College Algebra Week 2Chapter 2-3

The University of Phoenix

Inst. John Ensworth

This week we move on to LINEAR EQUATIONS

• What is a linear equation?

• Why is it scary?

• It just means you have ONE variable in it, and that variable isn’t squared or cubed or anything crazy.

This kind of equation is your friend!

Section 2.1 Page 66

The Linear Equation

• It has one variable : x

ax + b = 0

Where a and b are real numbers and a is not equal to 0.

A photo album of linear equations

• 2x+3=0

• x+8=0

• 3x=7

• 2x+5=9-5x

• 3+5(x-1)=-7+x

• 3=8x+5

• Etc.

How to legally play with linear equations.

• Most of the magic we work with are really obvious tricks.– Remember what happens if we multiply by

things that equal 1? e.g. 5/5 10/10 ?

• Adding the SAME THING to both sides of an equation does not change the equation…

The Addition Property of Equality

• So if you start with a = b

• You don’t change anything if you take a new number or variable and add it to both sides!

• a+c = b+c

The GOAL

• We want x all by itself on one side and the rest of the junk on the other (so we can use our calculator or fingers).

• The technical term for this is: solving for x

EXAMPLE 1 pg 66

• Solve x-3= -7

x – 3 = - 7

x-3 + 3 = -7 + 3

x + 0 = - 4

x = - 4

The theme

• ISOLATE THE ‘X’ !!!

Then you can check your answer

• We got x = -4 so plug it into the first equation:

• x- 3 = -7

• -4-3 = -7

• -7 = -7 CHECK! Side note: {-4} is solution set to the

equation.

EXAMPLE 2

• Solve 9 + x = -2

9 + x = - 2

9 + x – 9 = -2 –9

x + 0 = -11

x = -11 Solution set {-11}

Checking… 9 – 11 = -2 = -2 CHECK!

EXAMPLE 3

• Solve 1/2 = - 1/4 + y (who cares if the variable is x or y or c or whatever!)

1/2 = - 1/4 + y

1/2 + 1/4 = -1/4 + y + ¼

2/4 + 1/4 = y + 0

3/4 = y S.S. {3/4}

Checking ½ = - ¼ + ¾ = 2/4 = ½ = ½ check!

The next trick…multiplication

• That is nice if the stuff we want to move to the non-x side is just added or subtracted from the ‘x’, but what if the number is a coefficient of ‘x’? What if it looks like

3x= 9 ?!

The multiplication property of equality

• If, again, you have a equation like

a = b

And you multiply both sides by c, then you haven’t changed the equation!

ac=bc

Again, c can’t be 0.

EXAMPLE 4

• Solve x/2 = 6Note, you can’t subtract or add

anything to get x by itself.

So multiply both sides by 2! (Which is the INVERSE of the offending number).

2*x/2 = 2*62/2 * x = 12 x = 12 SS = {12}

A quick check…

• Plugging 12 into x/2 = 6– 12/2 = 6 = 6 good thing!

Example 5

• -5 w = 30• -5 is in the way! -1/5 is the inverse of it!

(-5w)/ (-5) = 30 / (-5)

(-5/-5) w = -6

w = -6 The solution set is then {-6}

EXAMPLE 6

• Fractions as coefficients aren’t a problem either!

• 2/3 p = 40• 2/3rd offends, kill it with 3/2nd

(3/2) * 2/3 p = 40 * (3/2)

1 * p = (40*3)/2 = 20*3 = 60

p = 60 our solution set is {60}

Example 7

• Are we bothered by –1 as a coefficient?

• Certainly not!

- h = 12

The inverse of –1 is 1/-1 = -1

Multiply both sides by –1!

(-1) –h = (-1) 12

h = -12 SS = {-12}

Example 8Sometimes things are messier…

• What if you have a more goofed up equation?

-9 + 6y = 7 y There are dumb y’s on both sides… so first we

need to get the y’s on ONE side, THEN do what is needed to get y alone.

-9 + 6y –6y = 7y – 6y -9 = y !

