1.1 Linear Equations Basic Terminology of Equations ▪ Solving Linear Equations ▪ Identities, Conditional Equations, and Contradictions ▪ Solving for a.

1.1 Linear Equations Basic Terminology of Equations ▪ Solving Linear Equations ▪ Identities, Conditional Equations, and Contradictions ▪ Solving for a Specified Variable (Literal Equations) 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

1.1 Example 1 Solving a Linear Equation (page 85)
Solve Solution set: {6} Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Solve . Solution set: {–10}
1.1 Example 2 Clearing Fractions Before Solving a Linear Equation (page 85) Solve Solution set: {–10} Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Example 3(a) Identifying Types of Equations (page 86)
Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. This is a conditional equation. Solution set: {11} Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Example 3(b&c) Identifying Types of Equations (page 86)
Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. This is an identity. Solution set: {all real numbers} This is a contradiction. Solution set: ø Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Example 4 Solving for a Specified Variable (page 87)
Solve for the specified variable. Solve for the specified variable. (a) d = rt, for t (b) , for k Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Example 4 Solving for a Specified Variable (cont.)
Solve for the specified variable. Solve for the specified variable. (c) , for y Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Example 5 Applying the Simple Interest Formula (page 88)
Caden borrowed $2580 to buy new kitchen appliances for his home. He will pay off the loan in 9 months at an annual simple interest rate of 6.0%. How much interest will he pay? Caden will pay $ in interest. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Summary Linear Equations Do now: #5,10,15,20,23
Solving Linear Equations Identities, Conditional Equations, and Contradictions Identities – all real numbers Conditional Equations – get a solution Contradictions – no solution Solving for a Specified Variable (Literal Equations) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Applications and Modeling with Linear Equations
1.2 Applications and Modeling with Linear Equations Solving Applied Problems ▪ Geometry Problems ▪ Motion Problems ▪ Mixture Problems ▪ Modeling with Linear Equations Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.2 Example 1 Find the Dimensions of a Square (page 91)
The length of a rectangle is 2 in. more than the width. If the length and width are each increased by 3 in., the perimeter of the new rectangle will be 4 in. less than 8 times the width of the original rectangle. Find the dimensions of the original rectangle. x + 2 peri of new is 4 less than 8xwidth of orig x 4x + 16 = 8x – 4 20 = 4x 5 = x x + 5 x + 3 The width of the original rectangle is 5 in and the length is 7 in.

1.2 Example 2 Solving a Motion Problem (page 92)
Krissa drove to her grandmother’s house. She averaged 40 mph driving there. She was able to average 48 mph returning, and her driving time was 1 hr less. What is the distance between Krissa’s house and her grandmother’s house? 40x = 48(x – 1) 40x = 48x – 48 x -8x = -48 Use d = rt x - 1 x = 6 6 is not my answer, because we are looking for the distance between the houses The distance from Krissa’s home to her grandmother’s home is 240 mi.

1.2 Example 2b Solving a Motion Problem
It took Red Riding Hood 58 minutes to walk to Grandma’s house, while the Wolf who travels 14 mph faster than Red Riding Hood only took 8 minutes. What is the distance they each traveled, assuming they started at the same point. R T D Red Wolf R 58/60 R + 14 8/60 Distance: 2.24(58/60) = mi 58R = 8R + 112 They each traveled miles to Grandma’s house. 50R = → R = 2.24

