in this example and the constants therefore to the left because we need to keep the variable’s numeric coefficient a positive one!

d) 1.1 2v = 3.1 3v

You will also be asked to translate problems and then solve them in this section. Here are a few examples for a brief review. Translation is SO important!

Example: Translate and solve.a) The difference of x and 7/5 is -13.

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b) The sum of 1/3 y and 6 is equal to the difference of twice 2/3 y and 5.

Ok, after all that, let’s make a point about the difference between solving an equation and simplifying an expression! Don’t forget that if the problem has an equal sign it is an equation and you will solve it! However, if the problem has no equal sign it is an expression and you will only simplify it!

Example: Simplify 2x + 2(x 2) 5

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Example: Solve 2x (x 2) 5 = 7

Note: Your authors want the solution set for each equation, meaning when you get variable = #, you give the answer as {#}. This is a personal preference of the authors that goes along with more advanced mathematics. It is important that you realize that they are equivalent and you may give your answers to me in either form. You will notice that I will most often use variable = #, when presenting the answers in class.§2.2 The Multiplication Property of Equality

OutlineDefinitions Multiplication Property of Equality Isolate the VariableMultiplication Property of Equality Multiply both sides by reciprocal of numeric coefficient

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Multiplication by a reciprocal is division – So it's the same to multiply by reciprocal as dividing Choose the easiest multiply by reciprocal or divide Divide by whole numbers or decimals Multiply by reciprocal for fractions or –1, whose reciprocal is –1!

Homework p. 104-106 #1-6all, #8-22even, #24-66mult.of3, #69-72all

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 = bca, b & c are real numberc 0

We will be using this to solve such problems as: 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! Using the idea of isolating the variable and the multiplication property of equality,

we can arrive at a solution for x!

Example: Solve (isolate the variable) using the Multiplication Property of Equality [Multiply both sides by the reciprocal of the numeric coefficient – for fractions and –1 (the reciprocal of –1 is –1 ) this is the best way to think of it and for whole numbers and decimals you may choose to think of the equivalent operation – division by the numeric coefficient!]

a) 16x = 48

b) x/3 = 36 (Recall that x/3 is equivalent to 1/3 x)

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c) 14x = 0

d) 0.5z = 25

e) - p = 5

f) -2/5 m = -2/3

Really, this section is a breeze. There is one fact and that is pretty basic. It is the next section where we put it all together that we have to really have our thinking caps on!§2.3 More on Solving Linear Equations

OutlineDefinitions Clearing an Equations of Fractions Identity Contradiction Null Set () All Real Numbers () Empty Set ()Solving Linear Equations in 1 Variable

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Putting it all together – Simplification, Addition Property and then Multiplication Property Removing Fractions

Removing Decimals

Homework p. 112-114 #2-46even, #47-52all

Now, let’s get a little more complicated with some material from section four. We have all the tools that we need to solve any problem, we just have to apply them in the correct order!

Solving Linear Equations in One VariableStep 1: Simplify both sides of the equationStep 2: Move all variables to one side of the equation using addition prop.Step 3: Move all constants to the opposite of equation from variables

using addition prop.Step 4: Remove numeric coefficient of variable by multiplying by its

reciprocal in order to solve the equation (multiplication prop.)Step 5: Check your solution

Example: Solve 12x 5 = 23 2xStep 1: Both sides of equation are simplifiedStep 2: Move all variables to one side (left)

Step 3: Move all numbers to the other side (right)

Step 4: Isolate the variable (use multiplicative prop. of equality)

Step 5: Check your solution from step 4

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Note: Steps 2 & 3 are interchangeable. These steps both use the additive property of equality.

Example: Solve 2x + 10 + 14x = 2x 18Step 1: Simplify each side of the equation

Step 2: Move all variables to one side(left)

Step 3: Move all #’s to the other side(right)

Step 4: Isolate x

Step 5: Check

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Example: Solve 5(2x 1) 6x = 4Step 1: Simplify each side of the equation (this includes

distributive property)

Step 2: Move all variables to one side (left)

Step 3: Move all #’s to the other side (right)

Step 4: Isolate x

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Step 5: Check

Ok, now that we have practiced all the basic steps for solving a linear equation in one variable, let’s complicate the situation with some decimals and fractions. There are 2 ways to deal with fractions and decimals. The first way is to go about your business of solving and not worry about them (in other words, just work with them) or you can get rid of them as part of your simplification step. Here’s the low down on how to get rid of each, with a few examples following each.