Wow, the solution set is there … {-9}

Set to Exercises 2.1

• Go to the section you have the hardest time working with– Definitions Q 1 –6– Solving equations with + and - Q 7 – 30– Solving equations with * and / Q 31 – 54– Doing it with –1 in play Q 55 – 62– Solving it with x on both sides Q 63 – 70– Random mix of anything Q 71 - 90

Section 2.2

• Now we graduate to equations that need both addition/subtraction and multiplication/division in the same solution.

• Not a problem… right?

Rules for Happiness

• First do whatever adding and subtracting you can do…

• THEN do whatever multiplication and division you can do.

• THEN you should be done!

Example 1• Solve 3r – 5 = 0

• Step one… addition or subtraction?

3r –5 + 5 = 0 + 5

3r = 5

• Step two … division or multiplication?

(1/3) 3r = (1/3) 5

r = 5/3 Our solution set is {5/3}

Example 2

• Solve -2/3 x + 8 = 0

Add/Subt. : -2/3 x + 8 – 8 = 0 – 8 -2/3 x = - 8

Mult/Div.: (-3/2) –2/3 x = (-3/2) (–8) 1*x = (-3)(-8)/2 x = (-3)(-4) = 12

Our solution set is {12}Checking: (-2/3) (12) + 8 = -24/3 + 8 = -8 + 8 = 0 check

Example 3Now with x’s elsewhere and +&- *&/

• Solve 3w –8 = 7w • Group the variable’s 3w –8 –3w = 7w –3w

-8 = 4w• What’s needed? Get ride of that 4 sticking to

the w! -8 (1/4) = (1/4) 4w

-8/4 = w-2 = w Our Solution set = {-2}

Example 4

• Solve ½ b – 8 = 12• Add 8 to both sides first

½ b – 8 + 8 = 12 + 8½ b = 20

• THEN multiply both sides by the inverse of ½ . Which is 2!

2* ½ b = 2 * 20b = 40 Our solution set is {40}

CHECK it… ½ (40 ) – 8 = 20 – 8 =12 = 12 CHECK!

Example 5 - a tad bit more complex

• The form NOW is ax+b=cx+d• No big change…• Addition first and get the x’s together then proceed• Solve 2m-4=4m-10

2m -4m –4 = 4m –4m –10 -2m -4 = -10

-2m -4 +4 = -10 +4-2m = -6(- ½ ) –2m = (- ½ ) (-6)

m = 3 Our solution set is {3}

Checking it…

• 2m-4=4m-10 and m=3 right?

• 2(3) –4 = 4(3) –10

• 6 –4 = 12 –10

• 2 = 2 YES!

Stepping back in time…Example 6

• Remember expanding stuff? (Chapter 1)

You MAY need to do it FIRST, then solve for x.

We’ll work with 2(q-3) +5q = 8(q-1)

On the next slide…

Workin’ on the railroad2(q-3) +5q = 8(q-1)First multiply the parenthesis out, then do the normal

stuff…2q-2(3) + 5q = 8q –1(8)2q –6 + 5q = 8q –8Combine the q’s on the right : 7q –6 = 8q –8Then subtract 7q from both sides 7q –7q –6 = 8q – 7q – 8-6 = q –8 Then add 8 to both sides -6 +8 = q –8 +8 give us2 = q Our solution set is {2}

Don’t forget to check it later!

Cook Booking It

1. Remove parentheses and combine like terms on each side

2. Use addition property to get like terms on the same side as each other (things with x on one side, numbers only on the other)

3. Use the multiplication property to get the x alone

4. If x is negative, use the –1 multiplication trick to make x positive

5. Check your solution to see if you got it!

Section 2.2 Exercises

• Do the hardest first!– Definitions Q 1-4– Simple one variable solutions Q 5 – 20– One variable but on both sides Q 21- 44– Multiply out first… Q 45 – 52– Numbers and variables on both sides Q53-90

Section 2.3 Equations with more than one simple answer (or no

answer)• Identities = stupid equations in which all

numbers are the answers

x/2 = ½ x x = xx+x = 2x 3x + 3 = 3(x+1)x+1 = x+1 yadda yadda5/x = 5/x (note x can’t equal 0)

Is it an identity?

• Simplify both sides…

Example 1 7-5(x-6)+4=3-2(x-5)-3x+28

Kill the ()’s 7-5x+30+4=3-2x+10-3x+28

Group like stuff -5x+41 = -5x+41 Hurray!