1.2 Example 3 Solving a Mixture Problem (page 93)
How many liters of a 25% anti-freeze solution should be added to 5 L of a 10% solution to obtain a 15% solution? 25% anti-fr 5 x liters 15% anti-fr 10% anti-fr 10 5 liters x = 2.5 2.5 liters of the 25% solution should be added. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.2 Example 4 Solving an Investment Problem (page 94)
Last year, Owen earned a total of $1456 in interest from two investments. He invested a total of $28,000, part at 4.8% and the rest at 5.5%. How much did he invest at each rate? X + Y = 28,000 0.048X Y = 1456 Owen invested $12,000 at 4.8% & $16,000 at 5.5%. Y = 28,000 – X 0.048X (28,000 – X) = 1456 0.048X – 0.055X = 1456 –0.007X = -84 X = 12,000 Y = 16,000 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.2 Example 4B Solving an Investment Problem
Amy has earned $903 from two investments. The amount invested at 5.6% is triple that invested at 9%. How much does she have invested in each part? 3x x @5.6% @9% @9% = $3500 @5.6% = $10500 0.056(3x) x = 903 0.168x x = 903 0.258x = 903 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.2 Example 6(a) Modeling Health Care Costs (page 96)
The projected per capita health care expenditures in the United States, where y is in dollars, x years after 2000, are given by the linear equation What were the per capita health care expenditures in 2007? Let x = 2007 – 2000 = 7, then find the value of y. In 2007, per capita health care expenditures were $7440. When will the per capita expenditures reach $8220? Per capita health care expenditures are projected to reach $8220 nine years after 2000, or in 2009. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.2 Summary Applications and Modeling with Linear Equations
Solving Applied Problems Geometry Problems Motion Problems: D = RT Mixture Problems Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.2 Example 5 Modeling the Prevention of Indoor Pollutants (page 95)
A range hood removes contaminants at a rate of F liters of air per second. The percent P of contaminants that are also removed from the surrounding air can be modeled by the linear equation where 10 ≤ F ≤ 75. What flow F must a range hood have to remove 70% of the contaminants from the air? Not in fall of 2015 Substitute 70 for P into the given equation and solve for F The flow rate must be approximately L per second to remove 70% of the contaminants. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Ex add on 2016 State if each is:
real, complex, pure imaginary, and/or non-real complex 5 + 3i 5 3i complex, non-real complex real, complex complex, pure imaginary, non-real complex Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Example 1 Writing √–a as i√a (page 104)
Write as the product of a real number and i. (a) (b) (c) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Multiply or divide. Simplify each answer.
1.3 Example 2 Finding Products and Quotients Involving Negative Radicands (page 105) Multiply or divide. Simplify each answer. (a) (b) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Multiply or divide. Simplify each answer.
1.3 Example 2 Finding Products and Quotients Involving Negative Radicands (cont.) Multiply or divide. Simplify each answer. (c) (d) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Write in standard form a + bi.
1.3 Example 3 Simplifying a Quotient Involving a Negative Radicand (page 105) Write in standard form a + bi. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Example 4 Adding and Subtracting Complex Numbers (page 106)
Find each sum or difference. (a) (4 – 5i) + (–5 + 8i) = –1 + 3i (b) (–10 + 7i) – (5 – 3i) = –10 + 7i – 5 + 3i = – i Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Example 5 Multiplying Complex Numbers (page 107)
Find each product. (a) (5 + 3i)(2 – 7i) = 10 – 35i + 6i – 21i2 FOIL = 10 – 29i + 21 = 31 – 29i (b) (4 – 5i)2 =(4 – 5i)(4 – 5i) = 42 – 2(4)(5i) + (5i)2 = 16 – 40i + 25i2 = 16 – 40i – 25 = –9 – 40i Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Example 5 Multiplying Complex Numbers (cont.)
Find the product. With a difference of squares, you don’t have to FOIL. You only have to do F & L because the outer and inner will always have a sum of 0. (c) (9 – 8i)(9 + 8i) = 92 – (8i)2 = 81 – 64i2 = = 145 = i Reminder: the i key on the calculator for the ACT exam Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Example 6 Simplifying Powers of i (page 107)
Simplify each power of i. (a) (b) Write the given power as a product with ieven power (a) (b) On the white board find i34, i35, i36, i37 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Example 7(a) Dividing Complex Numbers (page 108)
Write in standard form a + bi. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.3 Summary Complex Numbers Do now: #5,10,15,20,21
Basic Concepts of Complex Numbers Operations on Complex Numbers If the denominator has a binomial with an i or √, multiply the top and bottom by the conjugate Reminder: i4 = 1 and i2 = -1 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Quadratic Equations Solving a Quadratic Equation ▪ Completing the Square ▪ The Quadratic Formula ▪ Solving for a Specified Variable ▪ The Discriminant Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 1 Using the Zero-Factor Property (page 111)
Solve 10x² + 5x – 4x – 2 = 0 Multiply F & L= -20x² Diff of Middle = +1x 20 1*20 2*10 4*5 5x(2x + 1) – 2(2x + 1) = 0 or or Solution set: or Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 2 Using the Square Root Property (page 112)
Solve each quadratic equation. Since each has x² and no x term use the square root property when solving (a) (b) (c) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 3 Using the Method of Completing the Square
Steps to solving quadratics by completing the square. Rewrite the equation so that the coefficient of x² is one. Re-write the equation such that the constant is alone on one side of the equation and place the “plus the blanks” on both sides of the equation. Square half the coefficient of x, and add this square to both sides of the equation. In the blank, (½ of b)² Factor the resulting trinomial as a perfect square and combine terms on the other side. Use the square root property to complete the solution. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 3 Using the Method of Completing the Square, a = 1 (cont.)
Solve by completing the square. x² + 10x + ___ = 20 + ___ Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Solve by completing the square.
1.4 Example 4 Using the Method of Completing the Square, a ≠ 1 (page 113) Solve by completing the square. x² + 3/2x + ___ = -5/4 + ___ Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 5 Using the Quadratic Formula (Real Solutions) (page 115)
Solve by the quadratic formula. Write the equation in standard form. a = 1, b = 6, c = –3 Quadratic formula 48 = 16•3 Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Solve by quadratic formula.
1.4 Example 6 Using the Quadratic Formula (Nonreal Complex Solutions) (page 115) Solve by quadratic formula. Write the equation in standard form. a = 4, b = –3, c = 5 Quadratic formula Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 7 Solving a Cubic Equation (page 116)
Factoring sum/diff of cubes A3 – B3 = (A – B)(A² + AB + B²) A3 + B3 = (A + B)(A² – AB + B²) Solve or -75 = -1•25•3 Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Solve for r. Use ± when taking square roots.
1.4 Example 8(a) Solving for a Quadratic Variable in a Formula (page 116) Solve for r. Use ± when taking square roots. Goal: Isolate r. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Example 9(a) Using the Discriminant (page 118)
Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. There is one distinct rational solution. There are two distinct nonreal complex solutions. There are two distinct irrational solutions. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.4 Summary Quadratic Equations Do now: #5,10,15,20,23,25
Solving a Quadratic Equation Factoring Square Root Property Completing the Square The Quadratic Formula Solving for a Specified Variable The Discriminant Negative: 2 complex roots Zero: 1 real rational root Positive Perfect square: 2 real rational roots Not a perfect square: 2 real irrational roots Quadratic formula Discriminant Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Applications and Modeling with Quadratic Equations
1.5 Applications and Modeling with Quadratic Equations Geometry Problems • Using the Pythagorean Theorem • Height of Projected Object • Modeling with Quadratic Equations Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