Simplifying: Getting Rid of Fractions (Clearing the Equation of Fractions)Step 1: Find LCD of all fractions (on both left and right)Step 2: Multiply all terms by LCDStep 3: Simplify left & right sides of the equation

Example: Solve 1/2 x 6/12 = 1

Example: Solve 4/3 (5 w) = - w

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Example: Solve x/2 1 = x/5 + 2

Simplifying: Getting the Decimals OutStep 1: Examine all of the decimals in the equation, and determine which has the most

decimal places. Count those decimal places.Step 2: Multiply every term in the entire equation by a factor of 10 with the same

number of zeros as the number counted in step 1.Step 3: Simplify both sides of the equation

Example: 2.7 x = 5.4

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Example: Solve 4(y + 2.81) = 2.81

Note: In a problem involving the distributive property, it may be less confusing for you to simplify the distributive property before doing the first step.

Example: Solve 0.60(z 300) + 0.05z = 0.70z 0.41(500)

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Finally, we must discuss equations that have no solution or infinitely many solutions. An equation that has no solution yields an untrue statement, in other words, when you go through the process of solving such an equation you will get a false statement such as 0 = 5. This is a contradiction (an equation with no solution) and its solution set is written as a null set (symbolized with ). An equation with infinitely many solutions yields a true statement, in other words, when you go through the process of solving such an equation you will get a true statement such as 5 = 5. This is called an identity (an equation with infinite solutions and its solution set is all real numbers. I will allow you to write rather than {all real numbers}, but note that the book indicates the set notation as the correct answer.

Example: Solve 2x + 5 = 8x 2(3x + 5)

Example: Solve 2(x 5) = 8x 2(3x + 5)

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§2.4 An Introduction to Applications of Linear Equations

OutlineDefinitions Consecutive Integers Consecutive Even Integers Consecutive Odd Integers Degree Right Angle Straight Angle Supplementary Complementary Degrees in a Triangle() Sum of 's in a Number Problems

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More Complex Number Problems Deeper TranslationPrecursor to Mixture Problems (Really more number problems) Still translation with multiplicationConsecutive Integer Problems Definition Consecutive Integers x and x + 1 Even or Odd Integers Defined x and x + 2Angle Problems Straight Angle Relation to Supplementary Angles Right Angle Relation to Complementary Angles Sum of Angles in a Triangle 1 + 2 + 3 = 180

Homework p.122-128 #3, #4-32even, #33-36all, #38-40all, #42-58even

Number ProblemsNumber problems are no more than the translation problems that we have already been doing, except that the variable will not be predefined! You now need to be sure that you define the variable!

Unknown # = x (or whatever you choose)Equation You will get this from translation

You will solve the equation and find the unknown number using your skills for solving algebraic equations.

Example: Four times some number is the same as that number added to 15. Find the number.

Example: One number is twice another less 3. If their sum is 27 find each number.

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More Complex Number ProblemsThese number problems involve not one but two unknowns that are defined in terms of one another. This definition of one unknown in terms of the other will involve a translation of a sum. Finally, the two unknowns will sum to a given total, completing the similarity to the simple number problem.

First Unknown = xSecond Unknown = First Unknown + #Total = First Unknown + Second Unknown

Example: The Ringers play 5 more games than the Setters during a regular season. If together they play a total of 51 games how many does each team play?

Example: My husband and I took our 2 children to watch a movie. It cost a total of $23, to view the movie. If the price one child's ticket is $3 less than that of one adult ticket, how much was the price of a child's ticket?

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Precursor to Mixture ProblemsThe problems in this section are really a form of a more complex mixture problem that we will encounter later, but they also have similarities with the number problems. For instance like the more complex number problems they involve two unknown things that are defined in terms of one another and they also involve a sum of these two unknowns. However, what makes them different and is the most important thing to remember about any mixture problem is that they involve a sum of products. The products will involve an unknown and some portion OF that unknown. In this first very simple case the unknown is one of the products and some portion of that is the other (since "of" means multiplication, this is really not much more than a 2 step number problem, with the special translation step to arrive at the second number in the sum).

First Unknown = xSecond Unknown = Some Portion of the First unknown

= (#, Fraction, Percent, Decimal, Etc.) xTotal = First Unknown + Second Unknown

Example: For every cup of margarine in a chocolate chip cookie recipe, there are twice that many cups of flour. If there are a total of 9 cups of flour and margarine in the restaurant style recipe, how many cups of each are there in this recipe?

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Example: In my tamale pie recipe there is ½ as much cumin as there is salt. If I make a recipe using the basic ingredient ratio there is a total of ¼ teaspoon (tsp.) of both cumin and salt. How much salt must I use?

Consecutive Integer Problems These are of the same form as number problems, but with the twist that you must correctly define the integers using a variable and an expression(s) for the consecutive integers. Here is the trick.

Consecutive Integers – Define 1st as x and the next as x + 1, next as x + 2, etc.Consecutive Even/Odd Integers – Define 1st as x and next as x + 2, next as x + 4,

and so on.

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Example: Two consecutive odd integers sum to 56. What are the integers?