Another definition

• A Conditional Equation

• Is ANY equation that has at least one real answer (or more) but is NOT an identity

• Everything we’ve done so far is an example of a Conditional Equation

• Another example x2=4 2 and –2 both work. (Solution set = {2,-2} )

And another definition

• Inconsistent Equations

• They have NO answer.x=x+1 x-x = x-x+1 0 = 1 Woa! Yikes!

0 * x + 6 = 7 0 + 6 = 7 6 = 7 Buzz.

Etc.

Finding an inconsistent equation

• Example 2 Solve 2-3(x-4)=4(x-7)-7x

2- 3x+12 = 4x –28 –7x

-3x + 14 = -3x –28 Hey! That can’t be!

one more step -3x +3x + 14 = -3x +3x –28

14 = -28 Nope, it’s just off.

And now for another combination… least common

denominators making life easier• It seems a bit out of place, but let’s combine

what you learned about LCD’s and simplifying equations… it IS a good thing.

• Example 3 …

A bit of forethought makes things smoother

• Solve y/2 –1 = y/3 + 1

• We know that we will eventually have to add the y terms together. But they have different denominators.

• We know that 2*3 = 6 will give us a common denominator. So let’s multiply both sides by 6 first!!!

Common denominators• y/2 –1 = y/3 + 1

• 6(y/2-1) = 6(y/3+1)

• Expand (6/2) y –6 = (6/3)y +6

• 3y –6 = 2y + 6

• Next 3y –2y –6 = 2y –2y +6 y-6=6

• y-6 + 6 = 6 + 6 y = 12 SS = {12}

• Checking: 12/2-1 =? 12/3 + 1 6-1 =?4+1 5=5 Cool!

What if you have decimals? Break out the calculator.

• Nothing new here, the numbers just look a bit different.

• Solve 0.3p + 8.04 = 12.6

• 0.3p+ 8.04 –8.04 = 12.6 – 8.04

• 0.3p = 4.56

• (1/0.3) 0.3 p = (1/0.3) 4.56

• p= 15.2 The Solution Set is {15.2}

The same problem multiplying by 10, 100, 1000 or something

like that• Solve 0.3p + 8.04 = 12.6• Note that the greatest number of places to the right

of a decimal place is 2 which can be erased if we multiply by 100

• So 100(0.3p + 8.04) = 100(12.6)• 30p + 804 = 1260 subtract 804 from both sides• 30p = 1260 – 804 = 456 then divide by 30• (1/30) 30p = (1/30) 456 • p= 15.2 THE SAME ANSER!

Another with helpful factors of 10

• Solve 0.5x + 0.4(x+20) = 13.4• Expand 0.5x + 0.4x + 8 = 13.4• Combine 0.9x + 8 = 13.4• X 10 kills all decimals• 10(0.9x +8) = 10(13.4)• 9x + 80 = 134• 9x = 134 – 80 = 54• 9x= 54 divide both sides by 9• x= 6 The solution set = {6}

Practice section 2.3!

• Do it do it do it!– Definitions Q 1-6– Looking for conditional equations, inconsistent

equations or identities Q 7 – 26– Killing fractions with LCD’s Q 27- 38– Decimals Q 39 – 52– Anything goes mixed bag Q 53- 86

On to Section 2.4 Formulas

• Don’t be fooled, this isn’t any harder except we sometimes don’t have ANY obvious numbers… we have letters and we choose one variable to be OUR variable for solving.

• You just leave the answer with the letters however they fall… but the tricks are all the same as what we have already used.

The trick…

• Pretend all the OTHER variables are just numbers to get out of the way.

Definition

• A formula or literal equation is an equation involving two or more variables. We solve it for a certain variable.

• Making formulae work for us is a main reason to even worry about solving for any particular variable! This can be useful!

Example 1

• Solve D=RT for T…

• D/R = (R*T)/R divide both sides by R to get T alone

• D/R = T

• Swap sides T = D/R done!

Example 2

• The Celsius to Fahrenheit equation…• C = 5/9 ( F –32) for F• We need to get rid of the 5/9 with 9/5• (9/5) C = (9/5) (5/9) (F-32)• (9/5) C = F-32 then add 32 to both sides• 9/5 C + 32 = F (-32 + 32)• 9/5 C + 32 = F switch sides• F= 9/5 C + 32 done!

