Unit 2.

Unit 2

Unit 2 Section 3.7 Section 1.1-1.2 Section 3.8 Section 1.3
Review Slides Section Section 1.3 Section 1.4 Section 2.5 Section Section 3.3 Section 3.4 Section

1.1 – Evaluate Expressions 1.2 – Order of Operations
Warm-Ups Vocabulary Notes Examples

Warm-Up Lesson 1.1, For use with pages 2-7
Perform the indicated operation. ANSWER 8 ANSWER 23.1

Perform the indicated operation. – 5.9 ANSWER 5.7 4. Julia ran miles last week and miles this week. How many more miles did she run last week? 2 3 5 6 ANSWER 1 mi 5 6

Warm-Up Lesson 1.2, For use with pages 8-13 Evaluate the expression.
1. a when a = 1.3 ANSWER 7 2. b3 when b = 4 ANSWER 64

Warm-Up Lesson 1.2, For use with pages 8-13 Evaluate the expression.
3. The number of weeks it takes you to read a novel is given by , where n is total pages in the novel and p is pages read per week. How long will it take you to read a 340-page novel if you read 85 pages per week? n p ANSWER 4 weeks

Vocabulary Variable A letter used to represent 1 or more numbers Algebraic Expression Contains numbers, variables, and operations NO EQUALS SIGN!!! Power Repeated Multiplication Base Number that is multiplied repeatedly Exponent How many times the Base is multiplied Order of Operations Order to evaluate expressions involving multiple operations

Notes Any time you see a formula and numbers, what do you do?? Plug in what you know and solve for what you don’t! Exponents: am= Exponent a * a * a … * a Base Multiply “a” together “m” times

Notes Order of Operations – How many steps? Write it down on your whiteboard. P – Parenthesis E – Exponents M and D – Multiplication and Division IN ORDER FROM LEFT TO RIGHT!! A and S – Addition and Subtraction IN ORDER FROM LEFT TO RIGHT! Within Parenthesis, the order of operations still applies!

Examples GUIDED PRACTICE Evaluate the expression. 9. x3 when x = 8
10. k3 when k = 2.5 SOLUTION 9. x3 = 83 10. k3 = 2.53 = = (2.5)(2.5)(2.5) = 512 = 6.25

( ) GUIDED PRACTICE for Example 4 Evaluate the expression.
d4 when d = 1 3 SOLUTION d4 = ( ) 1 3 4 = 1 3 = 1 81

The volume of the storage cube is 2744 cubic inches.
EXAMPLE 5 Evaluate a power Storage cubes Each edge of the medium-sized pop-up storage cube shown is 14 inches long. The storage cube is made so that it can be folded flat when not in use. Find the volume of the storage cube. SOLUTION V = s3 Write formula for volume. = 143 Substitute 14 for s. = 2744 Evaluate power. The volume of the storage cube is 2744 cubic inches.

Evaluate the expression.
GUIDED PRACTICE for Examples 2 and 3 Evaluate the expression. (3 + 9) = 4 (12) Add within parentheses. = 48 Multiply. 6. 3(8 – 22) = 3 (8 – 4) Evaluate power. = 3 (4) Subtract within parentheses. = 12 Multiply. 7. 2[( 9 + 3) ] = 2 [(12) 4] Add within parentheses. = 2 [ 3 ] Divide within brackets. = 6 Multiply.

Evaluate the expression when y = 8.
GUIDED PRACTICE for Examples 2 and 3 Evaluate the expression when y = 8. y2 – 3 8. = 82 – 3 Substitute 8 for y. = 64 – 3 Evaluate power. = 61 Subtract.

Evaluate the expression when y = 8.
GUIDED PRACTICE for Examples 2 and 3 Evaluate the expression when y = 8. 12 – y – 1 9. = 12 – 8 – 1 Substitute 8 for y. = 4 – 1 Subtract. = 3 Subtract.

GUIDED PRACTICE for Examples 2 and 3 10y+ 1 10(8) + 1 10. = y + 1
8 + 1 10. Substitute 8 for y. = 80 + 1 8 + 1 Evaluate product. 81 = 9 Add. = 9 Divide.

Warm-Up – 1.3 1. Evaluate 2[54 (42 + 2)]. 6 ANSWER 5x
2. Evaluate when x = 3. 5x x + 2 ANSWER 3

3. Eight students each ordered 2 drawing kits and 4 drawing pencils. The expression 8(2k + 4p) gives the total cost, where k is the cost of a kit and p is the cost of pencil. Find the total cost if a kit costs $25 and a pencil costs $1.25. ANSWER $440

Vocabulary – 1.3 Rate Compares two quantities measure in DIFFERENT UNITS! Unit Rate Has a denominator of 1 - “Something over one” Equation Math sentence with an EQUALS.

Notes 1.3 CONVERTING ENGLISH SENTENCES TO MATHLISH SENTENCES: There are 3 steps to follow: Read problem and highlight KEY words. Define variable (What part is likely to change OR What do I not know?) Write Math sentence left to Right (Be careful with Subtraction and sometimes Division!)

Notes 1.3 LOOK FOR WORDS LIKE: is, was, total EQUALS
Less than, decreased, reduced, SUBTRACTION - BE CAREFUL! Divided, spread over, “per”, quotient DIVISION More than, increased, greater than, plus ADDITION Times, Of, Product MULTIPLICATION

EXAMPLE 1 Translate verbal phrases into expressions Verbal Phrase Expression a. 4 less than the quantity 6 times a number n 6n – 4 b. 3 times the sum of 7 and a number y 3(7 + y) c. The difference of 22 and the square of a number m 22 – m2

GUIDED PRACTICE for Example 1 1. Translate the phrase “the quotient when the quantity 10 plus a number x is divided by 2” into an expression. ANSWER 1. Expression 10 + x 2

EXAMPLE 4 Find a unit rate A car travels 110 miles in 2 hours. Find the unit rate. 110 miles 2 hours = 1 hour 55 miles 2 hours 2 110 miles 2 The unit rate is 55 miles per hour, or 55 mi/h ANSWER

EXAMPLE 5 Solve a multi-step problem Cell Phones Your basic monthly charge for cell phone service is $30, which includes 300 free minutes. You pay a fee for each extra minute you use. One month you paid $3.75 for 15 extra minutes. Find your total bill if you use 22 extra minutes. SOLUTION STEP 1 Calculate the unit rate. 15 3.75 = 0.25 1 = $.25 per minute

EXAMPLE 5 Solve a multi-step problem STEP 2 Write a verbal model and then an expression. Let m be the number of extra minutes. m Use unit analysis to check that the expression m is reasonable. minute dollars dollars + minutes + dollars + dollars = dollars Because the units are dollars, the expression is reasonable.

   GUIDED PRACTICE for Examples 2 and 3
Check whether the given number is a solution of the equation or inequality. Equation/Inequality Substitute Conclusion 2. 9 – x = 4; x=5 9 – ? = 4 = 4 5 is a solution.  3. b + 5 < 15; b=7 < ? 12<15 7 is a solution.  4. 2n ;n=9 > – 2(3) > – ? is a solution. > – 

GUIDED PRACTICE for Examples 4 and 5 STEP 2 Write a verbal model and then an expression. Let m be the number of extra minutes. m Use unit analysis to check that the expression m is reasonable. minute dollars dollars + minutes + dollars + dollars = dollars Because the units are dollars, the expression is reasonable.

Warm-Up – 1.4 – Write Inequalities
1. Write an expression for the phrase: 4 times the difference of 6 and a number y. ANSWER 4(6 – y) 2. A museum charges $50 for an annual membership and then a reduced price of $2 per ticket. Write an expression to represent the situation. Then find the total cost to join the museum and buy 9 tickets. 50 + 2t, where t is the number of tickets; $68 ANSWER

Warm-Up – 1.4 – Write Inequalities
Write an equation for the sentence: Find the quotient of the sum of 10 and a number and the quantity of the difference of the Number and 2 ANSWER (10 + x) / (x-2)

Vocabulary – 1.4 Dimensional Analysis (or Unit Analysis) Inequality
Math sentence that contains <, >, ≤ , ≥, or ≠ Solution of Equation or Inequality Number or numbers that make the statement (sentence) true. Dimensional Analysis (or Unit Analysis) Keeping units with calculations

Notes 1.4 WRITING INEQUALITIES
Similar to writing Equations and Expressions. Look for the following clues To check and see if your inequality is correct, pick 3 numbers and check them. Smaller Larger The number itself

Examples 1.4

EXAMPLE 1 Write equations and inequalities Verbal Sentence Equation or Inequality a. The difference of twice a number k and 8 is 12. 2k – 8 = 12 b. The product of 6 and a number n is at least 24. 6n ≥ 24 c. A number y is no less than 5 and no more than 13. 5 ≤ y ≤ 13

GUIDED PRACTICE for Example 1 1. Write an equation or an inequality: The quotient of a number p and 12 is at least 30. ANSWER P 12 > – 30

EXAMPLE 4 Solve a multi-step problem Mountain Biking The last time you and 3 friends went to a mountain bike park, you had a coupon for $10 off and paid $17 for 4 tickets. What is the regular price of 4 tickets? If you pay the regular price this time and share it equally, how much does each person pay?