The dimensions of the metal piece are 35 cm x 70 cm.
1.5 Example 1 Solving a Problem Involving the Volume of a Box (page 122) A box with volume 7500 cm3 is to be formed from a sheet of metal whose length is twice the width. Equal size squares measuring 10 cm on a side are to be cut from the corners of the metal sheet in order to form the box. What are the dimensions of the original piece of metal? (x – 20)(2x – 20)(10) = 7500 (2x – 20)(x – 20) = 750 2(x – 10)(x – 20) = 750 (x – 10)(x – 20) = 375 We want the dimensions of the metal x2 – 30x =375 Length cannot be negative. Reject x = –5 x2 – 30x – 175 = 0 (x + 5)(x – 35) = 0 The dimensions of the metal piece are 35 cm x 70 cm. or

1.5 Example 2 Solving a Problem Involving the Pythagorean Theorem (page 123)
A piece of property is in the shape of the right triangle. The longer leg is 10 m shorter than twice the length of the shorter leg, and the hypotenuse is 20 m longer than the longer leg. Find the lengths of the sides of the property. The lengths of the sides of the property are 80m, 150m, and 170m. or Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

After how many seconds will the projectile be 100 m above the ground?
1.5 Example 3(a) Solving a Problem Involving the Height of a Projectile (page 124) If a projectile is shot upward from the ground with an initial velocity of 73.5 m per sec, neglecting air resistance, its height (in meters) above the ground t seconds after the projection is given by After how many seconds will the projectile be 100 m above the ground? Find the value(s) of t so that the height s is 100. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Use the quadratic formula to solve for t.
1.5 Ex 3(a) Solving a Problem Involving the Height of a Projectile (cont.) Use the quadratic formula to solve for t. a = 4.9, b = –73.5, c = 100 Both solutions are acceptable since the projectile reaches 100 m twice, once as it rises and once as it falls. The projectile will be 100 m above the ground at 1.51 seconds and at seconds. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