Example 3

• Solving for x when it’s on both sides (you’ve already done this) but now you have some other letters going along for the ride like numbers.

Example 3

• 5x – b = 3x + d subtract 3x from both sides

• 5x –3x -b = d 2x –b = d

• Add b to both sides

• 2x = d + b

• Divide both sides by 2

• x= (d+b)/2 done!

Note on speed…

• Note, I’m dropping some steps in solving equations that you should be getting smoother at. I’m labeling the steps, but not showing ever tiny part.

Example 4

• x + 2y = 6 Solve it for y

• Subtract x from both sides 2y = 6 – x

• Multiply both sides by ½

• ( ½ ) 2y = ( ½ ) ( 6-x)

• y= (6-x)/2 is good BUT you can also do it thus…

• y= ½(6-x) = 6/2 –x/2 = 3-x/2 = -x/2 + 3

Example 5• Solve 2x-3y = 9 but make it look like y=mx+b

(the slope of a line, to be used a LOT later!). m and b are what ever numbers they turn out to be.

• We want y alone on the left… so subtract 2x from both sides.

• -3y = 9-2x Then divide both sides by –3• y = 9/(-3) – 2x/(-3)• y= -3 + (2/3)x then swap the number and the

x• y= (2/3)x –3 looks like y= mx+b m=2/3, b=-3

Finding the value once you’ve solved the equation

• In example 5 we found that 2x-3y-9 becomes y= (2/3)x –3

• If we’re told that x=6, what is y?

• y= (2/3) 6 –3 = 12/3 – 3 = 4 –3 = 1

• That means on this line (more on that later) if x = 6 then y = 1.

Example 7 The simple interest formula

Solve I=Prt You know the simple interest $120 (I), the principle is $400 P) over 2 years (t). What is the rate (r) ?

Prt = I divide both sides by P and t Prt/(Pt) = I/ (Pt)r= I/Pt = 120/(400*2) = 0.15 or 15%

fun no?

Example 8

• How about using the perimeter equation for a rectangle? What is the Length (L) if the Perimeter (P) is 36 feet and the width (W) is 6 feet?

• P=2L+2W• P-2W= 2L• 2L= P-2W divide by 2• L= (P-2W)/2 = (36-2(6))/2 = 12 feet

Section 2.4 Formula Practice

• Give ‘em a test drive…– Definitions again Q1 – Q 6– Working with formulae Q 7 – 24– Letters instead of numbers in equations Q 25-32– Making things look like y=mx+b Q 33- 50– The same thing but with x=2 Q 51-60– Working with word problems of all the above

• Q 61- 87 USEFUL!

Section 2.5 – Putting this into words

• This is the last section in Chapter 2 (then we hop to 3.1-3.3)

• This section expands on the verbal work you did in 1.6

Verbal Phrases… Math as a language again… Addition first

• Addition:– The sum of a number and 8 x+8– Five is added to a number x+5– Two more than a number x+2– A number is increased by 3 x+3

More talk - Subtraction

• Subtraction– Four is subracted from a number x-

4– Three is less than a number x-3– The difference between 7 and a number x-7– A number is decreased by 2 x-2

Multiplication

• Multiplication– The product of 5 and a number 5x– Twice a number 2x– One half a number ½ x– Five percent a number 0.05x

Division

• Division– The ratio of a number to 6

x/6– The quotient of 5 and a number 5/x– Three divided by some number 3/x

Algebraic Expressions in Words

• Example 1– The sum of a number and 9 x+9– Eighty percent of a number 0.80w– A number divided by 4 y/4– The result of a number subtracted from 5 5-z– Three less than a number a-3

Adding wording for pairs of numbers

• If you sort of know how two numbers are related to one another, then you can construct one of those equations we’ve worked with that has two x’s in it.

• Like addresses… “that house is three houses down the street from this house.” And I know the address of this house, so I can figure out the other house’s number.