Solve a multi-step problem
EXAMPLE 4 Solve a multi-step problem SOLUTION STEP 1 Write a verbal model. Let p be the regular price of 4 tickets. Write an equation. Regular price Amount of coupon  Amount paid = p – =

EXAMPLE 4 Solve a multi-step problem STEP 2 Use mental math to solve the equation p – 10 =17. Think: 10 less than what number is 17? Because 27 – 10 = 17, the solution is 27. ANSWER The regular price for 4 tickets is $27. STEP 3 $27 4 people Find the cost per person: = $6.75 per person. ANSWER Each person pays $ 6.75.

Write and check a solution of an inequality
EXAMPLE 5 Write and check a solution of an inequality Basketball A basketball player scored 351 points last year. If the player plays 18 games this year, will an average of 20 points per game be enough to beat last year’s total? STEP 1 SOLUTION Write a verbal model. Let p be the average number of points per game. Write an inequality. p > Number of games Points per game • Total points last year =

EXAMPLE 5 Write and check a solution of an inequality STEP 2 Check that 20 is a solution of the in equality18p > 351. Because 18(20) = 360 and 360 > 351, 20 is a solution.  ANSWER An average of 20 points per game will be enough.

Warm-Up – 2.5 ? 1. –15 + (–19) + 16 = –18 ANSWER 2. 6(–x)(–4) = ? 24x
1. –15 + (–19) + 16 = ? ANSWER –18 2. 6(–x)(–4) = ? ANSWER 24x

Lesson 2.5, For use with pages 96-101
3. –9(–2)(–4b) = ? 4. 4(x + 3)= ? ANSWER –72b ANSWER 4x + 12 4. Kristin paid $1.90 per black-and-white photo b and $6.80 per color photo c to have some photos restored. What was the total amount A that she paid if she had 8 black-and-white and 12 color photos restored. ANSWER $96.80

Vocabulary – 2.5 like terms equivalent expressions
Look alike! – Same variables! Constant A number w/o a variable simplest form All like terms combined simplifying the expression Combining all the like terms equivalent expressions Expressions that are equal no matter what “x” is Term A “part” of an Alg. Expression separated by + or - Coefficient The number in front of a variable

Notes - 2.5 QUICK REVIEW Four Fundamental Algebraic Properties
Commutative Addition – a + b = b + a Multiplication – a * b = b * a Associative Add – (a + b) + c = a + (b + c) Mult - (a * b) * c = a * (b * c) Distributive – MOST IMPORTANT!! a (b + c) = ab + ac Identity Add = a + 0 = a Mult = a * 1 = a

Notes - 2.5 a (b + c) = ab + ac NOTES Distributive Property
I can only combine things in math that ????? LOOK ALIKE!!!!!!! In Algebra, if things LOOK ALIKE, we call them “like terms.” BrainPops: The Associative Property The Commutative Property The Distributive Property

EXAMPLE 1 Apply the distributive property Use the distributive property to write an equivalent expression. a. 4(y + 3) = 4y + 12 b. (y + 7)y = y2 + 7y c. n(n – 9) = n2 – 9n d. (2 – n)8 = 16 – 8n

Distribute a negative number
EXAMPLE 2 Distribute a negative number Use the distributive property to write an equivalent expression. a. –2(x + 7)= – 2(x) + – 2(7) Distribute – 2. = – 2x – 14 Simplify. b. (5 – y)(–3y) = 5(–3y) – y(–3y) Distribute – 3y. = – 15y + 3y2 Simplify.

Distribute a negative number
EXAMPLE 2 Distribute a negative number c. –(2x – 11) = (–1)(2x – 11) of 21 Multiplicative property = (– 1)(2x) – (–1)(11) Distribute – 1. = – 2x + 11 Simplify.

EXAMPLE 3 Identify parts of an expression Identify the terms, like terms, coefficients, and constant terms of the expression 3x – 4 – 6x + 2. SOLUTION Write the expression as a sum: 3x + (–4) + (–6x) + 2 Terms: 3x, – 4, – 6x, 2 Like terms: 3x and – 6x; – 4 and 2 Coefficients: 3, – 6 Constant terms: – 4, 2

Use the distributive property to write an equivalent expression.
GUIDED PRACTICE for Examples 1, 2 and 3 Use the distributive property to write an equivalent expression. (x + 3) = 2x + 6 2. – (4 – y) = – 4 + y Distributive – 1 3. (m – 5)(– 3m) = m (– 3m) –5 (– 3m) Distributive – 3m = – 3m2 + 15m Simplify. 4. (2n + 6) = 1 2 1 2 2n 1 2 Distribute = n + 3 Simplify.

GUIDED PRACTICE for Examples 1, 2 and 3 Identify the terms, like terms, coefficients, and constant terms of the expression – 7y + 8 – 6y – 13. SOLUTION Write the expression as a sum: – 7y + 8 – 6y – 13 Terms: – 7y, 8, – 6y, – 13 Like terms: – 7y and – 6y , 8 and – 13; Coefficients: – 7, – 6 Constant terms: 8, – 13

EXAMPLE 5 Solve a multi-step problem Your daily workout plan involves a total of 50 minutes of running and swimming. You burn 15 calories per minute when running and 9 calories per minute when swimming. Let r be the number of minutes that you run. Find the number of calories you burn in your 50 minute workout if you run for 20 minutes. EXERCISING SOLUTION The workout lasts 50 minutes, and your running time is r minutes. So, your swimming time is (50 – r) minutes.

EXAMPLE 5 Solve a multi-step problem STEP 1 Write a verbal model. Then write an equation. C = r (50 – r) Amount burned (calories) Burning rate when running (calories/minute) Running time (minutes) Swimming time (minutes) = + • Burning rate when swimming (calories/minute) C = 15r + 9(50 – r) Write equation. = 15r – 9r Distributive property = 6r + 450 Combine like terms.

EXAMPLE 5 Solve a multi-step problem STEP 2 Find the value of C when r = 20. C = 6r + 450 Write equation. = 6(20) = 570 Substitute 20 for r. Then simplify. ANSWER You burn 570 calories in your 50 minute workout if you run for 20 minutes.

Warm-Up – 3.1 Simplify using the Distributive Property.
Solve using mental math. 1. x + 2 = 17 -3(2x – 5) ANSWER 15 ANSWER -6x + 15 2. x 6 = 4 2. 2x ( 4x – 12) ANSWER 24 ANSWER 8x2 – 24x

Warm-Up – 3.2 Solve the equation.
3. Simplify the expression 3(x + 2) – 4x + 1. ANSWER –x + 7 4. There are three times as many goats as sheep in a petting zoo. Find the number of sheep if the total number of goats and sheep is 28. ANSWER 7 sheep

Warm-Up – 3.2 Evaluate (½ ) * (2/1) (1/4) * (4/1) (4/7) * (7/4)
4. Notice any patterns??? What can you conclude about multiplying a fraction times its reciprocal? 58

Vocabulary – 3.1-3.2 Inverse Operations The opposite operation Input
Numbers that we plug into an equation (frequently represented by “x”) AKA the “domain” AKA the “independent variable” Output Numbers we get out of an equation (frequently represented by “y”) AKA the “range” AKA the “dependant variable”

Notes – 3.1-3.2 Order of Operations is what???
The goal of solving EVERY algebra equation you will EVER see for the rest of your life is … GET THE VARIABLE BY ITSELF! “How do I do that, Mr. Harl?” Four things to remember: Anything I do to one side of an equation, I MUST DO TO THE OTHER SIDE!! Do the opposite operations as necessary Simplify (if possible) SADMEP

Examples 3.1 x = 4 2 7 – Solve SOLUTION 7 2 – ( ) x ) = 2 7 – x = – 14
Multiply each side by the 7 2 – reciprocal, ( ) x ) = 2 7 – 4 x = – 14

Examples 3.1 for Example 5 Solve the equation. Check your Solution.
w = 10 5 6 13. SOLUTION 5 Multiply each side by the 6 reciprocal, 5 ( ) w ) = 6 10 w = 12

Examples 3.1 Solve an equation using correct operation
Solve – 6x = 48. – 65 = – 5y. = 5 x 4 = 13 z -2

Examples 3.2 Solve 8x – 3x – 10 = 20 8x – 3x – 10 = 20 5x – 10 = 20
Write original equation. 5x – 10 = 20 Combine like terms. 5x – = Add 10 to each side. 5x = 30 Simplify. = 30 5 5x Divide each side by 5. x = 6 Simplify.

Examples 3.2 Solve a two-step equation Solve + 5 = 11. x 2 + 5 = x 2
Write original equation. + 5 – 5 = x 2 11 – 5 Subtract 5 from each side. = x 2 6 Simplify. = x 2 2 6 Multiply each side by 2. x = 12 Simplify. ANSWER The solution is 12. Check by substituting 12 for x in the original equation.

Solve a two-step equation
EXAMPLE 1 Solve a two-step equation + 5 = x 2 11 CHECK Write original equation. 11 + 5 = 12 2 ? Substitute 12 for x. 11 = 11  Simplify. Solution checks.