Complementary and Supplementary Angle Problems

A degree is the basic unit of measure in angles. It is defined as 1/360th of a rotation in a

whole circle. A degree can be denoted by a tiny circle above and to the right of a number (). A right angle (denoted with a square inside the angle; see below) is a 90 angle and a straight angle is a 180 angle. Complementary and supplementary angles are related to right and straight angles respectively. Just one more note – the word angle can be written short hand with the symbol .

Complementary angles sum to 90. (I remember complementary is 90 because it comes before 180 and "C" is comes before than "S" in the alphabet.)

Example: The following angles are complementary.

Supplementary angles sum to 180. (Likewise, I remember supplementary is 180 because it comes after 90 and "S" is comes after than "C" in the alphabet.)

Example: The following angles are supplementary.

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x = 30

y = 60

x + y = 90; 30 + 60 = 90

x + y = 180;53 + 127 = 180

Sometimes we will be given the information that two angles are either supplementary or complementary, but we will only be given the measure of one angle and we will have to use algebra to find the missing angle.

Example: Find the missing angle:

Example: If the sun is at a 45 angle from the eastern horizon, what is its angle from the western horizon (assuming that there is a straight line between the two).

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x = 53 y = 127

x = 73

y = ?

Example: If one angle is 15 more than twice is complement, find the other.

While we are on the topic of angles let's briefly discuss a topic that comes up in some exercises – the number of degrees in a triangle. We have discussed the three-sided polygon before in terms of its perimeter, but we have not discussed any other properties of this figure. Since there are three sides that make up a triangle, each pair meet at an angle, and although each angle in any given triangle is unique there are some basic rules that they must adhere to. The most basic is that the sum of all angles in a triangle (degrees in a triangle) is always 180. This means that no angle can ever measure 180 or more.

Example: In a triangle, the second angle is 5 less than the twice the smallest and the largest is 10 more than four times the

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smallest. Find the largest angle's measure.

§2.5 Formulas and Applications from Geometry

OutlineDefinitions Trapezoid Rectangular Solid Sphere Pyramid Circumference Area Perimeter Volume Pi Base of Triangle Radius Diameter Big Base Little Base Height of Triangle Vertical 's Opposite 's Adjacent 'sBasic Formulas Rectangles P = 2L + 2W A = LW V = LWH

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Circle C = 2r = d A = r2

V = 4/3 r3

Triangle P = S1 + S2 + S3

A = ½ bh V = 1/3 BH Trapezoid A = ½ (B + b)h Temperature Conversion FC

C = 5/9 (F 32) CF

F = 9/5 C + 32Solving Formulas for a Variable Step 1: Simplify (by either multiplying out or multiplying both sides by a reciprocal) Step 2: Addition Property of Equality Step 3: Multiplication Property of EqualityOpposite(Vertical) & Adjacent Angles Opposite 's are equal Adjacent 's sum to 180

Homework p. 133-136 #3,4,5,8,9,12, #15-33mult.of3, #38, #45-50all, #51-72mult.of3, & #73

This section is a specialized section dealing with Geometry formulas and formulas in general, their algebraic manipulation and word problems containing them. A formula is an algebraic equation that contains 2 or more variables. Geometry formulas in this section will refer to area, perimeter, volume. The area of a geometric figure is the amount of surface that it covers measured in square units. If you break this down into a very basic visual exercise you would need to count the number of 1-unit squares that could be fit into a figure. The perimeter of a figure is the distance around the outside edge. In the case of a circle the perimeter is called the circumference. The volume of a figure is the amount of substance it can contain measured in cubic units. Following are some of the basic formulas needed for this section. Each contains special variables which I will briefly define. For more formulas refer to the rear inside cover of your text.

Rectangle L = lengthW = widthH = height

Area A = LWPerimeter P = 2L + 2WVolume of a Rectangular Solid V = LWH

Circle r = radius (the distance from the center to any side)d = diameter (the distance from one side to another through the

middle; twice the radius) = pi (a Greek letter representing the constant that defines the ratio of

the circumference to the diameter; 3.14 or 22/7)

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Circumference C = 2r = dArea A = r2

Volume of a Sphere V = 4/3 r3

Triangle b = base (the side to which the height meets at a right angle)h = height (the perpendicular distance from a vertex to an opposite

side)B = base of Pyramid (the area of the base rectangle)H = height of Pyramid (the height from the vertex to rectangle)

Perimeter P = S1 + S2 + S3

Area A = ½ bhVolume of a Pyramid V = 1/3 BH

Trapezoid B = big base [the longest perpendicular () side]b = little base (the smaller side)h = height (the distance from B to b or vice versa)

Area A = ½(B + b)h

Temperature Conversion

C = Degrees Celsius (the scale based upon 0 freezing & 100 boiling)

F = Degrees Fahrenheit (the scale most commonly used in the US. 32 is freezing and 212 is boiling)

F = 9/5 C + 32C = 5/9(F 32)

Now let's practice with a few of the less basic ones.