I know…

• I have two numbers… one number is 10 less than the other…

• I can write the first number as x

• And the other is 10 more than it, then it is x+10

Example 1

• Write two numbers that differ by 12 x and x-12 Check x – (x-12) = x-x+12 = 0 +12

• Two numbers with a sum of –8 x and -8-x

Check x+ (-8-x) = x-x-8 = 0 –8 = -8The trick, write down x. Write down the answer +

the opposite of the variable x (-x). Then show if you are adding or subtracting the values.

Example 2A related subject

• If you know the overall angle in some geometric shape…you can do this same ‘two things added’ verbal trick.

• But now we know the overall angle!

Ex 2 continued

• See page 96… a)• The right angle has 90 degrees. Fact.

• A smaller angle that is x and is part of the overall 90 degree angle. The rest of the stuff in that 90 degree angle is ?. So the remainder is 90 – x or ? = 90-x

• Nutty simple? If not, you are overthinking!

Ex 2 b and c

• (b) So if you have in (b) 180 degrees and a smaller part of that called x. Then the rest of the stuff in the 180 degree angle is 180-x

• (c) The interior angles of all triangles equals 180 degrees. So if you know one angle (30 degrees) and another angle is x, then the rest of it is: 180 –x –30 or 150 – x

Yet another in the same vein…

• Example 4 - describing sets of numbers a) Write three consecutive integers, the smallest

of which is w w, w+1, w+2

b) Write three consecutive even integers, the smallest of which is z

z, z+2, z+4

MORE Writing

• How about using known formulae?– Example 5 a– The distance if the rate is 30mph and the time is

T hours– We go find the formula D=RT for

distance=rate times time– Plug in the known value… R=30mph– So D=30T done!

Example 5b

• The discount rate if the rate is 40% and the original price is p dollars

• We go get the formula Discount = Rate of discount * Original Price

• Since the discount is the rate times the original price, an algebraic expression for the discount = 0.40p dollars

Page 97 is your repository of equations for these problems!

• See this page for many normal, everyday formulae.

• Use these for the quizzes and all!

Example 6 Words and Formulae

• a) Find two number that have a sum of 14 and a product of 45

• First, x is one of the numbers. Just say it.

• The other number is 14-x (the answer and the opposite of the first number)

• Their product is 45 : x(14-x) =45 done!

Ex 6b

• The coat is on sale for 25% off the list price. If the sale price is $87, then what is the list price?

• Our variables. x is the origional price (just say it)

The amount of discount is 0.25x• Which formula on pg 97?

Original price – discount = selling price (the third one)

• Plug it in: x –0.25x = 87 done!

Example 6c

• What percent of 8 is 2?

• If x is the percentage, then we want to know the ratio 2/8

• x=2/8 or 8x = 2

• Don’t forget to multiply the answer by 100 to make it a percent in the end.

Example 6d

• The value of x dimes and x-3 quarters is $2.05• We’ll say the value of x dimes at 10 cents each is

10x cents.• The value of x-3 quarters at 25 cents each is 25(x-

3)• The sum of them is 10x+25(x-3) = 205

(the $2.05 was turned into cents as well! Unit must match!)

Work on Section 2.5 Questions for a bit!

• Go for the hardest first and get help!– Definitions Q1 to Q6– Translating words to equations Q 7 to 30– Playing with the same idea but with angles Q31 to 33– Words describing sets of numbers to expressions Q 35 to

42– Words describing expressions to expressions Q 43-66– Words describing equations to equations Q67 - 96

CHAPTER 3All the same things with

inequalities instead of = signs

• It’s a shift in what ‘equals’ means.

• This allows you to practice all the tricks and games you’ve learned so far.

• Remember the number line? – Is –5 greater or less than 4?

• But it starts with some new definitions…

3.1 Inequalities• The Basics

• < Is less than =! Is not equal to• <= Is less than or equal• > Is greater than • >= Is greater than or equal to

We are comparing numbers(or sides of an expression)

• We are comparing them then asking if it is true or not (much of the time).

• True or false? -10 < -5

Example 1

• a ) is -5 < 3 ? True

• b) is -9 > -6 ? False

• c) is -3 <= 2 ? True

• d) is 4 >= 4 ? True

We can mark it on the number line!

• Remember the solution set? We put the answer in { }’s

• Any solution set can be written on the number line

• For instance, All x’s such that x is less than three… {x|x<3} Note the “ ) “

Another example

• This is the number line graph all x’s such that x is greater OR equal to 1.