Solve the equation. Check your solution. 1. 5x + 9 = 24
GUIDED PRACTICE for Example 1 Solve the equation. Check your solution. x + 9 = 24 SOLUTION 5x + 9 = 24 Write original equation. 5x + 9 – 9 = 24 – 9 Subtract 9 from each side. 5x = 15 Simplify. 5x 3 = 15 Divide each side by 5 x = 3 Simplify.

GUIDED PRACTICE for Example 1 ANSWER
The solution is 3. Check by substituting 3 for x in the original equation. CHECK 5x + 9 = 24 Write original equation. 24 = ? Substitute 3 for x. 24 = 24 Simplify. Solution check.

Solve the equation. Check your solution. 3. – 1 = –7 z 3 SOLUTION
GUIDED PRACTICE for Example 1 Solve the equation. Check your solution. 3. – 1 = –7 z 3 SOLUTION – 1 = z 3 – 7 Write original equation. – = z 3 – 7 + 7 Add 7 to each side. 6 = z 3 Simplify. 6 = 3 z Multiply each side by 3. 18 = z Simplify.

GUIDED PRACTICE for Example 1 ANSWER
The solution is 18. Check by substituting 18 for z in the original equation. – 1 = z 3 – 7 CHECK Write original equation. 18 3 – 7 – 1 = ? Substitute 18 for z. – 1 = – 1 Simplify. Solution checks.

Warm-Up – 3.3 1. Simplify the expression 9x + 2(x – 1) + 7 ANSWER
Solve the equation. 2. 5g – 7 = 58 ANSWER 13

Warm-Up – 3.3 1. Simplify the expression -3x - 2(x + 5) - 5 ANSWER
Solve the equation. 2. -3x + 12 = -3 ANSWER X=5

Warm-Up – 3.3 Solve the equation. 2 3. (x ) = 18 3 ANSWER 27 4.
A surf shop charges $85 for surfing lessons and $35 per hour to rent a surfboard. Anna paid $225. Find the number of hours she spent surfing. ANSWER 4 h

Vocabulary – 3.3 Reciprocal Inverse of a fraction

Notes – 3.3 – Solve Multi-Step Eqns.
What is the goal of solving every Alg. equation? GET THE VARIABLE BY ITSELF! Four things to remember: Anything I do to one side of an equation, I MUST DO TO THE OTHER SIDE!! Do the opposite operations as necessary Simplify (if possible) – Distributive property, combine like terms, etc SADMEP  SSADMEP

Examples 3.3 Solve 7x + 2(x + 6) = 39. SOLUTION
When solving an equation, you may feel comfortable doing some steps mentally. Method 2 shows a solution where some steps are done mentally.

Examples 3.3 METHOD 1 METHOD 2 Do Some Steps Mentally Show All Steps
7x + 2(x + 6) = 39 7x + 2(x + 6) = 39 7x + 2x + 12 = 39 7x + 2x + 12 = 39 9x + 12 = 39 9x + 12 = 39 9x + 12 – 12 = 39 – 12 9x = 27 9x = 27 x = 3 = 9x 9 27 x = 3

EXAMPLE 2 GUIDED PRACTICE for Examples 1, 2, and 3 Solve the equation. Check your solution. 2w + 3(w + 4) = 27 2. 2w + 3(w + 4) = 27 2w + 3w + 12 = 27 5w + 12 = 27 5w + 12 – 12 = 27 – 12 5w = 15 = 5w 5 15 w = 3

GUIDED PRACTICE for Examples 1, 2, and 3 Check 2w + 3(w +4) = 27
Write original equation. 2(3) + 3(3 + 4) = 27 ? Substitute 3 for w. 6 + 3(7) = 27 ? Simplify. = 27 ? Multiply. 27 = 27 Simplify solution checks.

EXAMPLE 2 GUIDED PRACTICE for Examples 1, 2, and 3 Solve the equation. Check your solution. 6x – 2(x – 5) = 46 3. 6x – 2(x – 5) = 46 6x – 2x + 10 = 46 4x + 10 = 46 4x + 10 – 10 = 46 – 10 4x = 36 = 4x 4 36 x = 9

GUIDED PRACTICE for Examples1, 2, and 3 Check 6x – 2(x – 5) = 46
6(9) – 2(9 – 5) = 46 ? 54 – 2(4) = 46 ? 54 – 8 = 46 ? 46 = 46

Example – 3.3 Solve the equation. (x + 7) 3. 2 3 = 8 ANSWER X = 5 4.

Warm-Up – 3.4 Lesson 3.4, For use with pages 154-160
Solve the equation. 1. 2m – 6 + 4m = 12 ANSWER 3 2. 6a – 5(a – 1) = 11 ANSWER 6

Lesson 3.4, For use with pages 154-160
Solve the equation. 3. A charter bus company charges $11.25 per ticket plus a handling charge of $.50 per ticket, and a $15 fee for booking the bus. If a group pays $297 to charter a bus, how many tickets did they buy? ANSWER 24 tickets

Vocabulary – 3.4 Identity Equation that is true for ALL input values

Notes – 3.4 What is the goal of solving every Alg. equation?
GET THE VARIABLE BY ITSELF! Four things to remember: Anything I do to one side of an equation, I MUST DO TO THE OTHER SIDE!! Do the opposite operations as necessary Simplify (if possible) SADMEP  SSADMEP If you have variables on BOTH sides of the equation, get rid of one of them. USUALLY easiest to get rid of the smallest one!!

Examples 3.4 Solve an equation with variables on both sides
Solve 7 – 8x = 4x – 17. 7 – 8x = 4x – 17 Write original equation. 7 – 8x + 8x = 4x – x Add 8x to each side. 7 = 12x – 17 Simplify each side. 24 = 12x Add 17 to each side. 2 = x Divide each side by 12. ANSWER The solution is 2. Check by substituting 2 for x in the original equation.

Examples 3.4 Solve an equation with grouping symbols 1 Solve 9x – 5 =
Write original equation. 9x – 5 = 4x + 15 Distributive property 5x – 5 = 15 Subtract 4x from each side. 5x = 20 Add 5 to each side. x = 4 Divide each side by 5.

GUIDED PRACTICE for Examples 1 and 2 1. 24 – 3m = 5m. 24 – 3m = 5m
Write original equation. 24 – 3m + 3m = 5m + 3m Add 3m to each side. 24 = 8m Simplify each side. 3 = m Divide each side by 8. ANSWER The solution is 3. Check by substituting 3 for m in the original equation.

GUIDED PRACTICE for Examples 1 and 2 2. 20 + c = 4c – 7 .
Write original equation. 20 + c – c = 4c – c – 7 Subtract c from each side. 20 = 3c – 7 Simplify each side. 27 = 3c Add 7 to each side. 9 = c Divide each side by 3. ANSWER The solution is 9. Check by substituting 9 for c in the original equation.

EXAMPLE 3 Solve a real-world problem CAR SALES A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new cars sold be twice the number of used cars sold?

EXAMPLE 3 Solve a real-world problem SOLUTION Let x represent the number of years from now. So, 6x represents the increase in the number of new cars sold over x years and – 4x represents the decrease in the number of used cars sold over x years. Write a verbal model. 67 78 + 6x = 2 ( (– 4 x) )

Solve a real-world problem
EXAMPLE 3 Solve a real-world problem 78 + 6x = 2(67 – 4x) Write equation. 78 + 6x = 134 – 8x Distributive property x = 134 Add 8x to each side. 14x = 56 Subtract 78 from each side. x = 4 Divide each side by 14. ANSWER The number of new cars sold will be twice the number of used cars sold in 4 years.

EXAMPLE 3 Solve a real-world problem CHECK You can use a table to check your answer. YEAR 1 2 3 4 Used car sold 67 63 59 55 51 New car sold 78 84 90 96 102

Identify the number of solutions of an equation
EXAMPLE 4 Identify the number of solutions of an equation Solve the equation, if possible. a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5) SOLUTION a. 3x = 3(x + 4) Original equation 3x = 3x + 12 Distributive property The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

EXAMPLE 4 Identify the number of solutions of an equation 3x – 3x = 3x + 12 – 3x Subtract 3x from each side. 0 = 12 Simplify. ANSWER The statement 0 = 12 is not true, so the equation has no solution.

EXAMPLE 1 EXAMPLE 4 Identify the number of solutions of an equation b. 2x + 10 = 2(x + 5) Original equation 2x + 10 = 2x + 10 Distributive property ANSWER Notice that the statement 2x + 10 = 2x + 10 is true for all values of x.So, the equation is an identity, and the solution is all real numbers.

GUIDED PRACTICE for Example 4 8. 9z + 12 = 9(z + 3) SOLUTION
Original equation 9z + 12 = 9z + 27 Distributive property The equation 9z + 12 = 9z + 27 is not true because the number 9z + 12 cannot be equal to 27 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

The statement 12 = 27 is not true, so the equation has no solution.
GUIDED PRACTICE for Example 4 9z – 9z + 12 = 9z – 9z + 27 Subtract 9z from each side. 12 = 27 Simplify. ANSWER The statement 12 = 27 is not true, so the equation has no solution.