Example: Find the area of the following trapezoid.

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2 ft.

3 ft.

1.5 ft.

Example: Find the temperature in Celsius, to the nearest degree, if it is 113 F.

Example: Find the temperature in Fahrenheit, to the nearest degree, if it is 14 C.

Example: Find the volume of a sphere with the diameter of 2 m.

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The next item of importance in this section is solving a formula for one of its variables. This process seems difficult, but it uses the same steps as solving any algebraic equation. (Addition Property to Isolate Variable, Multiplication Property to Further Isolate Variable) I find that it makes it more manageable to think of all other variables as numbers until I have manipulated them away from the variable I want to solve for. Another trick to make it more manageable is to use a colored pen or pencil to denote the variable that you wish to solve for.

Example: Solve the following for z.G = az

Example: Solve the following for x.x + by = c

Example: Solve the above (x + by = c ) for y.

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Example: Solve the following for g.½(g + a) = 9

Note: There are several ways to solve this problem. In keeping with the method used to solve any equation, you can simplify it by multiplying it out on the left and proceeding from there. You may also choose to multiply both sides by the reciprocal of the ½ to free the variable. Either method will yield equivalent solutions although they may look different.Finally, we must discuss some more facts about angles. We already know that supplementary angles sum to 180 and complementary angles sum to 90. We must now discuss when two lines intersect (cross) and form opposite (vertical) angles and adjacent angles. Opposite (vertical) angles are equal. Adjacent angles are supplementary – they sum to 180. This can be proven via methods from geometry, but you will have to take my word. The following diagrams show opposite (vertical) and adjacent angles.

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B C AC

AD

B & C are Opposite 's A & C are Adjacent 's

Example: What other angles are opposite? Adjacent?

Example: In the following diagram find the measure of the missing angles.

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(3x + 5)(7x 3)A

Example: What is the measure of A in the above problem?

§2.6 Ratios and Proportions

OutlineDefinitions Ratio Proportion Means Extremes Cross Multiply Rate Cross Product Unit Price Similar Means-Extremes Property\Ratios Definition Writing a to b, a:b, a/b Always lowest terms Never Mixed Number Always one # to anotherRates (Unit Prices) Similarity to Ratios One # to another Never a mixed #

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Differences The denominator is always 1 Divide numerator by denominator to simplify and make denominator 1Proportions Mean & Extremes Property Cross Products Solving UsingSimilar Triangles Proportionality of Using Proportions to find missing sides

Homework p. 141-144 #3, #3-12mult.of3, #14,20,24,25, #27-48mult.of3, #56, #57-63all

A ratio is a quotient of two numbers where the divisor isn't zero. A ratio is stated as:

a to b or a : b or a where a & b are whole numbers and b0 b

A ratio is always written as one whole number to another although it may not start out being written in this form. A ratio is really a special fraction and we should always simplify it by putting it in lowest terms (recall that this means that all common factors have been divided out). The only difference between a ratio and a fraction is that a ratio is never simplified to a whole number or a mixed number. A ratio must always be the quotient of two whole numbers. Another important thing to remember is that ratios do not have units, because both numerator and denominator have the same units so they cancel and a ratio appears unitless.

Writing a Ratio CorrectlyStep 1: Write one number divided by another number (order will be dependent upon your

problem, pay attention to the word to or the word quotient or its equivalent to help you out)Step 2: Make sure that both numbers are in the same units. Convert if necessary.Step 3: Divide out (cancel out) all common factorsStep 4: Rewrite the ratio making sure that it is written as one whole number to another

Example: Write the following correctly, as ratiosa) 8 to 32

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b) 50 to 150

Example: What is the ratio of the following?a) 20 min. to 1 hour

b) $1.50 to 25¢

A special case of ratios is called rates. Rates are sometimes, as in the case of your book, referred to as unit pricing. Rates (unit prices) are ratios in the sense that they are one number to another, but a rate is written as any number to one. We find a rate using the same principle as finding a correct ratio, except that we divide out the number that is in the denominator. This can create a number that is not a whole number, which is fine in a rate. A rate is not written as a whole number to another, it is simply written as an integer. The ratio form of a rate is shown in its units.

Writing a Rate (Unit Price)Step 1: Write as one number to another, also writing the units in this formStep 2: Divide the numerator by the denominator (round only when told)Step 3: Write the rate with its ratio like units

The following three examples are true examples of rates. The fourth and fifth examples are unit pricing. Unit pricing can help us in shopping at the grocery store.

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Example: Write 28 miles to 6 gallons as a rate.

Example: Find the average speed of a car that travels 292 miles in 4 hours.

Note: Average speed is a rate, found by the rate equation r = d / t, which can also be written as d = r t, which is know as the distance equation. r = rate, d = distance & t = time.