• {x|x>=1} Note the “ [ “ this time for the equal

Watch the (‘s and [‘s and the way they point.

The basic number line again

----|----|----|----|----|----|----|----|----|----|----|----

-5 -4 -3 -2 -1 0 1 2 3 4 5

Example 2

• a)

• x > -5

----(----|----|----|----|----|----|----|----|----|----|--->

-5 -4 -3 -2 -1 0 1 2 3 4 5

Example 2b

• b)

• x<=2

<---|----|----|----|----|----|----|----]----|----|----|----

-5 -4 -3 -2 -1 0 1 2 3 4 5

Solving Linear Inequalities

The official properties…

• If you multiply both sides of an inequality by a negative number you must switch the inequality sign

In SIMPLE terms

• All the rules are the same as you worked with equalities EXCEPT if you multiply or divide by a negative number

• Multiply or divide by a negative number and you switch the > to a < or the reverse.

Example 3 Seeing how it works…

• a) 2x –7 < -1

2x-7 +7 < -1 + 7

2x < 6

(1/2) 2 x < (1/2) 6

x < 3 or {x|x<3}

Example 3b

• 5- 3x < 11

• 5 –5 –3x < 11 –5 -5 both sides

• -3x < 6

• (-1/3) 3x < (-1/3) 6 * -1/3 both sides

• x > -6/3 = -2 SWITCHED < to >

• x > -2 or {x|x > -2}

Example 4

• (8+3x)/-5 >= -4 simplify and graph• So we solve for x again… NOTHING NEW (except

watching for that multiply by –1 thing)• (-5)*(8+3x)/(-5) <= -4 * (-5) so we do it first!

and switch the inequality

• 8+3x <= 20• 3x <= 20-8 (subtract 8 from both sides)• 3x <= 12 finish the right side• x <= 12/3 divide both sides by 3• x <= 4 graph on next page…

Ex 4 continued

x <= 4

<---|----|----|----|----|----|----|----|----|----]----|----

-5 -4 -3 -2 -1 0 1 2 3 4 5

Example 5 now with fractions for more flavor!

• 1/2 x – 2/3 <= x+ 4/3• If we multiply the whole thing by the least common

denominator ( 2*3=6) then life will be so so so so much more easy

• And remember, you trained for this moment!• 6(1/2x – 2/3) <= 6(x+ 4/3)• 6/2 x - 12/3 <= 6x + 24/3 no switch of the <=• 3x – 4 <= 6x + 8• 3x-6x –4 <= 8 subtract 6x from both sides• -3x <= 8+4 add 4 to both sides• x >= 12/-3 divide by –3 and switch the inequality• x >= -4

Graphing Ex 4

• X >= -4

---|----[----|----|----|----|----|----|----|----|----|--->

-5 -4 -3 -2 -1 0 1 2 3 4 5

We’ll skip the 2nd half of 3.1 and go right to part of the exercises

• How do you get to Carnegie Hall?• Do Inequalities…• Definitions Q 1 – Q 6• True or False – the Game Q 7 – Q 20• Write and graph it Q 21 – 28• Use that alternate notation pg 123 Q 29-36• Supply the <, <=,>, >= Q37-44• Solve and Graph (like examples) Q 45 -72

Section 3.2 The next step with inequalities

• We take two simple inequalities (section 3.1) and link them with “and” or “or”.

• We can use simple logic…

• We can defined a range!

Example 1 True or False? AND

• a) 3>2 and 3 < 5

• The first one is true, the second one is true

• True AND true = true

• b) 6> 2 and 6<5

• The first one is true, the second is false

• True AND false = false

Example 2 True or False? OR

• a) 2 < 3 or 2 > 7• The first is true, the second is false• BUT if ONE is true with an OR, it is true• True OR False = True!

b) 4<3 or 4 >= 7• The first is false and the second is false• False OR False = False! (there isn’t any truth here

at all)

Example 3 True or False – testing a number

• a) x <6 and x < 9 we’ll test 5• 5 < 6 and 5 < 9• True and True = True!