GUIDED PRACTICE for Example 4 9. 7w + 1 = 8w + 1 SOLUTION – w + 1 = 1
Subtract 8w from each side. – w = 0 Subtract 1 from each side. ANSWER w = 0

GUIDED PRACTICE for Example 4 10. 3(2a + 2) = 2(3a + 3) SOLUTION
Original equation 6a + 6 = 6a + 6 Distributive property ANSWER The statement 6a + 6 = 6a + 6 is true for all values of a. So, the equation is an identity, and the solution is all real numbers.

Solve the equation. 4 x 8 = x – 6 3. ANSWER -4 ANSWER X = 12 ANSWER 8 ANSWER None 4x + 10 = 2(2x+5) ANSWER All real #’s

Vocabulary – 3.5-36 Proportional It grows at the same rate.
Cross Product Product of the numerator and denominator of proportions Scale Drawing/Model Smaller or larger replica of an actual object Scale or Scale Factor How MUCH bigger/smaller the scale drawing/model is.

Notes – 3.5-3-6 Most important part of setting up Proportions is?
SAME stuff on top AND SAME stuff on bottom! Use a word fraction to help. Two step process to solve proportions: Cross Multiply and Drop!! (NOT DIVIDE!!) Set cross products equal and Solve If two figures are similar – Remember the 5 s’s Same Angles Same Shape Scale Factor Sides are Proportional – Most important

Examples 3.5 Use the cross products property Solve the proportion = 8
6 15 = 8 x 6 15 Write original proportion. 8 15 = x 6 Cross products property 120 = 6x Simplify. 20 = x Divide each side by 6. The solution is 20. Check by substituting 20 for x in the original proportion. ANSWER

Standardized Test Practice
EXAMPLE 2 Standardized Test Practice What is the value of x in the proportion = ? 4 x 8 3 x – C 3 A – 6 B – 3 D 6 SOLUTION 4 x = 8 – 3 Write original proportion. 4(x – 3) = x 8 Cross products property 4x – 12 = 8x Simplify. – 12 = 4x Subtract 4x from each side. – 3 = x Divide each side by 4.

Write and solve a proportion
EXAMPLE 3 Write and solve a proportion Seals Each day, the seals at an aquarium are each fed 8 pounds of food for every 100 pounds of their body weight. A seal at the aquarium weighs 280 pounds. How much food should the seal be fed per day ? SOLUTION STEP 1 Write a proportion involving two ratios that compare the amount of food with the weight of the seal. x 280 = 8 amount of food 100 weight of seal

Write and solve a proportion
EXAMPLE 3 Write and solve a proportion STEP 2 Solve the proportion. 8 100 x 280 = Write proportion. = 100 x Cross products property 2240 = 100x Simplify. 22.4 = x Divide each side by 100. ANSWER A 280 pound seal should be fed 22.4 pounds of food per day.

Solve the proportion. Check your solution.
EXAMPLE 1 GUIDED PRACTICE for Examples 1,2, and 3 Solve the proportion. Check your solution. = 4 a 24 30 1. = 4 a 24 30 Write original proportion. = a 24 Cross products property 120 = 24a Simplify. 5 = a Divide each side by 24. The solution is 5. Check by substituting 5 for a in the original proportion. ANSWER

The value of x is 18. Check by substituting 18 for x in the
EXAMPLE 2 GUIDED PRACTICE for Examples 1,2, and 3 3 x = 2 – 6 2. 3 x = 2 – 6 Write original proportion. 3(x – 6) = 2 x Cross products property 3x – 18 = 2x Distrubutive property 18 = x Subtract 3x from each side. The value of x is 18. Check by substituting 18 for x in the original proportion. ANSWER

EXAMPLE 2 GUIDED PRACTICE for Examples 1,2, and 3 4 m 5 = m – 6 3. m 5
Write original proportion. m 4 = 5(m – 6) Cross products property 4m = 5m – 30 Simplify m = 30 Subtract 5m from each side.

EXAMPLE 4 Use the scale on a map Maps Use a metric ruler and the map of Ohio to estimate the distance between Cleveland and Cincinnati. SOLUTION From the map’s scale, 1 centimeter represents 85 kilometers. On the map, the distance between Cleveland and Cincinnati is about 4.2 centimeters.

Use the scale on a map EXAMPLE 4
Write and solve a proportion to find the distance d between the cities. = 4.2 d 1 centimeters 85 kilometers 1 d = Cross products property d = 357 Simplify. ANSWER The actual distance between Cleveland and Cincinnati is about 357 kilometers.

EXAMPLE 4 Use the scale on a map GUIDED PRACTICE for Example 4 Model ships 6. The ship model kits sold at a hobby store have a scale of 1 ft : 600 ft. A completed model of the Queen Elizabeth II is 1.6 feet long. Estimate the actual length of the Queen Elizabeth II.

The actual length of the Queen Elizabeth II is about 960 feet.
EXAMPLE 4 Use the scale on a map GUIDED PRACTICE for Example 4 SOLUTION Write and solve a proportion to find the length l of the Queen Elizabeth II. 1 600 = 1.6 l 1 . l = Cross products property l = 960 Simplify. ANSWER The actual length of the Queen Elizabeth II is about 960 feet.

Solve the proportion. 5. A tennis ball machine throws 2 balls every 3 seconds. How many balls will the machine throw in 18 seconds? ANSWER 12 balls

Solve the proportion. 3. 6 25 12 n+1 = ANSWER 49 4. y-1 8 7 13 = ANSWER Y ≈ 5.3

Vocabulary – 3.7 Percent Base Whole (of whatever it is!)

Notes – 3.7 Must be a decimal!
DO NOT SOLVE PERCENT PROBLEMS WITH PROPORTIONS!!! Percent Equation Part = Whole * % Percent of Change Remember the NOOOOO ratio! New – Original Original Must be a decimal!

Examples 3.7 Find a percent using the percent equation
What percent of 136 is 51? a = p% b Write percent equation. 51 = p% 136 Substitute 51 for a and 136 for b. 0.375 = p% Divide each side by 136. 37. 5 = p%. Write decimal as percent. ANSWER 51 is 37.5% of 136.

Find a part of a base using the percent equation
EXAMPLE 3 Find a part of a base using the percent equation What number is 15% of 88? a = p% b Write percent equation. = 15% 88 Substitute 15 for p and 88 for b. = Write percent as decimal. = 13.2 Multiply. ANSWER 13. 2 is 15% of 88.

Find a base using the percent equation
EXAMPLE 4 Find a base using the percent equation 20 is 12.5% of what number? a = p% b Write percent equation. 20 = 12.5% b Substitute 20 for a and 12.5 for p. 20 = 0.125% b Write percent as decimal. 160 = b Divide each side by ANSWER 20 is 12.5% of 160.

Find a part of a base using the percent equation
EXAMPLE 3 Find a part of a base using the percent equation What number is 15% of 88? a = p% b Write percent equation. = 15% 88 Substitute 15 for p and 88 for b. = Write percent as decimal. = 13.2 Multiply. ANSWER 13. 2 is 15% of 88.

GUIDED PRACTICE for Examples 2 and 3 3. What percent of 56 is 49? a
= p% b Write percent equation. 49 = p% 56 Substitute 49 for a and 56 for b. 0.875 = p% Divide each side by 56. 87. 5 = p%. Write decimal as percent. ANSWER 49 is 87.5% of 56.

Find a percent using the percent equation for Examples 2 and 3
GUIDED PRACTICE Find a percent using the percent equation for Examples 2 and 3 CHECK Substitute for p% in the original equation. 49 = p% 56 Write original equation. 49 = Substitute for p%. 49 = 49  Multiply. Solution checks.

GUIDED PRACTICE for Examples 2 and 3 4. What percent of 55 is 11? a
= p% b Write percent equation. 11 = p% 55 Substitute 11 for a and 55 for b. 0.2 = p% Divide each side by 55. 20 = p%. Write decimal as percent. ANSWER 11 is 20% of 55.

Find a percent using the percent equation for Examples 2 and 3
GUIDED PRACTICE Find a percent using the percent equation for Examples 2 and 3 CHECK Substitute 0.2 for p% in the original equation. 11 = p% 55 Write original equation. 11 = Substitute 0.2 for p%. 11 = 11  Multiply. Solution checks.

Find a part of a base using the percent equation Percent of Change
EXAMPLE 3 GUIDED PRACTICE Find a part of a base using the percent equation Percent of Change Find the percent of change for each variation Original Price = $140 and the New price = $189 % change = (N – O) / O Write percent change equation. = (189 – 140) / 140 Substitute 189 for New and 140 for Original. = 49 / 140 Simplify = 0.35 = 35% Convert to a percent. ANSWER $140 to $189 is a 35% increase

Find a part of a base using the percent equation Percent of Change
EXAMPLE 3 GUIDED PRACTICE Find a part of a base using the percent equation Percent of Change Find the percent of change for each variation Original Price = $70 and the New price = $59.50 % change = (N – O) / O Write percent change equation. = (59.50 – 70) / 70 Substitute for New and 70 for Original. = / 70 Simplify = = -15% Convert to a percent. ANSWER $140 to $189 is a 15% DECREASE in price.