t = 0 represents the start time.
1.5 Example 3(b) Solving a Problem Involving the Height of a Projectile (page 124) How long will it take for the projectile to return to the ground? The projectile returns to the ground when s = 0. or t = 0 represents the start time. It takes 15 seconds for the projectile to return to the ground. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1. 5 Example 4(a) Analyzing Sport Utility Vehicle (SUV) Sales
1.5 Example 4(a) Analyzing Sport Utility Vehicle (SUV) Sales (page 126) Based on figures from 1990–2001, the equation models sales of SUVs from 1990 to 2001,where S represents sales in millions, and x = 0 represents 1990, x = 1 represents 1991, etc. Use the model to determine sales in 2000 and Compare the results to the actual figures of 3.6 million and 3.7 million. For 2000, x = 10. million For 2001, x = 11. million For 2000, the prediction is equal to the actual figure of 3.6 million. For 2001, the prediction is greater than the actual figure of 3.7 million. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

SUV sales reached 3 million in 1998.
1.5 Example 4(b) Analyzing Sport Utility Vehicle (SUV) Sales (page 126) According to the model, in what year did sales reach 3 million? (Round down to the nearest year.) x = 0 represents 1990, x = 1 represents 1991, etc Reject the negative solution, and round 8.5 down to 8. The year 1998 corresponds to x = 8. SUV sales reached 3 million in 1998.

Add on 2016 Let x = the additional # of passengers over 65
The cost of a charter flight to Miami is $230 each for 65 passengers, with a refund of $5 per passenger for each passenger in excess of 65. How many passengers must take the flight to produce a revenue of $13,870? Let x = the additional # of passengers over 65 Revenue = (# passengers)(cost per ticket) $13870 = (65 + x) (230 – 5x) 13870 = – 325x + 230x – 5x2 5x2 + 95x – 1080 = 0 5(x2 + 19x – 216) = 0 5(x + 27)(x – 8) = 0 x = -27 or x = 8 So there are 8 additional passengers in excess of 65 = 73 passengers Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.5 Summary Applications and Modeling with Quadratic Equations Do now:
#4,8,12,16 Applications and Modeling with Quadratic Equations Geometry Problems Using the Pythagorean Theorem Height of Projected Object Modeling with Quadratic Equations Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Other Types of Equations and Applications
1.6 Other Types of Equations and Applications Rational Equations ▪ Work Rate Problems ▪ Equations with Radicals ▪ Equations Quadratic in Form Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Example 1 LCD: LCD is 6x Restrictions: 6X ≠ 0 X ≠ 0 Solve: Check: 2 x

Example 2 Solve: 3 2 LCD: LCD is 6X(X+1) Restrictions: 6X(X + 1) ≠ 0
3 2 Check restrictions

Example 3 Solve: LCD: LCD is 3(x+ 2)(x + 5) Restrictions:
Check restrictions

Example 4 Solve: Single fraction on each side I would cross multiply
Restrictions: (x – 1)(x + 1) ≠ 0 x ≠ 1, x ≠ -1 Check restrictions

Example 5 Solve: LCD: LCD is (3 – a)(3 + a) Restrictions:
Restriction: a ≠ 3 so the answer is no solution. 

1.6 Example 6 Solving a Work Rate Problem (page 135)
Lisa and Keith are raking the leaves in their backyard. Working alone, Lisa can rake the leaves in 5 hr, while Keith can rake them in 4 hr. How long would it take them to rake the leaves working together? Assign variables. In 1 hr, Lisa can do 1/5 of the job by herself. In 1 hr, Keith can do 1/4 of the job by himself. It will take Lisa and Keith hr working together to rake the leaves.

1.6 Example 7 Solving an Equation Containing a Radical (Square Root) (page 138)
Solve Now check. –2 is not a solution. Solution set: {6} Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Solve . or or Solution set:
1.6 Example 9 Solving an Equation Containing a Radical (Cube Root) (page 139) Solve Now check. or or Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.6 Example 10(a) Solving Equations Quadratic in Form (page 140)
Solve Now check. or or Solution set: {19, 259}