• b) 2x-9 <= 5 or -4x >= -12 plug in 5…• 10-9 <= 5 or –20 >= -12• 1 <= 5 or -20 >= -12• True or False = True (One true is enough with OR)

Graphing those compound inequalities

• The trick is to lightly sketch ONE inequality then the OTHER then , depending on the AND or OR, combine them to make the answer…

• Definitions: The intersection of two sets is the set of all numbers both sets share.

Intersections

• A intersects B is written as A ‘upside down U’ B ( I can’t reliably make an upside down U here… sorry)

• A = { 1,2,3} and B = {2,3,4,5}• Then A “upside down U” B = {2,3}• That’s all they share, that’s all that intersects• This is the AND of sets

Ex 4 – Graphing them

• x > 2 and x < 5

----|----|----|----|----|----|----|----(----|----|----|-->

-5 -4 -3 -2 -1 0 1 2 3 4 5

<--|----|----|----|----|----|----|----)----|----|----|----

-2 -1 0 1 2 3 4 5 6 7 8

----|----|----|----(----|----|----)----|----|----|----|----

-1 0 1 2 3 4 5 6 7 8 9

The Union

• The Union of two sets is the same as OR.

• It is the set of all numbers in A OR B

• We write A U B

• If A = {1,2,3} and B = {2,3,4,5}

• Then A U B = {1,2,3,4,5}

• It’s ALL of them!

Graphing them… Ex 5• x > 4 or x < -4

----|----|----|----|----|----|----|----|----|----(----|-->

-5 -4 -3 -2 -1 0 1 2 3 4 5

<--|----)----|----|----|----|----|----|----|----|----|----

-5 -4 -3 -2 -1 0 1 2 3 4 5

<--|----)----|----|----|----|----|----|----|----(----|-->

-5 -4 -3 -2 -1 0 1 2 3 4 5

Doing it quicker, Ex 6

• a) If we are given x<3 and x < 5 we need only shade what is less than 3 (where they overlap)

<--|----|----|----|----)----|----|----|----|----|----|----

-1 0 1 2 3 4 5 6 7 8 9

Ex 6b

• b) x>4 or x>0 shade everything since it’s OR

----|----|----|----|----|----(----|----|----|----|----|--->

-5 -4 -3 -2 -1 0 1 2 3 4 5

All or nothing

• Sometimes the set just includes every number on the number line and sometimes there are NO numbers that satisfy the intersection

Example 7 a) x< 2 and x >6 No Intersection

b) x<3 or x> 1 ALL NUMBERS are the answer

Back to solving first…

• Sometimes you need to solve the inequality like we did in 3.1 …

• Example 8 x+2 > 3 and x-6 < 7

• Step 1, get x alone x>1 and x<13 subtract 2 from both sides add 6 to both sides

----|----(----|----|----|----|----|----|----|----|----|----|----|----)----|----|---- 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ex 9 solving with a Union• 5-7x >= 12 OR 3x –2 < 7

• Solve for x

• -7x >= 7 OR 3x < 9

• x<= -7 OR x<3 (switched inequality!)

<--|----|----|----|----|----|----|----|----)----|----|----

-5 -4 -3 -2 -1 0 1 2 3 4 5

Compacting the Trash

• We read these things with the variable first (the brain likes it that way)

• So 1< x reads “x is greater than 1” (and is also x>1)

• With intersections (AND) we can compact things a bit more

• Why write x >1 and x <13 when this is better:• 1 < x < 13 “Called a compound inequality”

Other legal compounds

• You can use < < and > > and >= >= and <= <= but not < >

• 4 < x < 9

• -6 >= x >= -8

• 10 > x > 2

• But NOT: 5 < x < 2 or 5 < x > 7

Ex 10 Compounding your troubles

• -2 <= 2x –3 < 7• Solve for x• -2 +3 <= 2x < 7 +3 (+3 to all parts)• 1 <= 2x < 10• ½ <= x < 5 (multiply ½ to all parts)

----|----|----|----|----|----|--[--|----|----|----|----)----

-5 -4 -3 -2 -1 0 1 2 3 4 5

A fable problem Ex 11

• Fiana is getting a grade from 2 scores in a class. Her midterm was a 76 and her final is unknown (x). She can get a B if she is between 80 and 89 in the end. What scores can she get to earn a B?