Find a part of a base using the percent equation for Examples 2 and 3
GUIDED PRACTICE Find a part of a base using the percent equation for Examples 2 and 3 6. What number is 140% of 50? a = p% b Write percent equation. = 140% 50 Substitute 140 for p and 50 for b. = Write percent as decimal. = 70 Multiply. ANSWER 70 is 140% of 50.

83 EXAMPLE 5 Solve a real-world percent problem Type of Pasta Students
Survey Type of Pasta Students Spaghetti 83 Lasagna 40 Macaroni and cheese 33 Fettucine alfredo 22 Baked ziti 16 Pasta primavera 15 Other 11 A survey asked 220 students to name their favorite pasta dish. Find the percent of students who chose the given pasta dish. a. macaroni and cheese b. lasagna

Solve a real-world percent problem
EXAMPLE 5 Solve a real-world percent problem SOLUTION The survey results show that 33 of the 220 students chose macaroni and cheese. a. a = p% b Write percent equation. 33 = p% 220 Substitute 33 for a and 220 for b. 0.15 = p% Divide each side by 220. 15% = p% Write decimal as percent.

EXAMPLE 5 Solve a real-world percent problem ANSWER 15% of the students chose macaroni and cheese as their favorite dish.

Solve the proportion. = 6 n-1 -2 n+1 = ANSWER -1/2 4. y-1 y+2 -1 8 = ANSWER 2/3

1. Write an equation for “3 less than twice a is 24.” ANSWER 2a - 3 = 24 15 percent of what number is 78? ANSWER 520

A rectangular serving tray is 26 inches long and has a Serving area of 468 in.2 What is the width? 18 inches wide ANSWER Get “a” by itself in the following equation. 3(a + 1) = 9x ANSWER a = 3x – 1 or a = (9x – 3) / 3

Vocabulary – 3.8 Literal Equation
Equation (or formula) where the coefficients and constants have been replaced with letters.

Notes – 3.8 – Rewriting Eqns.
Difficult section! Remember: The goal of solving EVERY alg. Eqn?? Remember: What process did we use to get the variable by itself? Simplify SSADMEP Anything I do to one side, I must do to the other! Best to learn this with examples!

Examples 3.8 Solve a literal equation
Solve ax +b = c for x. Then use the solution to solve 2x + 5 = 11. SOLUTION STEP 1 Solve ax + b = c for x. ax + b = c Write original equation. ax = c – b Subtract b from each side. x c – b a = Assume a Divide each side by a. =

Solve a literal equation
EXAMPLE 1 Solve a literal equation STEP 2 Use the solution to solve 2x + 5 = 11. x = c – b a Solution of literal equation. 11 – 5 2 = Substitute 2 for a, 5 for b, and 11 for c. = 3 Simplify. The solution of 2x + 5 = 11 is 3. ANSWER

GUIDED PRACTICE for Example 1
Solve the literal equation for x . Then use the solution to solve the specific equation 1. Solve a – bx = c for x. SOLUTION STEP 1 Solve a – bx = c for x. a – bx = c Write original equation. – bx = c – a Subtract a from each side. x a – c b = Assume b Divide each side by – 1. =

Use the solution to solve 12 – 5x = –3.
GUIDED PRACTICE for Example 1 STEP 2 Use the solution to solve 12 – 5x = –3. x = a – c b Solution of literal equation. 12 – (–3) 5 = Substitute a for 12, –3 for c, and 5 for b. = 3 Simplify. The solution of 12 – 5x = –3 is 3. ANSWER

GUIDED PRACTICE for Example 1 2. Solve a x = bx + c for x. SOLUTION
STEP 1 Solve a x = bx + c for x. a x = bx + c Write original equation. a x – bx = c Subtract bx from each side. c x a – b = Assume a Divide each = side by a – b.

Use the solution to solve 11x = 6x + 20.
GUIDED PRACTICE for Example 1 STEP 2 Use the solution to solve 11x = 6x + 20. x = c a – b Solution of literal equation. 20 11 – 6 = Substitute a for 11, 20 for c, and 6 for b. = 4 Simplify. The solution of 11x = 6x is 4. ANSWER

Write 3x + 2y = 8 so that y is a function of x.
EXAMPLE 2 Rewrite an equation Write 3x + 2y = 8 so that y is a function of x. 3x + 2y = 8 Write original equation. 2y = 8 – 3x Subtract 3x from each side. 3 2 y = 4 – x Divide each side by 2.

Solve and use a geometric formula
EXAMPLE 3 Solve and use a geometric formula The area A of a triangle is given by the formula A = bh where b is the base and h is the height. 1 2 Solve the formula for the height h. a. Use the rewritten formula to find the height of the triangle shown, which has an area of 64.4 square meters. b. SOLUTION a. bh 1 2 A = Write original formula. 2A bh = Multiply each side by 2.

Solve and use a geometric formula
EXAMPLE 3 Solve and use a geometric formula 2A b h = Divide each side by b. Substitute 64.4 for A and 14 for b in the rewritten formula. b. h 2A b = Write rewritten formula. = 2(64.4) 14 Substitute 64.4 for A and 14 for b. = 9.2 Simplify. ANSWER The height of the triangle is 9.2 meters.

3 . Write 5x + 4y = 20 so that y is a function of x.
GUIDED PRACTICE for Examples 2 and 3 Write 5x + 4y = 20 so that y is a function of x. 5x + 4y = 20 Write original equation. 4y = 20 – 5x Subtract 5x from each side. 5 4 y = 5 – x Divide each side by 4.

a. Solve the formula for the width w. 4 .
GUIDED PRACTICE for Examples 2 and 3 The perimeter P of a rectangle is given by the formula P = 2l + 2w where l is the length and w is the width. a. Solve the formula for the width w. 4 . SOLUTION a p = 2l + 2w Write original equation. p – 2l = 2w Subtract 2l from each side. p – 2l 2 = w Divide each side by 2.

Substitute 19.2 for P and 7.2 for l in the rewritten formula b .
GUIDED PRACTICE for Examples 2 and 3 Substitute 19.2 for P and 7.2 for l in the rewritten formula b . w = p –2l 2 Write original equation. 19.2 – 2 (7.2) 2 = Substitute 19.2 for P and 7.2 for l. = 2.4 Simplify. The width of the rectangle is 2.4 feet

EXAMPLE 4 Solve a multi-step problem Temperature You are visiting Toronto, Canada, over the weekend. A website gives the forecast shown. Find the low temperatures for Saturday and Sunday in degrees Fahrenheit. Use the formula C = (F – 32) where C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit. 5 9

EXAMPLE 4 Solve a multi-step problem SOLUTION STEP 1 Rewrite the formula. In the problem,degrees Celsius are given and degrees Fahrenheit need to be calculated. The calculations will be easier if the formula is written so that F is a function of C. (F – 32) 5 9 C = Write original formula. Multiply each side by , the reciprocal of . 9 5 (F – 32) 9 5 C = F – 32 C 9 5 = Simplify. 9 5 C + 32 = F Add 32 to each side.

EXAMPLE 4 Solve a multi-step problem ANSWER 9 5 = The rewritten formula is F C + 32.

EXAMPLE 4 Solve a multi-step problem STEP 2 Find the low temperatures for Saturday and Sunday in degrees Fahrenheit. Saturday (low of 14°C) Sunday (low of 10°C) C + 32 9 5 F = F C + 32 9 5 = = (14)+ 32 9 5 = (10)+ 32 9 5 = = = 57.2 = 50 The low for Saturday is 57.2°F. ANSWER The low for Sunday is 50°F. ANSWER

Warm-Up – 6.2 x – 5 > – 3.5. Graph your solution. Solve ANSWER

Warm-Up – 6.2 5. p – 9.2 < – 5 x < 4.2 ANSWER

Warm-Up – 6.3 2. ≤ – 2 m 8 ANSWER m < – 16 –3x > 24. Solve
≤ – 2 m 8 ANSWER m < – 16 –3x > 24. Solve ANSWER x < – 8

Solve an inequality using division
EXAMPLE 3 Solve an inequality using division –3x > 24. Solve –3x > 24. Write original inequality. –3x –3 < 24 Divide each side by –3. Reverse inequality symbol. x < – 8 Simplify.

GUIDED PRACTICE for Examples 2 and 3 x 4. > 12 – 4 SOLUTION x
> 12 x – 4 SOLUTION x – 4 > 12 Write original inequality. – 4 12 < – 4 x Multiply each side by – 4. x <– 48 Simplify. ANSWER The solutions are all real numbers greater than are equal to – 48. Check by substituting a number greater than – 48 in the original inequality.

Vocabulary – 6.1-6.3 Equivalent Inequalities
Inequalities that remain true after operations are performed.

Notes – 6.1-6.3 – Solving Inequalities
What’s the goal? How do we get the variable by itself? What’s the beauty of math? Solving Inequalities is EXACTLY the same as solving equations, with TWO exceptions!! There is usually more than one solution. WHEN YOU MULTIPLY OR DIVIDE AN INQUALITY BY A NEGATIVE, REVERSE THE DIRECTION OF THE INEQUALITY! Graphing Inequalities Open dot means < or > and closed means ≤ or ≥ ALWAYS check answer. Pick numbers higher, lower, and equal to the inequality

Examples 6.2 x Solve < . Graph your solution. 5 4 x < 5. 4 x
Write original inequality. x 4 < 5 Multiply each side by 4. x < 20 Simplify. ANSWER The solutions are all real numbers less than 20. Check by substituting a number less than 20 in the original inequality.