1.6 Example 10(b) Solving Equations Quadratic in Form (page 140)
Solve Now check. or or Solution set:

Solve Now solve for x. Solution set:
1.6 Example 12 Solving an Equation that Leads to One Quadratic in Form (page 142) Now check. Solve Now solve for x. Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.6 Summary Other Types of Equations and Applications Do now:
#5,10,15,20,25,30,32 Other Types of Equations and Applications Rational Equations Work Rate Problems Equations with Radicals Equations Quadratic in Form Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Add on 2016 Set Notation Interval Notation Graph {xӀ x > 6} (6,∞)
(-∞,6] ] 6 {xӀ 4< x < 6} (4,6] ( 4 ] 6 {xӀ x<4 or x > 6} (-∞,4)U[6,∞) ) 4 [6 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.7 Example 4 Solving the Break-Even Point (page 148)
If the revenue and cost of a certain product are given by R = 45x and C = 30x , where x is the number of units produced and sold, at what production level does R at least equal C? Set R ≥ C and solve for x. The break-even point is at x = 350. This product will at least break even only if the number of units produced and sold is in the interval Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.7 Example 5 Solving a Quadratic Inequality (page 149)
Solve x2 – 2x – 15 < 0 (x – 5)(x + 3) < 0 The breaking points for the # line are 5 & -3 Use closed dots since the inequality symbol includes equality. -4 6 (-)(-) (-)(+) (+)(+) + < 0 false - < 0 true + < 0 false Solution set: [–3, 5] Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.7 Example 6 Solving a Quadratic Inequality (page 150)
Solve The breaking points for the # line are -1/3 & 4 3x2 – 11x – 4 > 0 (3x + 1)(x – 4) > 0 Use open dots since the inequality symbol does not include equality. -1 6 (-)(-) (+)(-) (+)(+) Solution set: + > 0 true - > 0 false + >0 true Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.7 Example 7 Solving a Problem Involving the Height of a Projectile (page 151)
If an object is launched from ground level with an initial velocity of 144 ft per sec, its height in feet t seconds after launching is s feet, where When will the object be greater than 128 ft above ground level? < Breaking pts: 8 & 1 4 12 (-)(-) (-)(+) (+)(+) The object will be greater than 128 ft above ground level between 1 and 8 seconds after it is launched. + < 0 false - < 0 true + <0 false

1.7 Example 8 Solving a Rational Inequality (page 151)
Solve Breaking points: num = 0 and den = 0 or or 2 4 6 (+)/(-) (+)/(+) (-)/(+) - > 0 false + > 0 true - > 0 false Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Solution set:

1.7 Example 9 Solving a Rational Inequality (page 153)
Solve Breaking points: num = 0 and den = 0 or or 2 3 (+)/(-) (+)/(+) (-)/(+) - < 0 true + < 0 false - < 0 true Solution set: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

Absolute Value Equations and Inequalities
1.8 Absolute Value Equations and Inequalities Absolute Value Equations ▪ Absolute Value Inequalities ▪ Special Cases ▪ Absolute Value Models for Distance and Tolerance Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.8 Example 2(a) Solving Absolute Value Inequalities (page 160)
Done … we don’t need inequalities for compass or college algebra Are needed for ACT Skip to summary Solution set: (–1, 4) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

The absolute value of a number is always nonnegative.
1.8 Example 4(a) Solving Special Cases of Absolute Value Equations and Inequalities (page 161) is equivalent to 7x + 28 = 0 Solution set: {–4} The absolute value of a number is always nonnegative. Therefore, is always true. Solution set: There is no number whose absolute value is less than –5. Therefore, is always false. Solution set: ø

Write each statement using an absolute value inequality.
1.8 Example 5 Using Absolute Value Inequalities to Describe Distances (page 162) Write each statement using an absolute value inequality. (a) m is no more than 9 units from 3. This means that m is 9 units or less from 3. Thus the distance between m and 3 is less than or equal to 9, or (b) t is within .02 unit of 5.8 . This means that t is less than .02 unit from 5.8. Thus the distance between t and 5.8 is less than .02, or Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.8 Example 6 Using Absolute Value to Model Tolerance (page 162)
Suppose y = 5x – 2 and we want y to be within .001 unit of 6. For what values of x will this be true? 5.999 < 5x – 2 < 6.001 Values of x in the interval (1.5998, ) will satisfy the condition. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.8 Summary Absolute Value Equations and Inequalities
Absolute Value Inequalities Special Cases Absolute Value Models for Distance and Tolerance Copyright © 2008 Pearson Addison-Wesley. All rights reserved.

1.1 Linear Equations Basic Terminology of Equations ▪ Solving Linear Equations ▪ Identities, Conditional Equations, and Contradictions ▪ Solving for a.

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1.1 Linear Equations Basic Terminology of Equations ▪ Solving Linear Equations ▪ Identities, Conditional Equations, and Contradictions ▪ Solving for a.

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