Example: A Ferris wheel completes one revolution in 20 seconds. If its circumference is 240 feet, what is the rate that a person travels when riding on the ride?

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Example: A large box of Tidal Wave detergent (263 oz.) costs $19.95 while the smaller box (96 oz.) costs $13.95. Which box is the better buy? (Calculator OK)

A proportion is a mathematical statement that two ratios are equal.

2 = 4 is a proportion 3 6

It is read as: 2 is to 3 as 4 is to 6

The numbers on the diagonal from left to right, 2 and 6, are called the extremes

The numbers on the diagonal from right to left, 4 and 3, are called the means

If we have a true proportion, then the product of the means equals the product of the extremes. This is called the means-extremes property. It actually comes from the fact that when we multiply both sides of an equation by the same number (in this case the LCD) we get an equivalent equation. It is a very useful property for finding a missing term (one

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of the four numbers in a proportion 2 – 1st term, 3 – 2nd term, 4 – 3rd term, 6 – 4th term, in the example above) in a proportion.

The products of the means and extremes are also called the cross products and finding the product of the means and extremes is called cross multiplying .

Example: Find the cross products of the following to show that this is a true proportion

24 = 3 64 8

Solving a ProportionStep 1: Find the cross products and set them equal.Step 2: Simplify the expressions on both the left and the right side of the equal sign.Step 3: Proceed as a linear equation by isolating the variable using addition property of

equality.Step 4: Further isolate the variable by multiplying by the reciprocal of the numeric

coefficient.

Example: Solve each of the following proportionsa) a = 21

4 28

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b) z = 7 3 8

c) x + 2 = 2 3 5

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d) m 8 = m 18 2

e) m + 1 = 5 m 5 9

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Truly, the most useful thing about ratios and proportions are their usefulness in word problems. We can solve distance problems, rate of pay problems, gas mileage problems, unit price problems, equal concentration mixture problems, etc. using this method.

The key is to set up equal ratios of one thing to another and write it out in words 1st.

Example: Using proportions, solve each of the following problems.a) If Alexis gets 24 mpg in her car how many miles can she

drive on 7 gallons of gas?

b) Joe gets a check for $225 for working 5 hours on Tuesday, if he works 3 hours on Wednesday, how much should he expect his check to be?

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d) A car travels 251 miles in 3 hours. If it continues at the same speed, how long would you expect it to take in order to travel 362 miles? (Calculator OK. Round up to nearest whole hour.)

One more way to apply our knowledge of proportions is in finding the lengths of missing sides in similar triangles. Similar triangles are triangles that have exactly the same angles with corresponding sides that are proportional (not the same length, but all sides between equivalent angles will form an equivalent fractions).

Given A = D, B = E, C = F in the following 2 triangles

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A

B

C D

E

F

Then AB : DE, BC : EF, AC : DFMeaning that AB = BC , AB = AC , BC = AC (read AB is to DE as

DE EF DE DF EF DF BC is to EF, etc.)

The only thing left to recall, is that in a proportion (which is what the above “meaning that” statements are called) the cross products are equal. Cross products are the diagonal numbers, across the equal sign, multiplied.

Cross Products are equivalent in any true proportion and this allows us to set up an equation to solve.

AB EF = BC DE

Example: Find the missing side using your new knowledge.

AB = ?, BC = 9, AC = 13 1/3 While AE = 9, EF = ? and AF = 10

Example: Find the missing side of these congruent triangles.

AB = 4.5 in.DE = 3 in.BC = 7.5 in.EF = 5 in.AC = ?DF = 4 in.

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A

B

C

E

F

A

B

E

C D F

Example: A tree, standing beside a building, is known to be 27 feet tall. If it casts a shadow that is 1 city block long and the building casts a shadow that is 3 1/3 blocks long, how tall is the building?

§2.7 More About Problem Solving

OutlineDefinitions Simple Interest Percent Equivalent FormsReview of Percentages Memorizing Fraction to Decimal Conversions (and vice versa) Definition of percent Solving Percentage problemsMixture Problems TableInterest Problems TableMoney Problems TableDistance Problems Formula

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Table

Homework p. 152-157 #1-6all, #16-34even, #46-54even

This section is about mixture problems. I mentioned before that we had started the baby steps for mixture problems. They are problems that involve multiplication and addition. There are three main types of mixture problems and we will go over a few examples of each. The three main types are mixtures of solutions, simple interest (I = PRT), money problems and rate (D = RT) problems. The mixture of solutions can look like mixtures of any type of things, simple interest and rate problems always looks like what they are and money problems can also look like value of an item times the number of the items!

Because I gave you percentage problems as homework but never covered them in class, I would like to do a review at this time. It is useful to memorize some decimal to fraction conversions. The following table is a list of very helpful conversions to memorize. Recall that you can calculate these at any time you wish by simply dividing the numerator by the denominator, but it saves a lot of time if you have them memorized. Here are some tricks that I use to remember the fraction to decimal conversions.