Examples 6.2 Solve the inequality. Graph your solution. 1. > 8 x 3
> 8 x 3 SOLUTION x 3 > 8. Write original inequality. x 3 > 8 Multiply each side by 8. x > 24 Simplify. ANSWER The solutions are all real numbers are greater than 24. Check by substituting a number greater than 24 in the original inequality.

GUIDED PRACTICE for Examples 2 and 3 6. 5v ≥ 45 SOLUTION 5v ≥ 45 5v 5
Write original inequality. 5v 5 ≥ 45 Divide both the side by 5. v ≥ 9 Simplify. ANSWER The solutions are all real numbers greater than are equal to 9. Check by substituting a number greater than 9 in the original inequality.

GUIDED PRACTICE for Examples 2 and 3 7. – 6n < 24 SOLUTION
Write original inequality. > – 6n 6 24 Divide both the side by 6. n > – 4 Simplify. ANSWER The solutions are all real numbers greater than are equal to – 4. Check by substituting a number greater than – 4 in the original inequality.

EXAMPLE 4 Standardized Test Practice A student pilot plans to spend 80 hours on flight training to earn a private license. The student has saved $6000 for training. Which inequality can you use to find the possible hourly rates r that the student can afford to pay for training? 80r A < – B C D SOLUTION The total cost of training can be at most the amount of money that the student has saved. Write a verbal model for the situation. Then write an inequality.

EXAMPLE 4 Standardized Test Practice 80 r < – 6000 ANSWER The correct answer is B. A D C B

EXAMPLE 5 Solve a real-world problem In Example 4, what are the possible hourly rates that the student can afford to pay for training? PILOTING

Solve a real-world problem
EXAMPLE 5 Solve a real-world problem SOLUTION 80 r ≤ 6000 Write inequality. 80r 80 ≤ 6000 Divide each side by 80. r ≤ 75 Simplify. ANSWER The student can afford to pay at most $75 per hour for training.

Examples 6.3

Solve a two-step inequality
EXAMPLE 1 Solve a two-step inequality 3x – 7 < 8. Graph your solution. Solve 3x – 7 < 8 Write original inequality. 3x < 15 Add 7 to each side. x < 5 Divide each side by 3. ANSWER The solutions are all real numbers less than 5. Check by substituting a number less than 5 in the original inequality.

EXAMPLE 1 Solve a two-step inequality CHECK 3x –7 < 8 Write original inequality. ? 3(0) – 7 < 8 Substitute 0 for x. –7 < 8 Solution checks.

Solve a multi-step inequality
EXAMPLE 2 Solve a multi-step inequality Solve – 0.6(x – 5) < 15 – –0.6(x – 5) < 15 – Write original inequality. –0.6x < 15 – Distributive property – 0.6x < 12 – Subtract 3 from each side. – x > –20 Divide each side by Reverse inequality symbol. –

Solve the inequality. Graph your solution.
GUIDED PRACTICE for Examples 1 and 2 Solve the inequality. Graph your solution. 2x – 1. < – 2x – < – Write original inequality. 2x 28 < – Add 5 to both side. x 14 < – Divide each side by 2. ANSWER The solutions are all real numbers less than equal to 14. Check by substituting a number less than 14.

GUIDED PRACTICE for Examples 1 and 2 Solve the inequality. Graph your solution. – 6y –16. 2. < – – 6y –16 < – Write original inequality. –6 –16 –5 < – y Divide the equation by 6. y > – Simplify. ANSWER The solutions are all real numbers less than or equal to 3.5. Check by substituting a number less than 3.5.

Solve a multi-step inequality
EXAMPLE 3 Solve a multi-step inequality 6x – 7 > 2x+17. Graph your solution. Solve 6x – 7 > 2x+17 Write original inequality. 6x > 2x+24 Add 7 to each side. 4x > 24 Subtract 2x from each side. x > 6 Divide each side by 4. ANSWER The solutions are all real numbers greater than 6.

Identify the number of solutions of an inequality
EXAMPLE 4 Identify the number of solutions of an inequality Solve the inequality, if possible. a x + 5 < 7(2x – 3) SOLUTION 14x + 5 < 7(2x – 3) a. Write original inequality. 14x + 5 < 14x – 21 Distributive property 5 < – 21 Subtract 14x from each side. ANSWER There are no solutions because 5 < – 21 is false.

Identify the number of solutions of an inequality
EXAMPLE 4 Identify the number of solutions of an inequality b x – 1 > 6(2x – 1) 12x – 1 > 6(2x – 1) Write original inequality. 12x – 1 > 12x – 6 Distributive property – 1 > – 6 Subtract 12x from each side. ANSWER All real numbers are solutions because – 1 > – 6 is true.

Solve the inequality,if possible. Graph your solution.
GUIDED PRACTICE for Examples 3 and 4 Solve the inequality,if possible. Graph your solution. 5x – x – 4. 4. < – SOLUTION 5x – x – 4. < – Write original inequality. 5x 3x + 8. < – Add 12 to each side. 2x 8 < – Subtract 3x from each side. x 4 < – ANSWER The solutions are all real numbers lesser than or equal to 4.

Identify the number of solutions of an inequality for Examples 3 and 4
GUIDED PRACTICE Identify the number of solutions of an inequality for Examples 3 and 4 Solve the inequality, if possible. Graph your solution. (m + 5) < 5m + 17 SOLUTION 5(m + 5) < 5m + 17 Write original inequality. 5m < 5m + 17 Distributive property 25 < 17 Subtract 5m from each side. ANSWER There are no solutions because 25 < 17 is false.

Solve the inequalities and graph them. 1. 8 > -2x + 10 X > 1 All real numbers greater than 1. ANSWER 6 < -5x – 4 2. X <= -2 All real numbers less than or equal to -2 ANSWER

Solve the inequality. 3. You estimate you can read at least 8 history text pages per day. What are the possible numbers of day it will take you to read at most 118 pages? ANSWER at most 15 days

Vocabulary – 6.4 Compound Inequality
Two separate inequalities joined by a conjunction (“and” or “or”)

Notes – 6.4– Solving Compound Ineq.
Solving more than one equality at the same time and putting them on one number line. SAME RULES and GOAL! Remember the rule about negatives!!! If I multiply or divide an inequality by a negative, the direction of the inequality must change. Graphing Or’s go “out” And’s go “in” Sometimes we write “and” inequalities in a shortcut If x is greater than 12 and less than 15 12 < x < 15

Examples 6.4

EXAMPLE 1 Write and graph compound inequalities Translate the verbal phrase into an inequality. Then graph the inequality. a. All real numbers that are greater than – 2 and less than 3. Inequality: – 2 < x < 3 Graph: b. All real numbers that are less than 0 or greater than or equal to 2. Inequality: x < 0 or x ≥ 2 Graph:

EXAMPLE 2 Write and graph a real-world compound inequality CAMERA CARS A crane sits on top of a camera car and faces toward the front. The crane’s maximum height and minimum height above the ground are shown. Write and graph a compound inequality that describes the possible heights of the crane.

EXAMPLE 2 Write and graph a real-world compound inequality SOLUTION Let h represent the height (in feet) of the crane. All possible heights are greater than or equal to 4 feet and less than or equal to 18 feet. So, the inequality is 4 ≤ h ≤ 18.

Solve a compound inequality with and
EXAMPLE 3 Solve a compound inequality with and Solve 2 < x + 5 < 9. Graph your solution. SOLUTION Separate the compound inequality into two inequalities. Then solve each inequality separately. 2 < x + 5 and x + 5 < 9 Write two inequalities. 2 – 5 < x + 5 – 5 and x + 5 – 5 < 9 – 5 Subtract 5 from each side. 23 < x and x < 4 Simplify. The compound inequality can be written as – 3 < x < 4.

Solve a compound inequality with and for Example 2 and 3
GUIDED PRACTICE Solve a compound inequality with and for Example 2 and 3 solve the inequality. Graph your solution. –7 < x – 5 < 4 4. SOLUTION Separate the compound inequality into two inequalities. Then solve each inequality separately. –7 < x – 5 and x – 5 < 4 Write two inequalities. –7 + 5 < x –5 + 5 and x – < 4 + 5 Add 5 to each side. –2 < x and x < 9 Simplify. The compound inequality can be written as – 2 < x < 9.

The solutions are all real numbers greater than –2 & less than 9.
EXAMPLE 3 GUIDED PRACTICE for Example 2 and 3 ANSWER The solutions are all real numbers greater than –2 & less than 9. – – – Graph:

GUIDED PRACTICE for Example 2 and 3 Solve the inequality. Graph your solution. 10 ≤ 2y + 4 ≤ 24 5. SOLUTION Separate the compound inequality into two inequalities. Then solve each inequality separately. 10 ≤ 2y + 4 and 2y + 4 ≤ 24 Write two inequalities. 10 – 4 ≤ 2y + 4 –4 and 2y + 4 – 4 ≤ 24 – 4 subtract 4 from each side. 6 ≤ 2y and 2y ≤ 20 3 ≤ y and y ≤ 10 Simplify. The compound inequality can be written as 3 ≤ y ≤ 10.