Ex 11 continued

• 80<= (x+76)/2 <= 89

• 160 <= x+76 <= 178 ( mult. all parts by 2)

• 84 <= x <= 102 (subt. 76 from all parts)

• That’s the range (can she get extra credit for the 102?). What color was her sweater?

Section 3.2 hoops to jump through

• Definitions Q 1 to 6

• Compound True or False Q 7 to 18

• Graph given compound inequalities Q19-30

• Solve THEN graph compound ineq. Q31-52

• Working with the short ( ) notation Q53-66

• Going from graph to inequality Q67-78

• Solve and write in short ( ) notation Q77-82

Section 3.3 All the fun WITH absolute values now…

• |x|=5 The solution set is {-5,5}

• |0| = 0 which is {0}

• |x| = -7 is an empty set (no answer with real numbers)

Example 1 positive numbers

• a) |x-7| =2

• First: Take away the | |’s (it gives us 2 answers!)

• x-7 = 2 or x-7=-2 (Just like –5, 5 above)

• x=9 or x =5 so our solution set is {5,9}

Example 1b

• b) |3x-5| = 7

• First: 3x-5 = 7 or 3x-5 = -7 3x=12 or 3x = -2

x= 4 or x = -2/3 our solution set is {-2/3, 4}

Example 2 Now with 0

• |2(x-6)+7| = 0

• We have only 1 answer, 0

• 2(x-6) + 7 = 0

2x-12+7=0

2x-5=0

2x=5

x=5/2 The solution set is {5/2}

Example 3 now with a negative answer

• -5|3x-7| + 4 = 14

• -5|3x-7| = 10

• |3x-7| = -2

• Buzzzt. No answer in the Real World.

Example 4 Can we have ||’s on both sides?

• Sure! We’re just crazy enough!

Example 4

• |2x-1| = |x+3|

• First, kill the ||’s

• 2x-1 = x+3 or 2x –1 = -(x+3)

• 2x=x+4 or 2x= -x -3 +1

• 2x = x+4 or 2x = -x –2

• x= 4 or 3x = -2

x = -2/3 SS {-2/3,4}

Absolute values are HIDDEN inequalities… believe that?

• |x| > 5

• Is the same as x >5 or x < -5

----|----|----|----)----|----|----|----|----|----|----|----|----|----(----|----|----|----

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

And a < is like an AND

• |x| <= 3

• -3 <= x <= 3

----|----|----|----|----|----[----|----|----|----|----|----]----|----|----|----|----|----

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Example 5

• |x-9| < 2 (Less than is like AND)

• This is like –2 < x –9 < 2

• Solve for x -2 +9 < x < 2 + 9 (+9 to all parts)

• 7 < x < 11

----|----|----|----|----|----|----|----|----|----(----|----|----|----)----|----|----|----

--2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Example 6

• |3x+5| >2 (greater than is like OR = Outside)

• 3x+5 > 2 OR 3x+5 < -2

• 3x> -3 or 3x < -7

• x > -1 or x < -7/3

<---|----|----|----|----|----|---)-|----(----|----|----|----|----|----|----|----|----|--->

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Example 7 Another < (AND)

• |5 –3x| <= 6-6 <= 5-3x <= 6-11 <= -3x <= 1 subtract 5-11/3 >= x >= -1/3 divide by 3SS = {-1/3,11/3}

----|----|----|----|----|----|----|----|--[-|----|----|----|--]-|----|----|----|----|---- -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Example 8 Special 0 (not K)

• 3+ |7-2x| >= 3

• |7-2x| >= 0

• BUT ALL real numbers you can plug in work because the ||’s mean everything inside is made Positive or >= 0

Example 9 And the opposite

• |5x-12| < -2

• But the stuff inside the ||’s is ALWAYS positive when you are using real numbers, this cannot happen!

• Impossible.

Section 3.3 Practice

• Definitions Q 1 to 6• Solving to get solution sets Q 7 to 36• Going from graph to inequality with an absolute value

Q 37 to 44• Is the abs. value the same as the inequality True or False

Q 45 to 52• Solve and graph the abs. value inequality Q 53 to 72• Solve each inequality write using interval notation ( )’s

Q 73 to 84