4ths – Money! 1 quarter is $0.25, 2 quarters is half a dollar, which is $0.50 and 3 quarters is $0.75

3rds – 1/3 is 0.333, 1st in 3rds so repeats 3’s

2/3 is 0.666, 2nd in 3rds so repeats 6’s (23)

5ths – 1/5, 10 5 is 2 so 0.22/5, 20 5 is 4 so 0.4 3/5, 30 5 is 6 so 0.64/5, 40 5 is 8 so 0.8

6ths – Think of the pattern is 3rds and remember that 6ths and 3rds are related

1/6 is half of 1/3 so 0.1666

2/6 is equal to 1/3 so 0.3333/6 is equal to ½ so 0.5

4/6 is equal to 2/3 so 0.666

5/6 is 0.8333 (it’s following the repeating 3’s pattern and must be bigger than 2/3’s)

8ths – The pattern is in last 2 digits, and is the most difficult1/8 is 0.125 (recall that 1/8 is ½ of ¼ )3/8 is 0.375 (I always have to divide 30 by 8 to get the 3 and then I know that 75 is the

ending)5/8 is 0.625 (I always have to divide 50 by 8 to get the 6 and then I know that 25 is the

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ending)7/8 is 0.875 (I always have to divide 70 by 8 to get the 8 and then I know that 75 is the

ending)9ths – It’s a repeating decimal of whatever the numerator is

At no time is it ever acceptable to write the * or any similar decimals with fractions in them! The bar is the only acceptable method of showing repeat!!!!!!!

Fraction Decimal½ 0.51/3 *0.3332/3 *0.666¼ 0.25¾ 0.751/5 0.22/5 0.43/5 0.64/5 0.81/6 0.16665/6 0.83331/8 0.1253/8 0.3755/8 0.6257/8 0.8751/9 0.1112/9 0.2224/9 0.4445/9 0.5557/9 0.7778/9 0.888

Note: You must know any repeating decimal’s equivalent fraction because there is no method for conversion back to its equivalent fraction!A percent is a part of a hundred.

There are 3 equivalent ways to write a percentage. They are as follows:As a fraction – A part of one hundred (this must be reduced, of course, if possible)As a decimal – Convert the fractional form to a decimal (see §5.6)As a percentage – Multiply the decimal form by 100, since a percentage is just the

numerator portion of the fractional representation when the denominator is 100!

We can write percentages as parts of 100, but since a percentage is a fraction in fractional form it needs to be reduced.

Example: 75% = 75 = 3 = 0.75 100 4

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All these are called equivalent forms, because they indicate the same thing.

The following is the only way that I will be teaching you to solve percentage problems. I do not like the other method of creating a proportion. I find that the other method is unreliable at best, not to mention confusing.

We will be seeing percentage problems in word problems. The following will be the forms that we will be seeing and their corresponding algebraic equation. The portion in italics is the portion that is unknown and thus represented by the x in the algebraic equation.

Forms Algebra EquationPercent of a whole is what part (%)(whole) = xWhat percent of the whole is the part (x%)(whole) = partPercent of what whole is the part (%)(x) = part

When approaching any percentage problem, you should write out the following form and use the key words “of” and “%” to fill in the appropriate blanks in the form. The form directly translates into an algebra problem since the word “of” means multiply and the word “is” means equals. The only two things that you must remember are the following:

1. Change the percentage to a decimal to solve the algebra problem2. If the answer that you are getting is the percentage, then the algebra problem

will give you the equivalent decimal, and you must change it to a percentage for your final answer.

Here is the form that you should always write down and fill in.

_____% of _____ is _____

Now let’s practice!Example: Solve the following problems showing form, algebraic

equation and solution.a) 25% of 180 is what?

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b) 16% of 100 is what?

c) What percent of 10 is 3?

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d) 18% of what is 125?

e) 8% of what is 62?

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f) What is 120% of 120?

Note: When the percentage is more than 100% the part will be larger than the whole! In every case where the percentage is less than 100% the part will never exceed the whole and you can use this as a quick check to see if you got an appropriate answer.

g) What percentage of 25 is 95?

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Note: The whole is less than the part so it is quite appropriate that the percentage is greater than 100%

The key to doing word problems with percentages is to remember the form and to put each problem in that form!

______% of ________ (whole) is ______ (part)

Example: It took 12% of the Miller’s monthly income to pay for daycare of their child. If the Miller’s income is $2500 per month, what is the cost of daycare per month?

Mixture ProblemsKeys to solving are remembering that these are percentage problems and building a table like the following!