GUIDED PRACTICE Solve a compound inequality with and for Example 2 and 3 ANSWER The solutions are all real numbers greater than 3 & less than 10. Graph:

GUIDED PRACTICE Solve a compound inequality with and for Example 2 and 3 Solve the inequality. Graph your solution. –7< –z – 1 < 3 6. SOLUTION Separate the compound inequality into two inequalities. Then solve each inequality separately. –7 < –z – 1 and –z – 1 < 3 Write two inequalities. –7 + 1< –z – 1 + 1 and –z – 1+1 < 3 + 1 Add 1 to each side. 6 < z and z > – 4 Simplify. The compound inequality can be written as – 4 < z < 6.

EXAMPLE 4 Solve a compound inequality with and Solve – 5 ≤ – x – 3 ≤ 2. Graph your solution. – 5 ≤ – x – 3 ≤ 2 Write original inequality. – ≤ – x – ≤ 2 + 3 Add 3 to each expression. – 2 ≤ – x ≤ 5 Simplify. – 1(– 2) – 1(– x) – 1(5) > – Multiply each expression by – 1and reverse both inequality symbols. 2 x –5 > – > – Simplify.

EXAMPLE 4 Solve a compound inequality with and – 5 ≤ x ≤ 2 Rewrite in the form a ≤ x ≤ b. ANSWER The solutions are all real numbers greater than or equal to – 5 and less than or equal to 2.

GUIDED PRACTICE for Examples 4 and 5 ANSWER The solutions are all real numbers greater than or equal to – 6 and less than 7. –7 –6 –5 –4 –3 –2 –

Warm-Up – 6.5

GUIDED PRACTICE for Examples 4 and 5 Solve the inequality. Graph your solution. 7. – 14 < x – 8 < – 1 SOLUTION – 6 < x < 7 The solutions are all real numbers greater than or equal to – 6 and less than 7. –7 –6 –5 –4 –3 –2 –

EXAMPLE 5 Solve a compound inequality with or Solve 2x + 3 < 9 or 3x – 6 > 12. Graph your solution. SOLUTION Solve the two inequalities separately. x < 3 or x > 6 ANSWER The solutions are all real numbers less than 3 or greater than 6.

Warm-Up – 6.5 Lesson 6.5, For use with pages 390-395 1.
For a = –12, find, –a and |a|. ANSWER 12, 12 2. Evaluate |x| – 2 when x = –3. ANSWER 1

Warm-Up – 6.5 Lesson 6.5, For use with pages 390-395 3.
The change in evaluation as a diver explored a reed was –0.5 feet, 1.5 feet, –2.5 feet, and 2.25 feet. Which change in evaluation had the greatest absolute value? ANSWER –2.5 ft

Vocabulary – 6.5 Absolute Value Distance from zero Always positive
Absolute Deviation Absolute value of the difference of two numbers Absolute value Equation Equation with Absolute Value signs Can have 0, 1, or 2 solutions

Notes – 6.5 –Solving Abs. Value Eqns
Solving Abs value eqns is a “two for the price of one” deal. |any expression| = solution means two things |any expression| = + solution AND |any expression| = - solution Treat abs value signs like parenthesis in SADMEP Get abs value expression by itself and then split it into two equations. Distributive property does NOT work over abs value symbols!!! USUALLY two solutions!!

Examples 6.5

EXAMPLE 1 Solve an absolute value equation Solve x = 7. SOLUTION The distance between x and 0 is 7. So, x = 7 or x = –7. ANSWER The solutions are 7 and –7.

EXAMPLE 1 GUIDED PRACTICE for Example 1 Solve (a) x = 3 and (b) x = 15 ANSWER The solutions are 3 and –3. ANSWER The solutions are 15 and –15.

Solve an absolute value equation
EXAMPLE 2 Solve an absolute value equation x – 3 = 8 Solve SOLUTION Rewrite the absolute value equation as two equations. Then solve each equation separately. x – 3 = 8 Write original equation. x – 3 = or x – 3 = –8 Rewrite as two equations. x = 11 or x = –5 Add 3 to each side. ANSWER The solutions are 11 and –5. Check your solutions.

Rewrite an absolute value equation
EXAMPLE 3 Rewrite an absolute value equation 3 2x – 7 – 5 = 4. Solve SOLUTION First, rewrite the equation in the form ax + b = c. 3 2x – 7 – 5 = 4 Write original equation. 3 2x – 7 = 9 Add 5 to each side. 2x – 7 = 3 Divide each side by 3.

Rewrite an absolute value equation
EXAMPLE 3 Rewrite an absolute value equation Next, solve the absolute value equation. 2x – 7 = 3 Write absolute value equation. 2x – 7 = or 2x – 7 = –3 Rewrite as two equations. 2x = 10 or x = 4 Add 7 to each side. x = or x = 2 Divide each side by 2. ANSWER The solutions are 5 and 2.

GUIDED PRACTICE for Examples 2 and 3 sSolve the equation. 2. r – 7 = 9
SOLUTION Rewrite the absolute value equation as two equations. Then solve each equation separately. r – 7 = 9 Write original equation. r– 7 = or r – 7 = –9 Rewrite as two equations. r = 16 or r = –2 Add 7 to each side. ANSWER The solutions are 16 and –2.

First rewrite the equation in the form ax + b = c
GUIDED PRACTICE for Examples 2 and 3 Solve the equation. 2 s = 18.9 3. SOLUTION First rewrite the equation in the form ax + b = c 2 s = 18.9 Write original equation. 2 s = 14.8 Subtract 4.1 from each side. s = 7.4 Divide each side by 2.

GUIDED PRACTICE for Examples 2 and 3 ANSWER Solue the absolute value equation. s = 7.4 or s = – 7.4 The solution are 7.4 & –7.4

First, rewrite the equation in the form ax + b = c.
GUIDED PRACTICE for Examples 2 and 3 Solve the equation. 4 t + 9 – 5 = 19. 4. SOLUTION First, rewrite the equation in the form ax + b = c. 4 t + 9 – 5 = 19 Write original equation. 4 t + 9 = 24 Add 5 to each side. t + 9 = 6 Divide each side by 4.

Solve the absolute value equation.
GUIDED PRACTICE for Examples 2 and 3 Solve the absolute value equation. t + 9 = 6 Write absolute value equation. t + 9 = 6 or t + 9 = –6 Rewrite as two equations. t = –3 or t = –15 addition & subtraction to each side

Warm-Up – 6.6

1. Solve |x – 6| = 4. ANSWER 2, 10 2. Solve |x + 5| – 8 = 2. ANSWER –15, 5

3. A frame will hold photographs that are 5 inches by 8 inches with an absolute derivation of 0.25 inch for length and width. What are the minimum and maximum dimensions for photos? ANSWER min: 4.75 in. by 7.75 in.; max: 5.25 in. by 8.25 in.

Warm-Up – 6.6 Lesson 6.6, For use with pages 398-403 1. Graph |x| = 4.

Vocabulary – 6.6 Absolute Value Inequality
Inequality with Absolute value symbols

Notes – 6.6 – Solving Abs. Value Inequalities.
Solving and Graphing Inequalities are still part of our “Two for One” sale!! You will still have to solve two problems with a conjunction! Because we have multiple inequalities (<, >, <=, >=) and multiple conjunctions (and, or), we need a way to figure out which conjunction to use. Remember this Greater than = greatOR Less than = Less thAND

Examples 6.6

EXAMPLE 1 Solve absolute value inequalities Solve the inequality. Graph your solution. a. – 6 x > SOLUTION a. The distance between x and 0 is greater than or equal to 6. So, x ≤ – 6 or x ≥ 6. ANSWER The solutions are all real numbers less than or equal to – 6 or greater than or equal to 6.

EXAMPLE 1 Solve absolute value inequalities b. < x 0.5 – SOLUTION The distance between x and 0 is less than or equal to 0.5.So, to – 0.5 ≤ x ≤ 0.5. ANSWER The solutions are all real numbers greater than or equal to – 0.5 and less than or equal to 0.5.

. EXAMPLE 4 GUIDED PRACTICE Find a base using the percent equation
for Example 1 Solve the inequality. Graph your solution. 1. x 8 < SOLUTION a. The distance between x and 0 is less equal to 8. So, – 8 ≤ x ≤ 8. –9 –8 –7 – 6 – 5 – 4 – 3 – 2 – . ANSWER The solutions are all real numbers greater than or equal to – 8 & less than or equal to 8.

Find a base using the percent equation for Example 1
GUIDED PRACTICE Find a base using the percent equation for Example 1 3. v > 2 3 SOLUTION a. The distance between x and 0 is less or greater than so, 2 3 v < – or v > 3 – 2 1 ANSWER The solutions are all real numbers greater than and less than 2 3 –

Solve an absolute value inequality
EXAMPLE 2 Solve an absolute value inequality Solve x – 5 ≥ 7. Graph your solution. x – – > Write original inequality. x – 5 7 < – or x – 5 7 – > Rewrite as compound inequality. x –2 < – or – > x 12 Add 5 to each side. ANSWER The solutions are all real numbers less than or equal to – 2 or greater than or equal to 12. Check several solutions in the original inequality.