Amount Solution Strength = Amount of Pure

Unknown Given % Calculate by multiplying acrossRelation to unknown or given Given % (Different from other) Calculate by multiplying across

Given or Sum of 1st two rows Given % (Different from others)

Calculate by multiplying across Equal to sum of 1st two rows

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Sum

Equals

Example: How many cubic centimeters (cc) of a 25% anitbiotic solution should be added to a 10 cc of a 60% antibiotic solution in order to get a 30% antibiotic solution?

Example: How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze?

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Interest Problems

Again the table set up with multiplication across and addition down works well. It is important to note that simple interest (I) is calculated by multiplying principle (P) by rate (R) by time (T). The equation looks like P R T = I. In our work, time will most often be 1 year, but we will put it into our table just in case at some point it is not!

P R T = Ix Given Rate Usually 1 year Given x

x + Some amount 2nd Given Rate Usually 1 year + 2nd Given (x + amt.)= Given Total Interest Made

Example: Joe invested some of his inheritance in a bond that yields 7% annually. He invested $900 more than he invested on the bond in a high risk investment that yields 12% annually. What was his total inheritance if he earned $165 on both in one year?

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Money Problems

The key to these type of problems is to realize that there must be a multiplication of the value of the coin and the number of coins to find the total value of each type of coin. These total values, of each type of coin, then usually sum to your total value or will have some relationship to one another. You will never be given the value of each type of coin of course because that is common knowledge – you must simply remember that you must use this common knowledge.

Coin Type Value of Coin # of Coins Total ValueNickels 0.05 or 5 x 5xQuarters 0.25 or 25 Relationship

between # of Nickels & Quarters is given

25(expression in # of coins)

5x + 25(?) = Total Value Given

Example: I have 28 coins in my pocket. There are quarters and dimes. If the total value is $4.60, how many of each type do I have?

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Example: In my change jar I have $3.85. I have 5 more quarters than nickels and 2 less dimes than nickels. How many of each type of coin do I have?

Distance ProblemsThis type of problems can be tricky because there are so many scenarios. We will be looking at distances and their sums. Sometimes the time will be unknown and sometimes the rate will be unknown. Really, the most important thing is to read the problem carefully and think logically about what is happening. I find a diagram of the scenario as helpful as the table (which by the way looks just like the table for all other mixture problems). Important keys to remember for this type of problem are the formula

D = R T also T = D/R

Distance = Rate Time#1 t #1 Unknown = t#2 t #2 Same Unknown = t

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#1t + #2t = Given D

Second ScenarioDistance = Rate Time

2R Unknown = R Known = 22(R + #) Unknown Relation to 1st = R + # Same Known = 22R + 2(R + #) = Given D

Example: Two cars are traveling toward one another at a rate of 55 mph and 70 mph. If they begin 500 miles apart, how long before they meet?

Example: Two hikers are 14 miles apart and walking toward each other. They meet in 2 hours. Find the rate of each hiker if one hiker walks 1 mile per hour faster than the other.

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§2.8 Solving Linear Inequalities

OutlineDefinitions Inequalities Simple Inequality Standard Form (of an inequality) Endpoint Compound Inequality Interval Notation Sense of the InequalitySimple Inequalities Definition Review of Inequality Symbols < less than & less than or equal to > greater than & greater than or equal to Graphing < or > Open Circle (or Parentheses)

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or Solid Dot (or Bracket) Standard Notation – Line goes in direction of arrow Interval Notation Relation to parentheses and brackets in graphing Addition Property & Solving Using Multiplication Property & Solving Using CAUTION: Reverse sense of inequality when mult. or dividing by a negativeCompound Linear Inequality (Three Part Linear Inequality) Definition Reading Solving Addition Property Multiplication Property CAUTION – Don't forget to reverse the sense of the inequality when multiplying by a negative Writing solutions using interval notationApplications Translation Type Mixture Type Can Be Any Previously Studied Type Consecutive Integers Geometry Problems Etc.Homework p. 166-169 #3-15mult.of3, #25, #34-88even

Inequalities are algebraic expressions that contain inequality symbols. The following are the inequality symbols:

Graphing of simple inequalities (contain only one inequality) -- Graph with an open circle and a line to the left, if in standard

form; Lial uses a parentheses – we won't! -- Graph with an open circle and a line to the right, if in standard

form -- Graph with a closed circle and line to the right, if in standard

form; Lial uses a bracket – we won't! -- Graph with a closed circle and a line to the left, if in standard

form

Just a note about the brackets and parentheses that we won't be using – their use comes from the interval notation, a method for writing a solution set that tells us which endpoints are included. An included endpoint is always enclosed by a bracket ("[" or "]") and an excluded endpoint is enclosed by a parenthesis ( "(" or ")" ). If an interval

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continues on to infinity, we use either for infinity or – for negative infinity. Either infinity symbol is enclosed by a parenthesis. An endpoint is exactly what it sounds like; it is the end of the solution set.

Standard form (of an inequality) means that the variable is on the left and the constant on the right. In order to achieve this simply place the variable on the left and constant on the right, being careful to keep the point of the inequality facing toward the number or variable as it was before the swap.