Solve an absolute value inequality
EXAMPLE 3 Solve an absolute value inequality Solve – 4x – < 9. Graph your solution. – 4x – < 9 Write original inequality. – 4x – 5 < 6 Subtract 3 from each side. –6 <–4x – 5 < 6 Rewrite as compound inequality. –1 <–4x < –11 Add 5 to each expression. 0.25 > x > –2.75 Divide each expression by – 4. Reverse inequality symbol. –2.75 < x < 0.25 Rewrite in the form a < x < b.

EXAMPLE 3 Solve an absolute value inequality ANSWER The solutions are all real numbers greater than –2.75 and less than 0.25.

GUIDED PRACTICE for Examples 2 and 3 Solve the inequality.
4. SOLUTION x + 3 > 8 Write original inequality. x + 3 < 8 or x + 3 > 8 Rewrite as compound inequality. x < – 11 or x > 5 Add – 3 to each side. ANSWER The solutions are all real numbers less than – 11 or greater than to 5 .

GUIDED PRACTICE for Examples 2 and 3 2w – 1 < 11 5. SOLUTION
Write original inequality. 2w – 1 < – 11 or 2w – 1 > 11 Rewrite as compound inequality. 2w < – 10 or 2w > 12 Add 1 to each side. w < – 5 or w > 6 Divide by 2 to each side ANSWER – 5 < w <– 6

GUIDED PRACTICE for Examples 2 and 3 6. 3 5m – 6 – 8 13 < –
SOLUTION 3 5m – 6 – < – Write original inequality. 3 5m – < – Add 8 to each side. |5m – 6| 7 < – Divide by each side by 3. – m < – Rewrite as compound inequality.

GUIDED PRACTICE for Examples 2 and 3 – 1 5m 13 < – – 0.2 m 2.6 <
Add 6 to each side. – m < – Simplify. ANSWER The solutions are all real numbers greater than or equal to – 0.2 and less than or equal to 2.6 .

EXAMPLE 4 Graph a linear inequality in one variables Graph the inequality y > – 3. SOLUTION STEP 1 Graph the equation y = – 3. The inequality is >, so use a solid line. STEP 2 Test (2, 0) in y > – 3. You substitute only the y-coordinate, because the inequality does not have the variable x. 0 >–3

EXAMPLE 4 Graph a linear inequality in one variables STEP 3 Shade the half-plane that contains (2, 0), because (2, 0) is a solution of the inequality.

EXAMPLE 5 Graph a linear inequality in one variables Graph the inequality x < – 1. SOLUTION STEP 1 Graph the equation x = – 1. The inequality is <, so use a dashed line. STEP 2 Test (3, 0) in x < – 1. You substitute only the x-coordinate, because the inequality does not have the variable y. 3 <–1

EXAMPLE 5 Graph a linear inequality in one variables STEP 3 Shade the half-plane that does not contains 3, 0), because (3, 0) is not a solution of the inequality.

GUIDED PRACTICE for Examples 4 and 5 5. Graph the inequality y > 1. SOLUTION STEP 1 Graph the equation y = 1. The inequality is <, so use a dashed line. STEP 2 Test (1, 0) in y < 1. You substitute only the y-coordinate, because the inequality does not have the variable x. 1> 1

GUIDED PRACTICE for Examples 4 and 5 STEP 3 Shade the half-plane that contains (1, 0), because (1, 0) is a solution of the inequality.

GUIDED PRACTICE for Examples 4 and 5 6. Graph the inequality y < 3. SOLUTION STEP 1 Graph the equation y = 3. The inequality is <, so use a dashed line. STEP 2 Test (3, 0) in y < 3. You substitute only the y-coordinate, because the inequality does not have the variable x. 3> 3

GUIDED PRACTICE for Examples 4 and 5 STEP 3 Shade the half-plane that contains (3, 0), because (3, 0) is a solution of the inequality.

GUIDED PRACTICE for Examples 4 and 5 7. Graph the inequality x < – 2. SOLUTION STEP 1 Graph the equation y = –2. The inequality is <, so use a dashed line. STEP 2 Test (2, 0) in y < – 2 . You substitute only the y-coordinate, because the inequality does not have the variable x. 2 <–2

GUIDED PRACTICE for Examples 4 and 5 STEP 3 Shade the half-plane that does not contains (2, 0), because (2, 0) is not a solution of the inequality.

Review Slides

Daily Homework Quiz For use after Lesson 3.2 Solve the equation. – 6 = a 4 – ANSWER 32 – r – 12 = 6 ANSWER 3 = 7y 2y + – ANSWER 4 –

Daily Homework Quiz For use after Lesson 3.2 The output of a function is 9 less than 3 times the input. Write an equation for the function and then find the input when the output is 6. 4. – ANSWER y = 3x 9; 1 – A bank charges $5.00 per month plus $.30 per check for a standard checking account. Find the number of checks Justine wrote if she paid $8.30 in fees last month. 5. ANSWER 11 checks

Daily Homework Quiz For use after Lesson 3.3 Solve the equation. g – 2 + g = 16 ANSWER 2 b + 2(b – 4) = 47 11 ANSWER 3. – 6 + 4(2c + 1) = –34 – 4 ANSWER

Daily Homework Quiz For use after Lesson 3.3 (x – 6) = 12 2 3 24 ANSWER 5. Joe drove 405 miles in 7 hours. He drove at a rate of 55 miles per hour during the first part of the trip and 60 miles per hour during the second part. How many hours did he drive at a rate of 55 miles per hour? 3h ANSWER

Daily Homework Quiz For use after Lesson 3.4 Solve the equation, if possible. 3(3x + 6) = 9(x + 2) 1. ANSWER The equation is an identity. 7(h – 4) = 2h + 17 2. ANSWER 9 8 – 2w = 6w – 8 3. ANSWER 2

Daily Homework Quiz For use after Lesson 3.4 4g + 3 = 2(2g + 3) 4. ANSWER The equation has no solution. Bryson is looking for a repair service for general household maintenance. One service charges $75 to join the service and $30 per hours. Another service charge $45 per hour. After how many hours of service is the total cost for the two services the same? 5. ANSWER 5 h

Daily Homework Quiz For use after Lesson 3.5 1. A chocolate chip cookie recipe calls for cups of flour and cup of brown sugar. Find the ratio of brown sugar to flour. 3 4 1 2 ANSWER 1 3 Solve the proportion. 2. a 7 9 21 = ANSWER 3

Daily Homework Quiz For use after Lesson 3.5 32 28 3. = m 14 16 ANSWER 4. A printer can print 12 color pages in 3 minutes. How many color pages can the printer print in 9 minutes? Write and solve a proportion to find the answer. ANSWER 12 3 = x 9 ; 36 color pages

Daily Homework Quiz For use after Lesson 3.6 10 35 = y 42 1. ANSWER 12 13 h = 26 16 2. ANSWER 8 5r 6 = 15 2 3. ANSWER 9

Daily Homework Quiz For use after Lesson 3.6 9 d + 3 6 17 = 4. ANSWER 22.5 A figurine of a ballerina is based on a scale of 0.5 in.:4 in. If the real ballerina used as a model for the figurine is 68 inches tall, what is the height of the figurine? 5. ANSWER 8.5 in

Daily Homework Quiz For use after Lesson 3.7 Solve the percent problem 1. What percent of 50 is 1 ANSWER 2% What percent of 128 is 48? 37.5% ANSWER What number is 16% of 45? ANSWER 7.2

Daily Homework Quiz For use after Lesson 3.3 is 12.5% of what number? ANSWER 96 Leonard has read 1001 pages out of of Tolstoy’s War and peace. What percent of the novel has he read? 5. ANSWER 68.75%

Daily Homework Quiz For use after Lesson 3.6 9 d + 3 6 17 = 4. X 2X - 3 10 17 = 4. ANSWER 22.5 ANSWER 10 X-1 3 2X+1 9 = 4. 14 12 X+11 18 = 4. ANSWER 10 ANSWER 4

Daily Homework Quiz For use after Lesson 3.8 Put the following in function form. 5X + 4Y = 10 1. ANSWER Y = 5/2 – (5/4)x 12 = 9X + 3Y 2. ANSWER Y = 4 – 3X 2 + 6y = 3x + 4 3. ANSWER y = ½ x + 1/3

Daily Homework Quiz For use after Lesson 3.8 Put the following in function form. 30 = 9x – 5y. 1. ANSWER Y = 9/5x - 6 Solve for w if V = l*w*h 2. ANSWER W = V/(lh) Solve for h in the following formula: S = 2B + Ph 3. ANSWER H = (S – 2B)/P

Warm-Up – X.X

Vocabulary – X.X Holder Holder 2 Holder 3 Holder 4

Notes – X.X – LESSON TITLE.
Holder

Examples X.X

Unit 2.

Similar presentations

Presentation on theme: "Unit 2."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Unit 2.

Similar presentations

Presentation on theme: "Unit 2."— Presentation transcript:

Similar presentations

About project

Feedback