Examples: Put the following in standard form and write them using interval notation.

a) 5 x

b) –2 y

c) 5002 m

Examples: Graph each of the following (on a number line)a) x 3

b) 2 y

c) -2 v

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d) t 5

Note: Parts b and c in the example are examples of inequalities written in nonstandard form. To read an inequality written in nonstandard form, we must read right to left, instead of our usual left to right. Thus it is easiest to always put inequalities in standard form before reading and trying to graph them.

A compound linear inequality (three part linear inequality) contains 2 inequalities connected by the word “and”. When joined by “and” it means the intersection of the sets. This means that we will be graphing where the two sets overlap. If there was no overlap then there would be no solution. The “and” is usually not explicitly written, and is usually written in the shorthand form

# < x < # or # > x > #.

Example: 15 x 20 is read as: x is greater than 15 and x is less than 20

In order to graph a compound inequality written in this form we must first graph the endpoints and then draw a line in between the endpoints as this is the intersection of a compound inequality joined by “and”. Make sure that you graph the endpoints correctly as either an open circle (parenthesis) for a strict inequality or a closed circle (bracket) for greater than/less than or equal to.

Graphing a Compound Linear Inequality (and)Step 1: Graph each endpoint. Open circle (parenthesis) for < or >. Closed circle (bracket)

for or .Step 2: Join endpoints with a line, after making sure that there is an intersection.

Example: Graph the following compound inequalities and put them in interval notation.

a) 8 x 9

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b) –2 x 5

c) 5 x -2

Solving Linear InequalitiesIf all we must do to solve a linear inequality is to add or subtract then they are solved in the exact same manner as a linear equation. Recall the steps to solving a linear equation covered in §2.1.

Example: Solve and graph the linear inequality. Put the solution in interval notation.

a) x + 2 7

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b) 2x + 7 9 + x

c) 2x 15 + x 15 + 2(x 1)

Solving a compound linear inequality is a little more complex because there are actually two problems. The first problem is from the middle to the left and the second if from the middle to the right. Remember that it is the intersection of these solution sets. Now this could be quite complicated if you solved every one as two separate problems, but thank heavens we have a shortcut. The shortcut is to treat the problem as having 3 parts instead of 2. Translation: What you do to the middle, do to both the left and the right! Equivalence in solving equations: What you do to one side do to the other! In terms of the addition property, add or subtract the same things from both the left and the right that you add or subtract from the middle to isolate the variable. Note: The variable must stay in the middle!

Example: Solve the following compound linear inequalities and answer in interval notation. Also graph each answer.

a) 5 < x 1 9

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b) -2 5 + x < 0

Multiplication Property of Equality & Linear InequalitiesIf we must multiply or divide by a positive number to get the solution, then the steps for finding the solution are still the same!

Example: Solve and graph the linear inequality. Give the answer in interval notation.

5x 15 + 2 15 + 3x

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Problem Area!!The problem occurs when I must multiply or divide by a negative number! Whenever I multiply or divide by a negative number, the sense of the inequality must reverse. The sense of the inequality means that the direction the inequality symbol faces changes.

Example: Solve and graph. Give the answer in interval notation.a) 3/4x 1 2x + ¼

b) -2x + 5 9

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Why Must Sense of Inequality Reverse?To see why it is true that the sense of the inequality must reverse, let’s rework the last problem keeping the numeric coefficient of the variable positive. Notice that you must read the inequality in nonstandard form (right to left) and the answer is the same as above.

Example: Solve by keeping the variable positive. -2x + 5 9

Multiplication Property and Compound InequalitiesJust as with addition property of equality, we treat a compound inequality as having 3 parts and what we do to the middle we do both the left and right. Thus when we wish to remove the numeric coefficient by multiplying by its reciprocal we must multiply both the left and the right sides by the reciprocal as well! Since there is no way to get around reversing the sense of the inequality when multiplying by a negative –be careful!

Example: Solve for the variable. Give the answer using interval notation.

a) 5 < 2x 5 9

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b) -0.5 0.1y + 0.25 < 0.23

ApplicationsThe most important comment that I have for you when solving word problems that involve linear inequalities is to point out 2 very important phrases and their mathematical interpretation!

At Most means At Least means

Example: Three times some number less five is greater than or equal to the result of five subtracted from the number. Find all such numbers.

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Example: John's phone has a monthly charge of $34.25 and each long distance call costs 5¢ per minute. If he has $45.50 in his budget per month for his phone bill, how many minutes of long distance can his use at most?

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Example: The length of a rectangle is three less than twice its width. If the perimeter is at most 53 cm, what are the possible values of the width?

Of course there are many other problems that can come into play here, but they all rely upon the skills developed in §2.4, & §2.7, so we won’t cover any more at this time.

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